147 free, high-quality past-paper-style questions on Motion in electromagnetic fields for IB Physics SL, each with a full mark scheme. Covers: Force on a current-carrying wire; Direction of the magnetic force; The velocity selector; and more. For when you have finished the past papers, or want to practise one topic at a time.D.3 at Higher Level →Read the D.3 notes →My revision plan →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.753 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1D-1A-19Force on a current-carrying wirePaper 1AEasy1 mark
A straight wire of length 0.40 m carries a current of 3.0 A at right angles to a uniform magnetic field of flux density 0.25 T.
What is the force on the wire?
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Step 1F = BIL sin θ with θ = 90°, so F = BIL = 0.25 × 3.0 × 0.40 = 0.30 N.
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Answer B
Answer: B · 1 stage of work, one mark
Every option, and why
AThis leaves out the current: BL = 0.25 × 0.40 = 0.10.
BCorrect: F = BIL = 0.25 × 3.0 × 0.40 = 0.30 N.
CThis divides by the length instead of multiplying by it.
DThis divides by B instead of multiplying: IL/B = 3.0 × 0.40/0.25 = 4.8.
2D-1A-23Direction of the magnetic forcePaper 1AEasy1 mark
A straight wire in the plane of the page carries a current towards the bottom of the page. A uniform magnetic field is directed out of the page.
What is the direction of the magnetic force on the wire?
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Step 1Fleming's left-hand rule: first finger out of the page (field), second finger towards the bottom of the page (current), so the thumb (force) points towards the left of the page.
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Answer A
Answer: A · 1 stage of work, one mark
Every option, and why
ACorrect: the force is perpendicular to both the current and the field, towards the left of the page.
BThis reverses the rule (for example by using the right hand).
CThe force is perpendicular to the field, so it cannot point along the field.
DThe force is perpendicular to the current, so it cannot act along the wire.
In a velocity selector the electric field strength is 4.5 × 104 V m−1 and the magnetic flux density is 0.030 T. The diagram shows the electric and magnetic forces on a negatively charged particle moving through the selector.
What speed must the particle have to pass through undeflected?
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Step 1Undeflected: qE = qvB, so the charge cancels and v = E/B.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2v = 4.5 × 104/0.030 = 1.5 × 106 m s−1.
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Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: v = E/B = 4.5 × 104/0.030 = 1.5 × 106 m s−1.
BThis multiplies E by B instead of dividing.
CThis is B/E, the inverted ratio.
DThe charge cancels from qE = qvB, which is what makes a velocity selector useful.
4D-1A-41Paths in electric and magnetic fieldsPaper 1AEasy1 mark
A positive ion enters a region of uniform field moving at right angles to the field lines. In one experiment the field is electric; in another it is magnetic.
Which row describes the path of the ion while it is in each field?
Uniform electric fieldUniform magnetic field
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Step 1In the electric field the force qE is constant in size and direction (like gravity on a projectile): a parabola. In the magnetic field the force qvB is always perpendicular to the velocity: a circular arc.
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Answer D
Answer: D · 1 stage of work, one mark
Every option, and why
AThese are the wrong way round.
BThe magnetic force changes direction as the velocity changes, so it cannot give a parabola.
CThe electric force acts on a charge whatever its direction of motion; it is the magnetic force that is zero for motion along the field.
DCorrect: a constant force gives a parabola; a force always perpendicular to the velocity gives a circular arc.
6D-1A-90Force on a moving chargePaper 1AEasy1 mark
A proton moves with constant speed through a uniform magnetic field. The angle θ between the velocity of the proton and the field can be varied from 0° to 180°.
Which graph shows how the magnitude F of the magnetic force on the proton varies with θ?
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Step 1F = qvB sin θ: zero at 0° and 180° (motion along the field) and a maximum at 90°.
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Answer D
Answer: D · 1 stage of work, one mark
Every option, and why
AThis is |cos θ|, which is largest when the proton moves along the field, where the force is in fact zero.
BThe force depends on the angle; only the component of velocity at right angles to the field contributes.
CThe force does not keep increasing with θ; it falls to zero again when the motion is antiparallel to the field.
DCorrect: F = qvB sin θ rises from zero to a maximum at 90° and falls to zero at 180°.
9IBP2S-032-01-01Direction of the magnetic forcePaper 1AEasy1 mark
A straight wire in a robotic arm's actuator carries a constant current in a uniform magnetic field. When the wire makes an angle of 30° with the field, the magnetic force on it has magnitude F. The wire is rotated so that it is perpendicular to the field. The current and the length of wire in the field are unchanged. What is the new magnitude of the force?
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The force is F = BIL sin θ. At 30°, sin 30° = 0.5, so F = 0.5BIL. At 90°, sin 90° = 1, so the force becomes BIL = 2F. The option 2F/√3 comes from using cos θ instead of sin θ. The option F/2 inverts the dependence. The option F assumes that orientation does not matter.
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Answer B
Answer: B
Common error: θ is the angle between the wire (the current direction) and the field. The force is zero when they are parallel, so use sin θ, not cos θ.
10IBP2S-032-01-02Force on a current-carrying wirePaper 1AEasy1 mark
Which combination of units is equivalent to the tesla?
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From F = BIL, we get B = F/(IL). The unit is therefore N/(A m) = N A−1 m−1. The other options either multiply by the current or the length instead of dividing, or put A in the numerator. In base units, T = kg s−2 A−1, not kg A s−2.
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Answer B
Answer: B
Exam tip: Rearrange a defining equation such as F = BIL to find the unit of a field quantity quickly.
11IBP2S-032-01-06Paths in electric and magnetic fieldsPaper 1AEasy1 mark
A proton from the solar wind moves at right angles to a uniform region of a planet's magnetic field. No other forces act. Which statement is correct?
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The magnetic force qvB is always perpendicular to the velocity. It therefore does no work, and the kinetic energy (and so the speed) stays constant. The path is a circle, not a parabola. A parabola is the path in a uniform electric field.
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Answer B
Answer: B
Exam tip: A force perpendicular to the velocity changes only the direction of motion, never the speed.
12IBP2S-032-00-01Motion in an electric fieldPaper 1AEasy1 mark
In an electrostatic crop sprayer, a negatively charged pesticide droplet passes between two large horizontal parallel plates. The upper plate is positive and the lower plate is negative. What is the direction of the electric force on the droplet?
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The field between the plates points from the positive plate to the negative plate, so it is vertically downwards. The force on a negative charge is opposite to the field, so it is vertically upwards, towards the positive plate. A uniform field still exerts a force F = qE. Only its magnitude is the same everywhere.
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Answer A
Answer: A
Common error: The field direction is the direction of the force on a positive charge. Reverse it for electrons and other negative charges.
13IBP2S-032-00-02Force on a moving chargePaper 1AEasy1 mark
A cosmic-ray proton travels exactly along a magnetic field line close to the Earth's magnetic pole. The field there is approximately uniform. Which statement describes the magnetic force on the proton?
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The magnetic force has magnitude F = qvB sinθ, where θ is the angle between v and B. Here the velocity is parallel to the field, so θ = 0 and sinθ = 0. The force is therefore zero and the proton continues in a straight line at constant speed. A magnetic force can never act along the velocity, because it is always perpendicular to v.
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Answer A
Answer: A
Exam tip: Before using qvB, check the angle between v and B. Only the component of velocity perpendicular to B produces a magnetic force.
14IBP2S-032-00-07Force on a moving chargePaper 1AEasy1 mark
In a dental X-ray tube, electrons start from rest and are accelerated through a potential difference of 30 kV. What kinetic energy does each electron gain?
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The energy gained is Ek = qV = 1.60 × 10−19 × 3.0 × 104 = 4.8 × 10−15 J. Distractors:
2.4 × 10−15 J wrongly includes a factor of ½.
4.8 × 10−18 J does not convert kV to V.
3.0 × 104 J leaves out the electron charge. The energy is 30 keV, not 30 kJ.
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Answer A
Answer: A
Common error: The energy gained is qV with no factor of ½. The ½ appears only in ½mv², when you go on to find the speed.
15IBP2S-032-03-01Parallel current-carrying wiresPaper 1AEasy1 mark
A domestic extension cable contains two long, straight, parallel conductors close together. When an appliance is switched on, the conductors carry equal currents in opposite directions. What is the effect of the magnetic forces between the two conductors?
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Each conductor sits in the magnetic field of the other, so each feels a force F = BIL at right angles to its length. For antiparallel currents this force pushes the conductors apart, so they repel. Parallel currents would attract. A zero net current does not cancel the force on each individual wire. The force is perpendicular to the wire, not along it.
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Answer B
Answer: B
Exam tip: Remember the rule: like currents attract, unlike currents repel. This is the opposite of the rule for electric charges.
16IBP2S-032-03-03Circular motion in a magnetic fieldPaper 1AEasy1 mark
In an electron microscope, an electron moves through a region of uniform magnetic field. Its velocity is perpendicular to the field. Which quantity remains constant while the electron is in the field?
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The magnetic force is always perpendicular to the velocity, so it does no work. The speed, and therefore the kinetic energy, stays constant. The direction of motion changes continuously. As a result, velocity and momentum (both vectors) change, and so does the direction of the centripetal acceleration.
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Answer C
Answer: C
Common error: A constant speed does not mean a constant velocity. In circular motion the velocity vector is always changing direction.
17IBP2S-032-03-08Force on a current-carrying wirePaper 1AEasy1 mark
A cable carrying a steady current is placed inside the uniform magnetic field of an MRI scanner. For which orientation of the cable is the magnetic force on it zero?
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F = BIL sin θ, where θ is the angle between the current and the field. When the cable is parallel to the field, θ = 0, so sin θ = 0 and F = 0. The force is greatest when the cable is perpendicular to the field. At 45° the force is non-zero (0.71BIL).
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Answer C
Answer: C
Common error: The angle in F = BIL sin θ is the angle between the wire and the field, not between the wire and the force.
18IBP2S-032-02-01Direction of the magnetic forcePaper 1AEasy1 mark
An electron beam in a welding machine travels horizontally due east through a region where the magnetic field is uniform and directed vertically downwards. What is the direction of the magnetic force on the electrons as they enter the field?
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The force is F = qv × B. For a positive charge moving east in a downward field, east × down points north. The electron is negative, so the force is reversed and points south. The force is always perpendicular to both v and B, so it cannot be vertical or along the line of motion.
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Answer B
Answer: B
Common error: Applying the right-hand rule correctly but forgetting to reverse the direction for a negative charge gives 'north'.
19IBP2S-032-02-02Force on a moving chargePaper 1AEasy1 mark
In a solar-wind detector on a space probe, a proton and an alpha particle enter the same uniform magnetic field with the same speed, moving perpendicular to the field. An alpha particle has four times the mass and twice the charge of a proton. What is \(\frac{\text{radius of alpha particle's path}}{\text{radius of proton's path}}\)?
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From qvB = mv²/r, r = mv/(qB). With v and B the same, r ∝ m/q. For the alpha particle m/q is 4/2 = 2 times that of the proton, so the radius ratio is 2.
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Answer C
Answer: C
Exam tip: Write r = mv/(qB) first, then cross out everything that is the same for both particles before comparing.
20IBP2S-032-02-07Parallel current-carrying wiresPaper 1AEasy1 mark
In an electric car, two long parallel cables run side by side. One carries current from the battery to the motor and the other carries the same current back to the battery. What is the magnetic force between the cables?
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Parallel wires with currents in the same direction attract. Wires with currents in opposite directions repel. Here the supply and return currents are opposite, so the cables repel. The forces act perpendicular to the cables.
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Answer B
Answer: B
Exam tip: The rule for parallel wires is the reverse of the rule for electric charges: like currents attract.
21IBP2S-032-04-01The velocity selectorPaper 1AEasy1 mark
An ion-implantation machine used in microchip manufacture has a velocity selector. The selector has crossed uniform fields: electric field strength 4.0×104 V m−1 and magnetic flux density 0.20 T. Which ion speed passes through undeflected?
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An ion passes undeflected when the electric and magnetic forces balance: qE = qvB. So v = E/B = 4.0×104 / 0.20 = 2.0×105 m s−1. Option A comes from multiplying E by B. Option B inverts the ratio. Option D assumes the speed equals E numerically and ignores B.
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Answer C
Answer: C
Exam tip: A quick unit check helps: (V m⁻¹)/T = m s⁻¹. Only E/B gives a speed.
22IBP2S-032-04-04Force on a moving chargePaper 1AEasy1 mark
In a hospital X-ray tube, electrons are accelerated from rest through a potential difference of 30 kV. What is the kinetic energy gained by each electron?
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The work done on the electron is W = qV = 1.60×10−19 × 3.0×104 = 4.8×10−15 J (equivalently, 30 keV). Option B omits the charge. Option C divides by e instead of multiplying. Option D wrongly includes a factor of ½.
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Answer A
Answer: A
Common error: The ½ belongs in ½mv², not in qV. The kinetic energy gained is the full qV.
23IBP2S-032-04-09Direction of the magnetic forcePaper 1AEasy1 mark
An electron is projected in the same direction as a uniform electric field. Which statement describes the motion of the electron immediately after it is projected?
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The electron is negatively charged, so the electric force on it acts opposite to the field direction. That force is therefore opposite to its velocity. The field is uniform, so the force and the acceleration are constant. The electron decelerates uniformly. (If the field region is long enough, it will eventually stop and then accelerate back the other way.) It cannot speed up, because that would need a force along the field, which is true only for a positive charge. It does not move in a circle, because a constant force parallel to the velocity gives straight-line motion. It cannot keep a constant velocity, because a resultant force acts on it.
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Answer B
Answer: B
Common error: Field lines show the direction of the force on a positive charge. For an electron, reverse the direction.
24D-1A-20Force on a current-carrying wirePaper 1AMedium1 mark
A straight wire carrying a current lies at 30° to a uniform magnetic field. The magnetic force on the wire is F.
The wire is turned until it is at right angles to the field. The current and the length of wire in the field do not change. What is the magnetic force on the wire now?
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Step 1F = BIL sin θ, where θ is the angle between the wire and the field: at 30°, F = BIL × 0.5.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2At 90° the force is BIL = F/sin 30° = 2F.
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Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThis uses cos θ, which is zero at 90°. The force is zero when the wire is parallel to the field, not perpendicular.
BThis multiplies by sin 30° again instead of dividing by it.
CThis uses F = BIL cos θ for the first position, giving F/cos 30° = 2F/√3.
DCorrect: BIL = F/sin 30° = 2F, the maximum force.
25D-1A-21Force on a moving chargePaper 1AMedium1 mark
A proton and an alpha particle move at right angles to the same uniform magnetic field. The speed of the alpha particle is one quarter of the speed of the proton.
What is (magnetic force on the alpha particle)/(magnetic force on the proton)?
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Step 1F = qvB, so F ∝ qv in the same field; the alpha particle has charge 2e.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Ratio = 2 × 1/4 = 0.5.
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Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThis allows for the lower speed but takes the charge of the alpha particle as e instead of 2e.
BCorrect: twice the charge and a quarter of the speed give 2 × 0.25 = 0.5.
CThis uses the mass ratio, 4, instead of the charge ratio, 2. The magnetic force does not depend on mass.
DThis allows for the doubled charge but not for the lower speed.
A particle of charge q and mass m moves with speed v at right angles to a uniform magnetic field of flux density B. It follows a circular path of radius r.
Which expression gives the charge-to-mass ratio q/m of the particle?
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Step 1The magnetic force provides the centripetal force: qvB = mv²/r.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2q/m = v/Br.
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Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: qvB = mv²/r rearranges to q/m = v/Br.
BThis is m/q, the inverted ratio.
CThis leaves v out of the magnetic force, writing qB = mv²/r.
DThis writes the centripetal force as mv²r instead of mv²/r.
32D-1A-58Circular motion in a magnetic fieldPaper 1AMedium1 mark
Singly charged positive ions of different masses leave a velocity selector with the same speed. They then enter a uniform magnetic field at right angles to the field and move in circular paths of radius r.
Which graph shows the variation of r with the mass m of the ion?
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Step 1The magnetic force provides the centripetal force: qvB = mv²/r, so r = mv/(qB).
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2q, v and B are the same for every ion, so r ∝ m: a straight line through the origin.
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Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: with the same charge, speed and field, r = mv/qB is proportional to m.
BThis takes r to be inversely proportional to m, as if a heavier ion were bent more; a heavier ion has more momentum and is bent less.
CThe radius is not fixed by the field alone; ions with the same speed but greater mass move in larger circles.
DThis is the result for ions accelerated through the same potential difference (equal kinetic energies, r ∝ √m); after a velocity selector the ions have the same speed.
33D-1A-87Charge in a uniform electric fieldPaper 1AMedium1 mark
An electron enters the uniform electric field between two parallel plates in a vacuum, moving parallel to the plates. x is the distance travelled by the electron parallel to the plates after entering the field and v⊥ is the magnitude of the component of its velocity at right angles to the plates.
Which graph shows how v⊥ varies with x while the electron is between the plates?
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Step 1The force eE is constant and at right angles to the plates, so v⊥ = at; the velocity component parallel to the plates is constant, so t = x/vx.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2v⊥ = (a/vx)x: a straight line through the origin.
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Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThis is the shape of the displacement at right angles to the plates (y ∝ x²), not of the velocity component.
BCorrect: v⊥ = at and t ∝ x, so v⊥ ∝ x.
CA constant force gives a constant acceleration, not a constant velocity component.
Dv ∝ √s applies to a particle accelerated from rest along the field through a distance s; here x is measured at right angles to the force.
34D-1A-89The velocity selectorPaper 1AMedium1 mark
Positive ions pass undeflected through a region in which a uniform electric field of strength E and a uniform magnetic field of flux density B are perpendicular to each other and to the velocity of the ions.
The electric field is switched off. The ions now move in a circular path of radius r in the magnetic field.
What is the charge-to-mass ratio q/m of the ions?
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Step 1Undeflected: qE = qvB, so v = E/B.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2In the magnetic field alone qvB = mv²/r, so q/m = v/Br = E/B²r.
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Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: q/m = v/Br with v = E/B.
BThis inverts the selector condition, using v = B/E.
CThis is m/q, the inverted ratio.
DThis writes the centripetal force as mv²r instead of mv²/r.
35D-1A-91Parallel current-carrying wiresPaper 1AMedium1 mark
A long horizontal wire X is fixed. A second long, thin wire Y, parallel to X and with a mass per unit length of 2.0 × 10−3 kg m−1, is free to move vertically between smooth guides. Y is held at rest 4.0 mm directly above X by magnetic force alone. The same current I flows in both wires.
Which row gives I and the directions of the currents?
Diagram NOT accurately drawn
I / ADirections of the currents
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Step 1Y must be repelled upwards, so the currents are in opposite directions.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2μ0I²/2πr = weight per unit length: I² = 2π × 4.0 × 10−3 × 2.0 × 10−3 × 9.81/μ0 = 392 A², so I = 20 A.
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Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThis uses the mass per unit length instead of the weight per unit length (no factor g).
BThe value is right, but currents in the same direction attract, which would pull Y down onto X.
CCorrect: opposite currents repel, and μ0I²/2πr = 0.0196 N m−1 gives I = 20 A.
DThis stops at I² = 392 A² and does not take the square root.
36D-1A-94Parallel current-carrying wiresPaper 1AMedium1 mark
Two long parallel wires X and Y are 0.15 m apart in a vacuum. X carries a current of 2.0 A and Y a current of 8.0 A, both out of the page. A third long parallel wire Z, carrying a current, is placed on the line between X and Y. The resultant magnetic force on Z is zero.
What is the distance of Z from X?
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Step 1The forces on Z from X and Y are opposite, so μ0IXIZ/2πr = μ0IYIZ/2π(0.15 − r); IZ cancels.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 22.0/r = 8.0/(0.15 − r), so r = 0.15 × 2.0/10.0 = 0.030 m.
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Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: the force per unit length is proportional to 1/r, so r : (0.15 − r) = 2.0 : 8.0 and r = 0.030 m.
BThis sets r/0.15 equal to 2.0/8.0, using the whole separation instead of the distance from Y.
CThis treats the force as an inverse-square law, so r : (0.15 − r) = √2.0 : √8.0.
DThis puts Z next to the larger current: it is 0.030 m from Y, not from X.
37D-1A-96Circular motion in a magnetic fieldPaper 1AMedium1 mark
In a simple model of the magnetron in a microwave oven, electrons move in circles at right angles to a uniform magnetic field, completing 2.45 × 109 revolutions per second.
What is the magnetic flux density?
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Step 1evB = mev²/r gives v/r = eB/me; the time for one revolution 2πr/v = 2πme/eB does not depend on the speed.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2f = eB/2πme, so B = 2π × 2.45 × 109 × 9.11 × 10−31/1.60 × 10−19 = 0.088 T.
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Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThis uses the mass of a proton instead of the mass of an electron.
BThis multiplies by 2π twice, confusing f with the angular velocity ω = 2πf and then adding 2π again.
CCorrect: B = 2πfme/e = 0.088 T.
DThis leaves out the 2π, taking B = fme/e (treating f as v/r).
38D-1A-97Charge in a uniform electric fieldPaper 1AMedium1 mark
An electron of mass m and charge −e is released from rest at a plate. It is accelerated by the uniform electric field between this plate and a parallel plate a distance d away, which is at a potential difference V above it.
What is the time taken by the electron to cross the gap?
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Step 1½mv² = eV gives the final speed v = √(2eV/m).
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2The acceleration is uniform, so the mean speed is v/2 and t = 2d/v = d√(2m/eV).
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Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: t = 2d/v with v = √(2eV/m).
BThis divides d by the final speed, as if the electron crossed the whole gap at its final speed.
CThis uses the mean speed correctly but leaves out the ½ in the kinetic energy, so v = √(eV/m).
DThis uses the final speed instead of the mean speed and also leaves out the ½ in the kinetic energy.
39IBP2S-032-01-03Charge-to-mass ratioPaper 1AMedium1 mark
In a mass spectrometer used in an anti-doping laboratory, two types of ion, X and Y, enter the same uniform magnetic field with the same speed, perpendicular to the field. The radius of the path of X is twice the radius of the path of Y. What is \(\frac{(q/m)_X}{(q/m)_Y}\)?
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Equating qvB to mv2/r gives r = mv/(qB), so q/m = v/(Br). With v and B the same for both ions, q/m ∝ 1/r. Doubling the radius therefore halves q/m, giving a ratio of 1/2. The option 2 wrongly takes q/m to be proportional to r. The options 1/4 and 4 wrongly square the ratio.
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Answer B
Answer: B
Common error: A larger radius means a SMALLER charge-to-mass ratio, because the ion is harder to turn.
40IBP2S-032-01-04Paths in electric and magnetic fieldsPaper 1AMedium1 mark
Ions start from rest and are accelerated through a potential difference V. They then enter a uniform magnetic field at right angles and move in a circle of radius r. The accelerating potential difference is increased to 4V. What is the new radius?
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From energy conservation, qV = ½mv2, so v ∝ √V. The radius is r = mv/(qB) ∝ v ∝ √V. Quadrupling V therefore doubles v, which doubles r. The option 4r assumes v ∝ V. The option r assumes the radius depends only on the field.
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Answer C
Answer: C
Exam tip: Chain the proportionalities: r ∝ v and v ∝ √V, so r ∝ √V.
41IBP2S-032-01-05The velocity selectorPaper 1AMedium1 mark
A beam on a spacecraft instrument contains positive ions of many different masses, charges and speeds. The beam passes through a region of perpendicular uniform electric and magnetic fields. Some ions emerge undeflected. Which quantity must be the same for all of the undeflected ions?
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An ion is undeflected when the electric and magnetic forces balance: qE = qvB. This gives v = E/B. The charge cancels, and mass does not appear at all. Every undeflected ion therefore has the same speed, whatever its q/m, kinetic energy or momentum.
✓ 1
Answer B
Answer: B
Common error: Crossed fields select speed, not charge-to-mass ratio. The q/m is measured afterwards, from the radius in a magnetic field.
42IBP2S-032-00-03Circular motion in a magnetic fieldPaper 1AMedium1 mark
In a research instrument, protons and alpha particles start from rest. Each is accelerated through the same potential difference and then enters the same uniform magnetic field at right angles to it. An alpha particle has twice the charge and four times the mass of a proton. What is \(\frac{\text{radius of alpha particle path}}{\text{radius of proton path}}\)?
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After acceleration, qV = ½mv², so v = √(2qV/m). The radius of the path is r = mv/(qB) = √(2mV/q)/B. The ratio of radii is therefore √((mα/qα)/(mp/qp)) = √(4/2) = √2. Distractors:
2 assumes both particles have equal speeds.
1 assumes both particles have equal kinetic energies. In fact the alpha particle gains twice the energy.
1/√2 is the inverted ratio.
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Answer C
Answer: C
Common error: The same accelerating p.d. does not give the same speed or the same kinetic energy. The energy gained is qV, so it depends on the charge.
43IBP2S-032-00-04Circular motion in a magnetic fieldPaper 1AMedium1 mark
In a synchrotron light source, electrons travel around a curved section of the ring at constant speed under the action of a uniform magnetic field perpendicular to their velocity. Ignore radiation losses. Which quantity changes as an electron moves around the curved section?
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The magnetic force is always perpendicular to the velocity, so it does no work. This means the speed and the kinetic energy stay constant. Since F = qvB, the magnitude of the force is also constant. The direction of the velocity changes continuously, however. Momentum is a vector, so the momentum changes.
✓ 1
Answer C
Answer: C
Exam tip: A force perpendicular to the velocity changes only the direction of motion. Look for the vector quantity among the options.
44IBP2S-032-00-05The velocity selectorPaper 1AMedium1 mark
A velocity selector in an ion-implantation machine has parallel plates 5.0 mm apart. A uniform magnetic field of 0.20 T acts perpendicular to the electric field. Ions travelling at 3.0 × 105 m s−1 pass through undeflected. What is the potential difference between the plates?
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For an undeflected ion, qE = qvB, so E = vB = 3.0 × 105 × 0.20 = 6.0 × 104 V m−1. The potential difference is V = Ed = 6.0 × 104 × 5.0 × 10−3 = 300 V. Distractors:
6.0 × 104 V is the field strength, not the p.d.
3.0 × 105 V uses d in mm without converting to metres.
1.2 × 107 V divides by d instead of multiplying.
✓ 1
Answer A
Answer: A
Common error: Balancing the forces gives the field strength E. You still need V = Ed to find the potential difference, with d in metres.
45IBP2S-032-00-06Circular motion in a magnetic fieldPaper 1AMedium1 mark
Protons for cancer therapy circulate in a uniform magnetic field perpendicular to their velocity. The speed of a proton is doubled, and its speed remains much less than the speed of light. What happens to the time it takes to complete one revolution?
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The magnetic force provides the centripetal force, so r = mv/(qB). The period is T = 2πr/v = 2πm/(qB), which does not depend on v. Doubling the speed doubles the radius, and the proton travels twice as far at twice the speed, so the period is unchanged. Option A wrongly assumes the path length stays the same. Option C confuses the period with the radius.
✓ 1
Answer B
Answer: B
Exam tip: Substitute r = mv/(qB) into T = 2πr/v to show that v cancels.
46IBP2S-032-00-09Direction of the magnetic forcePaper 1AMedium1 mark
An electron moves horizontally due east through a region where a uniform magnetic field points vertically upwards. What is the direction of the magnetic force on the electron at that instant?
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For a positive charge, the force is along v × B. With v pointing east and B pointing up, the right-hand rule gives south. The electron is negative, so the force on it is reversed and points north. Distractors:
South is the answer for a positive charge.
Upwards is impossible, because the force must be perpendicular to B.
West is impossible, because the force must be perpendicular to v.
✓ 1
Answer A
Answer: A
Common error: Use the hand rule for conventional current, then reverse the result for electrons. An electron moving east acts like a conventional current flowing west.
47IBP2S-032-03-02Parallel current-carrying wiresPaper 1AMedium1 mark
Two parallel copper busbars in an electrical substation carry equal currents. The force per unit length between them is F. The current in each busbar is doubled and the separation of the busbars is halved. What is the new force per unit length?
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F/L = μ0I1I2/(2πr). Doubling both currents multiplies the product I1I2 by 4. Halving r multiplies the force by 2. The new force is 4 × 2 = 8F. The option 4F ignores the change in separation. The option 16F wrongly treats the force as an inverse-square dependence on r.
✓ 1
Answer C
Answer: C
Common error: The force between long parallel wires varies as 1/r, not 1/r². Do not confuse it with Coulomb's law or Newton's law of gravitation.
48IBP2S-032-03-04Force on a moving chargePaper 1AMedium1 mark
A proton and an alpha particle from the solar wind enter the same uniform magnetic field with the same speed, both moving perpendicular to the field. The mass of the alpha particle is 4 times the mass of the proton and its charge is 2 times the charge of the proton. What is \(\frac{r_\alpha}{r_p}\), the ratio of the radius of the alpha particle's path to the radius of the proton's path?
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From qvB = mv²/r, the radius is r = mv/(qB). With v and B the same for both particles, r ∝ m/q. For the alpha particle, m/q = 4/2 = 2 times that of the proton, so rα/rp = 2. The answer 4 ignores the larger charge. The answer ½ inverts the ratio.
✓ 1
Answer C
Answer: C
Exam tip: Write r = mv/(qB) first, then decide which quantities are fixed. The ratio then follows directly from m/q.
49IBP2S-032-03-05The velocity selectorPaper 1AMedium1 mark
An ion implanter used in microchip manufacture contains a velocity selector, in which a uniform electric field and a uniform magnetic field are perpendicular to each other and to the ion beam. Which ions pass through the selector undeflected?
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An ion is undeflected when the electric and magnetic forces balance: qE = qvB, which gives v = E/B. The charge q cancels and the mass does not appear, so any ion with speed E/B passes straight through. For a negative ion, both forces reverse together and still cancel. A specific charge-to-mass ratio matters only in the deflecting region that comes after the selector.
✓ 1
Answer C
Answer: C
Common error: In a velocity selector, both forces are proportional to q, so the condition for no deflection is independent of the sign and size of the charge.
50IBP2S-032-03-06Motion in an electric fieldPaper 1AMedium1 mark
In an old analogue oscilloscope, an electron enters the uniform field between two parallel deflecting plates, moving parallel to the plates. The arrangement is repeated with the potential difference between the plates doubled. The electron enters at the same speed as before. In both cases it leaves the region between the plates without striking either plate. What happens to the time the electron spends between the plates?
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The electric field between the plates is perpendicular to the plates. The electric force on the electron therefore acts only at right angles to its initial motion. There is no force parallel to the plates, so the velocity component parallel to the plates stays constant. The time between the plates, t = (plate length)/(speed parallel to plates), is therefore unchanged. Doubling the p.d. doubles the field and the perpendicular acceleration. This doubles the sideways deflection, but the transit time does not change. Option A wrongly assumes the larger force speeds the electron up along the plates. Option C confuses the transit time with the deflection. Option D wrongly applies an inverse-square relationship.
✓ 1
Answer B
Answer: B
Exam tip: Treat motion in a uniform field like projectile motion. Analyse the components parallel and perpendicular to the field separately.
51IBP2S-032-03-09Direction of the magnetic forcePaper 1AMedium1 mark
Three long, straight, parallel wires X, Y and Z lie in a plane, in that order, with adjacent wires a distance d apart. Each wire carries current I. The currents in X and Z are in the same direction and the current in Y is in the opposite direction. Two wires a distance d apart, each carrying current I, exert a force per unit length F0 on each other. What is the resultant force per unit length on X?
Cross-section of the three long wires, all in one plane.
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Y is at distance d and carries an opposite current, so it repels X with F0, directed away from Y. Z is at distance 2d and carries the same current direction, so it attracts X with F0/2, directed towards Z (and therefore towards Y). The resultant is F0 − F0/2 = F0/2, away from Y. The option 3F0/2 wrongly treats both forces as repulsive.
✓ 1
Answer B
Answer: B
Exam tip: Work out each pair force separately, with its direction, before combining them. Remember that doubling the distance halves the force.
52IBP2S-032-03-10Paths in electric and magnetic fieldsPaper 1AMedium1 mark
A spacecraft ion thruster accelerates ions from rest through a potential difference V. Ion P is singly charged and ion Q is doubly charged. The two ions have the same mass. What is \(\frac{v_Q}{v_P}\), the ratio of the final speed of Q to the final speed of P?
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The work done by the field equals the kinetic energy gained: qV = ½mv², so v = √(2qV/m). With m and V the same for both ions, v ∝ √q. Doubling the charge therefore multiplies the speed by √2. The answer 2 confuses kinetic energy (which does double) with speed.
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Answer C
Answer: C
Common error: Doubling the charge doubles the kinetic energy gained, not the speed. Speed depends on the square root of the kinetic energy.
53IBP2S-032-02-03Circular motion in a magnetic fieldPaper 1AMedium1 mark
A cyclotron used to produce medical isotopes accelerates protons in a uniform magnetic field. A proton's speed doubles while the field stays constant. What happens to the time taken for the proton to complete one circular orbit?
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r = mv/(qB), so the radius doubles when v doubles. The period T = 2πr/v = 2πm/(qB) does not depend on v, so the period stays the same. This is why a fixed-frequency alternating pd works in a cyclotron.
✓ 1
Answer B
Answer: B
Common error: Treating the orbit as a fixed circle and concluding that a faster particle must take less time; the radius grows in proportion to the speed.
54IBP2S-032-02-04Circular motion in a magnetic fieldPaper 1AMedium1 mark
A charged dust grain in a planetary ring moves in a circle under the action of a uniform magnetic field alone. Which statement about the grain is correct?
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The magnetic force is always perpendicular to the velocity, so it does no work and the speed (and kinetic energy) stays constant. The direction of the velocity changes continuously, so momentum, a vector, changes.
✓ 1
Answer B
Answer: B
Exam tip: Remember that a force acting perpendicular to motion does zero work, but it still changes the direction of the momentum.
55IBP2S-032-02-05Force on a moving chargePaper 1AMedium1 mark
Electrons from the Sun approach the Earth's polar regions, where the Earth's magnetic field is approximately uniform over a short distance. An electron enters this region with a velocity at 40° to the field lines. What is the shape of the electron's path while it is in this region?
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The velocity component perpendicular to B produces circular motion around the field lines. The component parallel to B experiences no force, so it carries the electron steadily along the field. The combination is a helix whose axis is along the field.
✓ 1
Answer C
Answer: C
Common error: A parabola is the path in a uniform electric field, where the force is in a fixed direction. A magnetic force keeps changing direction with the velocity.
56IBP2S-032-02-06The velocity selectorPaper 1AMedium1 mark
A velocity selector in an ion-implantation machine uses perpendicular uniform electric and magnetic fields. It lets ions of speed v pass through undeflected. Which single change lets ions of speed 2v pass through undeflected instead?
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An ion passes undeflected when qE = qvB, so v = E/B. Doubling E alone doubles the selected speed. Doubling B alone halves it. Doubling both leaves it unchanged. The charge cancels, so changing it has no effect.
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Answer B
Answer: B
Common error: Thinking a larger charge changes the selected speed. Both forces are proportional to q, so q cancels.
57IBP2S-032-02-08Force on a current-carrying wirePaper 1AMedium1 mark
A horizontal overhead cable runs east–west and carries a current flowing due east. Consider only the vertical component of the Earth's magnetic field at this location, which points downwards. What is the direction of the force on the cable due to this component?
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F = IL × B with L pointing east and B pointing down. East × down gives north. The force must be perpendicular to both the current and the field, so it is horizontal and cannot be vertical.
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Answer A
Answer: A
Common error: Giving a vertical force: the force can never be parallel to the field component that produces it.
58IBP2S-032-04-02The velocity selectorPaper 1AMedium1 mark
A proton moving with speed v passes undeflected through a velocity selector. Three other particles enter the same selector, each moving with the same velocity v. I. an electron II. an alpha particle III. a neutron Which of these particles also pass through undeflected?
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For a charged particle, both forces are proportional to q, so the balance qE = qvB depends only on v. For the electron, the sign of q reverses both forces, so they still cancel. For the alpha particle, both forces double, so they still cancel. The neutron experiences no electric or magnetic force, so it is undeflected as well. All three pass through undeflected.
✓ 1
Answer D
Answer: D
Common error: A greater mass or charge does not upset the balance. A velocity selector selects by speed only, not by charge, mass or charge-to-mass ratio.
59IBP2S-032-04-03The velocity selectorPaper 1AMedium1 mark
In a velocity selector, positive ions moving at speed E/B travel in a straight line. A positive ion enters the same selector, along the same line, with a speed greater than E/B. What is its initial motion in the selector?
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The electric force qE does not depend on speed. The magnetic force qvB increases with speed. When v > E/B, the magnetic force exceeds the electric force, so the resultant force is in the direction of the magnetic force. The ion curves that way.
✓ 1
Answer C
Answer: C
Exam tip: Ask which force depends on v. Faster particles are dominated by the magnetic force; slower particles are dominated by the electric force.
60IBP2S-032-04-05Motion in an electric fieldPaper 1AMedium1 mark
An electron enters the uniform electric field between two parallel plates. It is moving parallel to the plates, midway between them. Gravity is negligible. Which statement is correct while the electron is between the plates?
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The field is uniform, so the force eE is constant and acts perpendicular to the plates. There is no force parallel to the plates, so that velocity component is constant. The perpendicular component increases at a constant rate. This gives a parabolic path. The speed increases because the force does work, so B is wrong. The acceleration is constant, so C is wrong. A circular path needs a force that stays perpendicular to the velocity, so D is wrong.
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Answer A
Answer: A
Exam tip: Treat a charge in a uniform electric field like a projectile: constant velocity in one direction and constant acceleration in the other.
61IBP2S-032-04-06Motion in an electric fieldPaper 1AMedium1 mark
In an industrial inkjet printer, charged ink droplets travel horizontally at constant speed between two parallel deflecting plates. A fixed potential difference is applied across the plates. The plate separation is doubled, while the potential difference and the plate length stay the same. What happens to the vertical deflection of a droplet as it leaves the plates?
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The field strength E = V/d halves, so the force qE and the acceleration halve. The time spent between the plates, L/v, is unchanged. The deflection ½at2 therefore halves. Option A wrongly treats the field as an inverse-square field. Option C assumes the p.d. alone sets the deflection.
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Answer B
Answer: B
Common error: The field between parallel plates depends on the separation through E = V/d. This is an inverse relationship, not an inverse square.
62IBP2S-032-04-08Motion in an electric fieldPaper 1AMedium1 mark
In a test of an electrostatic spray-painting system, a charged paint droplet of mass 4.9×10−15 kg is held stationary between two horizontal plates. The plates are 1.2 cm apart, with a potential difference of 600 V between them. How many excess electrons are on the droplet?
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For the droplet to be stationary, qE = mg with E = V/d = 600/0.012 = 5.0×104 V m−1. Then q = (4.9×10−15 × 9.8)/(5.0×104) = 9.6×10−19 C. This is 9.6×10−19/1.60×10−19 = 6 electrons. Option B uses d = 0.12 m. Option C uses V in place of E. Option D uses d = 1.2 m.
✓ 1
Answer A
Answer: A
Common error: Convert centimetres to metres before calculating E = V/d. A wrong conversion changes the answer by a factor of 10 or 100.
63IBP2S-032-04-10The velocity selectorPaper 1AMedium1 mark
A velocity selector in a research mass spectrometer currently selects ions of speed v. The operator wants it to select ions of speed 2v instead. Which change achieves this?
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The selected speed is v = E/B. Doubling E with B fixed doubles v. Option A halves v. Option C leaves v unchanged. Option D halves v.
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Answer B
Answer: B
Exam tip: Write v = E/B and check each option by simple proportion.
64D-1A-34Charge in a uniform electric fieldPaper 1AHard1 mark
An electron enters the uniform electric field between two parallel plates, moving parallel to the plates. It leaves the plates with a deflection y.
The potential difference between the plates is doubled and the entry speed of the electron is doubled. The electron still leaves the plates. What is (new deflection)/y?
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Step 1The time between the plates is t = L/v: doubling v halves t.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The acceleration a = eV/md doubles when V doubles.
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Step 3y = ½at², so the factor is 2 × (½)² = 0.5.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis allows for the shorter time but not for the doubled p.d.
BCorrect: y ∝ V/v², so the factor is 2/2² = 0.5.
CThis takes y ∝ V/v, so that the two changes cancel; the time enters squared.
DThis allows for the doubled p.d. but not for the shorter time between the plates.
67D-1A-56Force on a current-carrying wirePaper 1AHard1 mark
A horizontal overhead cable runs from west to east and carries a direct current I towards the east. At this location the Earth's magnetic field has flux density B and is directed northwards at 60° below the horizontal. The diagram is a side view seen from the east, so the cable is seen end-on.
Which row gives the magnitude and the direction of the magnetic force on a length L of the cable?
Diagram NOT accurately drawn
MagnitudeDirection
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Step 1The cable runs east–west and the field lies in the vertical north–south plane, so the field is at 90° to the current: F = BIL sin 90° = BIL.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2The force is perpendicular to both the current and the field. Fleming's left-hand rule: the horizontal component B cos 60° gives an upward force and the downward component B sin 60° gives a force towards the north, so the resultant BIL acts at 90° to B, 30° above the horizontal towards the north.
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Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThis uses only the horizontal component of the field; the vertical component B sin 60° is also perpendicular to the cable and adds a force towards the north.
BThis uses only the vertical component of the field, as for a cable running north–south; here the horizontal component is also perpendicular to the cable.
CCorrect: the whole field is perpendicular to the cable, so F = BIL, at 90° to B: 30° above the horizontal, towards the north.
DThe magnitude is right, but the direction is reversed, for example by using the right hand instead of the left.
68D-1A-57Charge in a uniform electric fieldPaper 1AHard1 mark
A proton and an alpha particle, with the same kinetic energy, each enter the uniform electric field between the same pair of parallel plates, moving at right angles to the field. Both particles leave the region between the plates.
What is (deflection of the alpha particle)/(deflection of the proton) as they leave the plates?
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Step 1Time between the plates of length L: t = L/v; acceleration at right angles to the plates: a = qE/m.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Deflection y = ½(qE/m)(L/v)² = qEL²/(2mv²) = qEL²/(4Ek).
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Step 3E, L and Ek are the same for both particles, so y ∝ q: the alpha particle (charge 2e) is deflected twice as much, and the ratio is 2.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis takes both particles to enter at the same speed and also ignores the charge, so y ∝ 1/m gives 1/4.
BThis takes both particles to enter at the same speed, so y ∝ q/m = 2e/4mp gives 1/2; equal kinetic energies do not mean equal speeds.
CThis assumes particles with equal kinetic energies are always deflected equally, overlooking that the force on the alpha particle, 2eE, is twice as large.
DCorrect: with equal kinetic energies y = qEL²/(4Ek) ∝ q, so the ratio is 2.
69D-1A-86Circular motion in a magnetic fieldPaper 1AHard1 mark
A cloud chamber is placed in a uniform magnetic field directed into the page. A charged particle moves through the gas in the chamber at right angles to the field and leaves a track. The radius of curvature of the track decreases steadily along its length, so the track spirals inwards.
Which statement explains why the radius decreases?
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Step 1The magnetic force qvB provides the centripetal force, so r = mv/qB: at constant q and B the radius is proportional to the momentum.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2The magnetic force is perpendicular to the velocity and does no work; the particle slows because it transfers energy to the gas molecules it ionizes, so mv and r fall.
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Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThe magnetic force is always perpendicular to the velocity, so it does no work; the energy is lost to the gas.
BCorrect: kinetic energy is transferred to the gas by ionization, so the momentum and hence r = mv/qB decrease.
CThe field is uniform and F = qvB actually decreases as the particle slows; a smaller radius comes from a smaller momentum, not a larger field.
DIonizing the gas removes electrons from gas atoms; the charge of the moving particle does not keep increasing.
70D-1A-88Direction of the magnetic forcePaper 1AHard1 mark
A long straight wire in the plane of the page carries a current I to the right. A narrow beam of electrons in a vacuum travels to the right, parallel to the wire and a short distance d above it, in the plane of the page.
Which row gives the direction of the magnetic field of the wire at the beam and the direction of the magnetic force on the electrons?
Diagram NOT accurately drawn
Magnetic field at the beamForce on the electrons
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Step 1Right-hand grip rule: with the current to the right, the field above the wire is out of the page.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2The conventional current of the electrons is to the left; F = qv × B with q negative (or Fleming's left-hand rule with the current to the left) gives a force up the page, away from the wire.
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Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThis uses the grip rule the wrong way round; with that field the force on the electrons would be towards the wire.
BThe field is right, but this treats the electrons as positive charges, like a current in the same direction, which would be attracted.
CThis makes two errors that cancel in the force: the field is reversed and the electrons are treated as positive.
DCorrect: the field above the wire is out of the page and the force on the negative electrons is away from the wire.
71D-1A-92Force on a current-carrying wirePaper 1AHard1 mark
A lightning stroke transfers negative charge from a cloud to the ground along a straight vertical channel. At one instant the current in the channel is 2.0 × 104 A. At this place the Earth's magnetic field has a flux density of 5.0 × 10−5 T and is directed towards the north at 60° below the horizontal.
Which row gives the magnitude and the direction of the magnetic force on each metre of the channel at this instant?
Force / NDirection
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Step 1Negative charge moving down is a conventional current upwards; the angle between the upward current and the field is 90° + 60° = 150°.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2F = BIL sin 150° = 5.0 × 10−5 × 2.0 × 104 × 1 × 0.50 = 0.50 N; only the horizontal (northward) component of the field contributes, and up × north gives a force towards the west.
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Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThe magnitude is right, but this takes the current to be downwards, in the direction in which the electrons move.
BCorrect: F = BIL sin 150° = 0.50 N, directed west.
CThis takes the angle between the channel and the field as 60° instead of 150° (or uses the vertical component of the field).
DThis treats the whole field as perpendicular to the channel and leaves out sin θ.
72IBP2S-032-01-09Force on a moving chargePaper 1AHard1 mark
An alpha particle and a proton each move in a circular path in the same uniform magnetic field. The alpha particle has four times the mass and twice the charge of the proton. Their speeds are different. What is \(\frac{\text{period of alpha particle}}{\text{period of proton}}\)?
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The period is T = 2πr/v, with r = mv/(qB). This gives T = 2πm/(qB), which does not depend on speed. The ratio is therefore (4/2) = 2. The option 4 ignores the charge. The option 1 assumes the period is the same for all particles. The option 1/2 inverts the ratio.
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Answer C
Answer: C
Exam tip: The period of circular motion in a magnetic field does not depend on the speed. Different speeds are irrelevant here.
73IBP2S-032-01-10Paths in electric and magnetic fieldsPaper 1AHard1 mark
An electron and a proton have the same kinetic energy. Each enters, separately, the same uniform electric field between parallel plates, moving at right angles to the field. Ignore gravity. Which statement compares their vertical deflections over the length of the plates?
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The deflection is y = ½at2, with a = qE/m and t = x/v. This gives y = qEx2/(2mv2). Since mv2 = 2Ek, we get y = qEx2/(4Ek). The mass cancels. With equal |q| and equal kinetic energy, the magnitudes are equal. The opposite charges give opposite directions. The 'mass ratio' options wrongly assume equal speeds.
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Answer A
Answer: A
Common error: Equal kinetic energy is not the same as equal speed. Express the deflection in terms of the given quantity before comparing.
74IBP2S-032-00-08Charge in a uniform electric fieldPaper 1AHard1 mark
In an electron-beam lithography machine, electrons enter the uniform electric field between two parallel deflecting plates. They enter moving parallel to the plates and do not strike them. On leaving the plates their vertical deflection is y. The p.d. between the plates is doubled and the entry speed of the electrons is also doubled. The length and separation of the plates are unchanged. What is the new deflection?
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Notes
The time spent between plates of length L is t = L/v. The acceleration is a = qE/m, where E = V/d. The deflection is therefore y = ½(qV/(md))(L/v)2, so y ∝ V/v2. Doubling V doubles y. Doubling v halves t, which reduces y by a factor of 4. The overall factor is 2/4 = ½, so the new deflection is y/2. The distractors come from these errors: y assumes the two changes cancel, treating y ∝ V/v. y/4 ignores the change in V. 2y ignores the change in transit time.
✓ 1
Answer B
Answer: B
Exam tip: Write a single expression for y in terms of all the variables before scaling. The deflection depends on the time squared, so it varies as 1/v².
75IBP2S-032-00-10Charge in a uniform electric fieldPaper 1AHard1 mark
In a vacuum, an electron is released from rest at the negative plate of a pair of parallel plates. It takes time t to reach the positive plate. The plate separation is then doubled, while the p.d. between the plates is kept the same. What is the new time taken to cross the gap?
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Marking point
Mark
Notes
The acceleration is a = eV/(md). From d = ½at², t = √(2d/a) = d√(2m/(eV)), so t ∝ d when V is constant. Doubling d therefore doubles t. A physical check: the final speed is unchanged, because the energy gained is eV in both cases. The average speed is therefore the same, and twice the distance takes twice the time. Distractors:
t assumes that the same final speed means the same time.
√2 t forgets that the acceleration halves.
4t squares the factor incorrectly.
✓ 1
Answer C
Answer: C
Exam tip: If V is constant but d changes, the field E = V/d changes too. Always recalculate the acceleration.
76IBP2S-032-03-07Circular motion in a magnetic fieldPaper 1AHard1 mark
In a cyclotron that produces isotopes for medical imaging, protons move in circular paths in a uniform magnetic field. A proton's speed is doubled, with the field unchanged. What happens to the radius of its path and to its period of revolution? Ignore relativistic effects.
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Marking point
Mark
Notes
r = mv/(qB), so the radius is proportional to speed and doubles. The period is T = 2πr/v = 2πm/(qB), which does not depend on speed. The period is unchanged: the proton covers twice the distance at twice the speed. This is why a cyclotron can use an alternating voltage of fixed frequency.
✓ 1
Answer A
Answer: A
Exam tip: Substitute r = mv/(qB) into T = 2πr/v to show that the period does not depend on v. This is a useful result to derive quickly.
77IBP2S-032-02-09Charge-to-mass ratioPaper 1AHard1 mark
In radiocarbon dating by mass spectrometry, singly charged ions of carbon-12 and carbon-14 are accelerated from rest through the same potential difference. They then enter the same uniform magnetic field perpendicular to their velocity. What is \(\frac{\text{radius of carbon-14 path}}{\text{radius of carbon-12 path}}\)?
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Mark
Notes
Both ions gain the same kinetic energy, Ek = qV, so their momentum is p = √(2mEk) ∝ √m. Since r = p/(qB) with q and B equal, r ∝ √m. The ratio is therefore √(14/12).
✓ 1
Answer B
Answer: B
Common error: Using r ∝ m assumes equal speeds. Ions accelerated through the same pd have equal kinetic energies, not equal speeds.
78IBP2S-032-02-10Direction of the magnetic forcePaper 1AHard1 mark
A particle detector has a uniform magnetic field directed vertically upwards. A charged particle moves in a horizontal plane. Viewed from above, its track is a clockwise spiral of steadily decreasing radius. What can be deduced about the particle?
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Mark
Notes
Viewed from above, B points towards the viewer. For a positive charge moving to the right, v × B points down the page, so the track bends clockwise. The charge is therefore positive. Since r = mv/(qB), a decreasing radius means a decreasing speed: the particle is losing energy to the detector material.
✓ 1
Answer A
Answer: A
Exam tip: Split the problem: use the sense of rotation to find the sign of the charge, and use the change in radius to find the change in speed.
79IBP2S-032-04-07Paths in electric and magnetic fieldsPaper 1AHard1 mark
A proton and an alpha particle are each accelerated from rest through the same potential difference. Each then enters the same uniform electric field between parallel deflecting plates, moving initially parallel to the plates. What is the ratio \(\frac{\text{deflection of alpha particle}}{\text{deflection of proton}}\) at the end of the plates?
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Notes
The deflection is y = ½(qE/m)(L/v)2 = qEL2/(2mv2). After acceleration through p.d. V, mv2 = 2qV. Substituting gives y = EL2/(4V). This is independent of both q and m, so the deflections are equal and the ratio is 1.
✓ 1
Answer A
Answer: A
Exam tip: Write the deflection in terms of kinetic energy before comparing particles. Here the charge and mass both cancel.
80D-1B-05Force on a current-carrying wirePaper 1BEasy9 marks
A student investigates the force on a current-carrying conductor. Two flat magnets are fixed to a steel yoke that rests on a top-pan balance of resolution 0.01 g. Seen from above, the north pole is on the student's left and the south pole on the student's right. A stiff copper rod is clamped horizontally so that it runs directly away from the student through the gap between the poles, without touching the magnets. The length of rod in the field is equal to the length of the pole faces, (5.0 ± 0.1) cm.
When there is a current I in the rod, the balance reading decreases by Δm. The uncertainty in each value of Δm is ±0.02 g.
I / A
Δm / g
0.50
0.20
1.00
0.37
1.50
0.58
2.00
0.76
2.50
0.94
3.00
1.15
Graph drawn to scale
(a)
Show that the force on the rod when I = 3.00 A is about 1.1 × 10−2 N.
(1)
(b)
Draw the line of best fit for the data.
(1)
(c)
Determine, using the gradient of your line, the magnetic flux density B between the poles.
(2)
(d)
Deduce the direction of the current in the rod.
(2)
(e)
A calibrated magnetic field sensor placed between the poles reads 0.075 T. Ignoring the uncertainty in the gradient, determine the absolute uncertainty in your value of B and deduce whether the two values agree.
(2)
(f)
Suggest one change to the apparatus that would reduce the percentage uncertainty in Δm for the smallest currents.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
F = Δm g = 1.15 × 10−3 × 9.81 = 1.13 × 10−2 N
✓ 1
Must see full substitution with Δm in kg OR answer to at least 3 s.f.
Part (b)
Single straight line through all the error bars, passing through or very close to the origin
✓ 1
Do not accept a line joining the points.
Part (c)
gradient = 0.379 g A−1 (= 3.8 × 10−4 kg A−1)
✓ 1
Accept 0.37–0.39 g A−1. Gradient from a large triangle.
B = gradient × g/L = 3.79 × 10−4 × 9.81/0.050 = 0.0744 T
✓ 1
Allow ECF from the gradient. Accept 0.073–0.077 T. Award [0] for MP2 if L is not in metres.
Part (d)
The reading decreases, so the force on the magnets is upwards; by Newton's third law the force on the rod is downwards
✓ 1
The field is from left to right (N to S), so by the left-hand rule / F = IL × B the current is directed away from the student
✓ 1
Award MP2 only if a downward force on the rod is used. OWTTE
Part (e)
ΔB/B = ΔL/L = 0.1/5.0 = 2 %, so ΔB = ±0.0015 T
✓ 1
Allow ECF from (c). Accept ±0.001 T or ±0.002 T.
B = (0.0744 ± 0.0015) T, i.e. 0.0730–0.0759 T, which includes 0.075 T, so the values agree
✓ 1
The conclusion must be consistent with the candidate's range.
Part (f)
Use magnets with longer pole faces / several magnets in a row (larger L) OR stronger magnets, so that Δm is larger compared with the 0.01 g resolution
✓ 1
Accept a balance of higher resolution (e.g. 0.001 g). Do not accept “repeat the readings” alone.
Answers: (a) 1.13 × 10−2 N · (c) B = 0.0744 T · (e) B = (0.0744 ± 0.0015) T (the remaining parts are explanations — see the table above)
81D-1B-17Force on a current-carrying wirePaper 1BEasy9 marks
A straight horizontal wire of length (40.0 ± 0.5) mm lies between the wide, flat pole pieces of an electromagnet, where the magnetic field is uniform and horizontal. The wire is fixed to a holder on a force sensor and can be turned in the horizontal plane so that it makes an angle θ with the field; the leads to the wire are twisted together outside the field.
The current in the wire is kept at (2.00 ± 0.01) A. The sensor measures the vertical force F on the wire with an uncertainty of ±0.2 mN.
θ / °
sin θ
F / mN
0
0.000
0.1
15
0.259
4.2
30
7.9
45
0.707
11.4
60
0.866
13.8
75
15.6
90
1.000
15.9
Graph drawn to scale
(a)
Complete the table.
(1)
(b)
Plot the two missing points on the graph and draw the line of best fit for the data.
(2)
(c)
Determine the gradient of the line.
(2)
(d)
Calculate the magnetic flux density B between the pole pieces.
(1)
(e)
The uncertainty in the gradient is ±2 %. Calculate the percentage uncertainty in B.
(1)
(f)
The manufacturer states that B = 0.205 T for this setting. Deduce whether the student's value agrees with this.
(1)
(g)
The wire becomes warm during the experiment. Suggest how the student could make sure that this does not affect the results.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
0.500 AND 0.966
✓ 1
Both needed. Accept 0.5 for sin 30°.
Part (b)
Both points plotted correctly to within half a small square
✓ 1
Allow ECF from (a).
Single straight line through all the error bars, passing through or very close to the origin
✓ 1
Part (c)
Gradient calculated from a large triangle
✓ 1
gradient = 15.9 mN
✓ 1
Accept 15.4–16.4 mN. Unit required.
Part (d)
B = gradient/(IL) = 1.59 × 10−2/(2.00 × 0.0400) = 0.199 T
✓ 1
Allow ECF from (c). Award [0] if L is not in metres.
Part (e)
2 % + 0.5 % + 1.25 % = 3.8 %
✓ 1
Accept 3.8 % or 4 %.
Part (f)
ΔB = 3.8 % × 0.199 = ±0.007 T; the range 0.192–0.206 T includes 0.205 T, so the values agree
✓ 1
Allow ECF from (d) and (e).
Part (g)
Check the ammeter before each reading and adjust a variable resistor (rheostat) so that the current stays at 2.00 A (the resistance of the wire increases as it warms) OR switch off between readings so the wire stays cool
✓ 1
OWTTE
Answers: (a) 0.500, 0.966 · (c) 15.9 mN · (d) B = 0.199 T · (e) 3.8 % (the remaining parts are explanations — see the table above)
82D-1B-42Force on a current-carrying wirePaper 1BEasy8 marks
A long solenoid has 600 turns wound uniformly over a length of 0.300 m and lies, with its axis horizontal, on a top-pan balance of resolution 0.001 g. A straight horizontal wire, held by a clamp that does not touch the solenoid, passes through two small holes across the middle of the solenoid, at right angles to its axis. The length of wire inside the solenoid is (40.0 ± 0.5) mm and the current in the wire is kept at (4.00 ± 0.02) A.
The current Is in the solenoid is varied and the change Δm in the balance reading is recorded, with an uncertainty of ±0.002 g. Theory gives the flux density inside a long solenoid as B = μ0nIs, where n is the number of turns per unit length.
Is / A
Δm / g
0.50
0.021
1.00
0.040
1.50
0.061
2.00
0.083
2.50
0.101
3.00
0.123
Graph drawn to scale
(a)
Show that the magnetic force on the wire when Is = 3.00 A is about 1.2 × 10−3 N.
(1)
(b)
Draw the line of best fit for the data.
(1)
(c)
Determine the gradient of your line.
(1)
(d)
Determine, using your gradient, the magnetic flux density inside the solenoid per ampere of solenoid current, B/Is.
(2)
(e)
Calculate the percentage uncertainty in your answer to (d), ignoring the uncertainty in the gradient, and deduce whether your value agrees with μ0n.
(2)
(f)
A second student places the wire close to one end of the solenoid. State and explain the effect on the gradient of the graph.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
F = Δm g = 0.123 × 10−3 × 9.81 = 1.21 × 10−3 N
✓ 1
Must see full substitution with Δm in kg OR answer to 3 s.f. «by Newton's third law the force on the solenoid equals the force on the wire»
Part (b)
single straight line through all the error bars, passing through or very close to the origin
✓ 1
Part (c)
gradient = 4.09 × 10−2 g A−1
✓ 1
Accept 3.9 × 10−2–4.2 × 10−2 g A−1. Unit not required.
Part (d)
B = F/(IL) so B/Is = gradient × g/(IL) with the gradient in kg A−1
Allow ECF from (c). Award [1] max for a power-of-ten error.
Part (e)
0.5 % + 1.25 % = 1.75 % «≈ 1.8 %»
✓ 1
Accept 2 %.
μ0n = 4π × 10−7 × 2000 = 2.51 × 10−3 T A−1, which lies within 2.46 × 10−3–2.55 × 10−3 T A−1, so the values agree
✓ 1
Allow ECF from (d).
Part (f)
the field near the end of a solenoid is weaker «about half» than in the middle, so the force for each value of Is is smaller and the gradient is smaller
✓ 1
OWTTE.
Answers: (a) 1.21 × 10−3 N · (c) 4.09 × 10−2 g A−1 · (d) 2.51 × 10−3 T A−1 · (e) 1.75 % (the remaining parts are explanations — see the table above)
83D-1B-06Force on a current-carrying wirePaper 1BMedium10 marks
A horizontal copper rod of mass (4.80 ± 0.01) g hangs from two light vertical conducting wires, forming a swing. A pair of large coils produces a uniform vertical magnetic field of flux density B in the region of the rod. The length of the rod between the two suspension wires is (0.120 ± 0.001) m.
When there is a current I in the rod it swings sideways and comes to rest with the suspension wires at an angle φ to the vertical, measured with a protractor to ±1°. In equilibrium, tan φ = BIL/(mg), where m is the mass and L the length of the rod. The graph shows tan φ against I with error bars.
I / A
φ / °
tan φ
1.0
5
0.087
2.0
10
0.176
3.0
15
0.268
4.0
20
0.364
5.0
24
0.445
6.0
28
0.532
Graph drawn to scale
(a)
Outline why the magnetic forces on the two suspension wires can be ignored.
(1)
(b)
Show that the absolute uncertainty in tan φ for I = 3.0 A is about 0.02.
(1)
(c)
Draw the line of best fit and the lines of maximum and minimum gradient for the data.
(2)
(d)
Determine the gradient of the line of best fit and its absolute uncertainty.
(2)
(e)
Determine B and its absolute uncertainty.
(2)
(f)
A calibrated magnetic field sensor gives B = 0.034 T. Comment on the student's result.
(1)
(g)
Suggest how the measurement of the deflection could be changed to reduce the uncertainty in the result.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The currents in the two suspension wires are in opposite directions, so the magnetic forces on them are equal and opposite (along the line of the rod) and cancel OR for small deflections the wires are almost parallel to the vertical field, so the force on them is very small
✓ 1
OWTTE
Part (b)
tan 16° − tan 15° = 0.2867 − 0.2679 = 0.0188
✓ 1
Accept ½(tan 16° − tan 14°) = 0.0187. Must see the calculation OR an answer to at least 2 s.f. (0.019).
Part (c)
Line of best fit passing through all the error bars, with points on both sides
✓ 1
Steepest and least steep lines that pass through all (or nearly all) the error bars
Accept ±0.005 to ±0.010 A−1. Allow ECF from the candidate's lines.
Part (e)
B = gradient × mg/L = 0.0893 × 4.80 × 10−3 × 9.81/0.120 = 3.50 × 10−2 T
✓ 1
Allow ECF from (d).
fractional uncertainty = 9.0 % + 0.2 % + 0.8 % ≈ 10 %, so B = (3.5 ± 0.4) × 10−2 T
✓ 1
MP2 is for adding fractional uncertainties AND matching the precision of value and uncertainty. Accept ignoring the (negligible) mass and length terms.
Part (f)
0.034 T lies within the range 3.1–3.9 × 10−2 T, so the result is consistent with the sensor value
✓ 1
Allow ECF from (e).
Part (g)
Measure the horizontal displacement of the rod with a ruler (or from a photograph against a grid) and calculate the angle from it, because ±1° is a large fractional uncertainty for small angles
✓ 1
Accept: use larger currents/larger angles so the fractional uncertainty in φ is smaller. Do not accept “use a more accurate protractor” without detail.
Answers: (b) 0.0188 · (d) (0.089 ± 0.008) A−1 · (e) B = (3.5 ± 0.4) × 10−2 T (the remaining parts are explanations — see the table above)
84D-1B-09Charge in a uniform electric fieldPaper 1BMedium11 marks
In an electron deflection tube, electrons accelerated through a potential difference Va = (3.00 ± 0.05) kV enter the region between two horizontal parallel plates, travelling horizontally midway between them. The plates are (50.0 ± 0.5) mm apart and the upper plate is at a potential difference Vp = (1.20 ± 0.02) kV above the lower plate.
The path of the beam is shown on a fluorescent screen marked with a grid. A student reads the upward deflection y of the beam at horizontal distances x from the point where the beam enters the field. Each value of y has an uncertainty of ±1 mm. Theory predicts y = kx², where k = Vp/(4dVa) and d is the plate separation.
x / cm
y / mm
x² / cm²
4
3
16
5
5
25
6
7
7
10
49
8
13
64
9
16
10
20
100
Graph drawn to scale
(a)
Explain why the beam is deflected upwards, towards the upper plate.
(2)
(b)
Complete the table.
(1)
(c)
Plot the two missing points on the graph and draw the line of best fit for the data.
(2)
(d)
Determine the gradient of the line, in mm cm−2.
(2)
(e)
Calculate the predicted value of k, in mm cm−2, and its absolute uncertainty.
(2)
(f)
Deduce whether the data support the theory.
(1)
(g)
Suggest one change to the settings that would reduce the percentage uncertainty in the values of y.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
the upper plate is at the higher potential, so the electric field between the plates is directed downwards, from the upper plate to the lower plate
✓ 1
electrons are negatively charged, so the electric force on them is opposite to the field, upwards
✓ 1
MP2 only scores if the direction of the field is correct or not stated. OWTTE
Part (b)
36 AND 81
✓ 1
Both needed; integers expected.
Part (c)
Both points plotted correctly to within half a small square
✓ 1
Allow ECF from (b).
Single straight line through all the error bars, passing through or very close to the origin
✓ 1
Part (d)
Gradient calculated from a large triangle (at least half the line)
✓ 1
gradient = 0.201 mm cm−2
✓ 1
Accept 0.190–0.210 mm cm−2.
Part (e)
k = 1.20 × 103/(4 × 0.0500 × 3.00 × 103) = 2.00 m−1 = 0.200 mm cm−2
✓ 1
Unit conversion required: 1 m−1 = 0.1 mm cm−2.
percentage uncertainty = 1.7 % + 1.7 % + 1.0 % = 4.3 %, so k = (0.200 ± 0.009) mm cm−2
✓ 1
Accept ±0.008 to ±0.009 mm cm−2. Allow ECF for the value of k.
Part (f)
The measured gradient lies within the predicted range 0.191–0.209 mm cm−2 and the line passes through the origin, so the data support the theory
✓ 1
Allow ECF from (d) and (e).
Part (g)
Increase Vp (within the limit of the plates) or decrease Va, so that the deflections y are larger compared with the ±1 mm reading uncertainty
✓ 1
Do not accept “use a finer grid” as a change to the settings.
Answers: (b) 36, 81 · (d) 0.201 mm cm−2 · (e) k = (0.200 ± 0.009) mm cm−2(the remaining parts are explanations — see the table above)
85D-1B-15Circular motion in a magnetic fieldPaper 1BMedium11 marks
A student uses a computer simulation of a mass spectrometer. Singly charged positive ions, initially at rest, are accelerated through a potential difference of 2.00 kV and then enter a uniform magnetic field of flux density 0.250 T at right angles to the field. Each ion travels through a semicircle and strikes a detector. The student measures the diameter D of each semicircle with the simulation's on-screen ruler, to ±1 mm.
The masses m of the ions, in unified atomic mass units (u), are taken from a published table of isotope masses. The student plots lg (D / mm) against lg (m / u); five points have been plotted.
Ion
m / u
D / mm
lg (m / u)
lg (D / mm)
lithium-7
7.016
136
0.846
2.134
carbon-12
12.000
179
1.079
2.253
oxygen-16
15.995
207
neon-20
19.992
230
1.301
2.362
argon-40
39.962
326
1.602
2.513
iron-56
55.935
385
copper-63
62.930
409
1.799
2.612
Graph drawn to scale
(a)
Complete the table.
(2)
(b)
Plot the two missing points on the graph and draw the line of best fit for the data.
(2)
(c)
Determine the gradient of the line.
(2)
(d)
Explain, by considering the motion of an ion, why theory predicts this value of the gradient.
(2)
(e)
An unknown singly charged ion gives D = (268 ± 1) mm under the same conditions. Determine its mass, in u, with its absolute uncertainty, and deduce whether the ion is aluminium-27 (26.982 u) or silicon-28 (27.977 u).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
lg (m / u) = 1.204 AND 1.748
✓ 1
Accept 3 or 4 d.p.
lg (D / mm) = 2.316 AND 2.585
✓ 1
Accept 3 or 4 d.p.; do not accept 2 d.p.
Part (b)
Both points plotted correctly to within half a small square
✓ 1
Allow ECF from (a).
Single straight line of best fit through (or very close to) all seven points
✓ 1
Part (c)
Gradient calculated from a large triangle (at least half the line)
✓ 1
gradient = 0.50
✓ 1
Accept 0.48–0.52. No unit.
Part (d)
qvB = mv²/r gives r = mv/(qB) AND qV = ½mv² gives v = √(2qV/m)
✓ 1
so r = √(2mV/q)/B ∝ m½ (same q, V and B): D ∝ m0.5, so the gradient is 0.5
✓ 1
OWTTE
Part (e)
m = qB²r²/(2V) = 1.60 × 10−19 × 0.250² × 0.134²/(2 × 2000) = 4.49 × 10−26 kg = 27.0 u
✓ 1
Accept the ratio method m = 12.000 × (268/179)² = 26.9 u, or a reading from the graph (lg D = 2.428).
Δm/m = 2 × 1/268 = 0.75 %, so m = (27.0 ± 0.2) u
✓ 1
The fractional uncertainty in D must be doubled. Allow ECF.
The range 26.8–27.2 u includes 26.982 u but not 27.977 u, so the ion is aluminium-27
✓ 1
Award MP3 only if the decision follows from a stated range.
Answers: (a) 1.204, 2.316; 1.748, 2.585 · (c) 0.50 · (e) m = (27.0 ± 0.2) u — aluminium-27 (the remaining parts are explanations — see the table above)
86D-1B-27Charge in a uniform electric fieldPaper 1BMedium11 marks
Electrons leave a hot filament with negligible speed and are accelerated through a potential difference V by the uniform electric field between the filament and a grid. They then travel at constant speed v through a field-free tube. Two detectors (0.500 ± 0.001) m apart record the time of flight t of each pulse of electrons between them, with an uncertainty of ±0.5 ns.
A student uses the data to test whether ½mev² = eV and to determine e/me. Four of the points of the graph of v² against V have been plotted.
V / V
t / ns
v / 106 m s−1
v² / 1013 m² s−2
100
85.9
5.82
3.39
200
61.2
8.17
6.67
300
50.2
9.96
400
43.7
11.4
13.1
500
39.2
12.8
600
35.9
13.9
19.4
Graph drawn to scale
(a)
Show that, if ½mev² = eV, the graph of v² against V should be a straight line through the origin with a gradient equal to 2e/me.
(1)
(b)
Complete the table and plot the two missing points on the graph.
(2)
(c)
Calculate the absolute uncertainty in v² for V = 600 V.
(2)
(d)
Draw the line of best fit for the data.
(1)
(e)
Determine e/me using the gradient of your line.
(2)
(f)
The percentage uncertainty in your answer to (e) is about the same as the percentage uncertainty in (c). Compare your answer to (e) with the accepted value of e/me calculated from the data booklet.
(1)
(g)
The student later finds that the timing electronics add a constant delay of 1.5 ns to every measured time. Explain how this delay affects the value of e/me obtained.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
v² = (2e/me)V, which has the form y = kx: a straight line through the origin with gradient 2e/me
Single straight line through the origin passing through or close to all six points
✓ 1
Accept a line that passes between the points with roughly equal scatter on each side.
Part (e)
gradient = 3.26 × 1011 m² s−2 V−1 (C kg−1)
✓ 1
Accept 3.16 × 1011–3.36 × 1011.
e/me = gradient/2 = 1.63 × 1011 C kg−1
✓ 1
Allow ECF from MP1.
Part (f)
accepted value 1.60 × 10−19/9.11 × 10−31 = 1.76 × 1011 C kg−1; the student's value is about 7 % smaller, more than its uncertainty of about 3 %, so the two do not agree
✓ 1
Allow ECF from (c) and (e).
Part (g)
Each measured time is too long, so each calculated speed (and v²) is too small
✓ 1
The effect is largest for the shortest times / largest V, so the gradient and hence e/me are too small — this accounts for the discrepancy in (f)
✓ 1
OWTTE
Answers: (b) 9.92, 16.3 × 1013 m² s−2 · (c) ±0.62 × 1013 m² s−2 · (e) 1.63 × 1011 C kg−1(the remaining parts are explanations — see the table above)
87D-1B-41Charge in a uniform electric fieldPaper 1BMedium12 marks
In a school oscilloscope tube, electrons accelerated from rest through a potential difference Va pass between two parallel deflecting plates and then travel in a straight line to a fluorescent screen. The potential difference between the plates is kept at Vp = (24.0 ± 0.5) V. A student reads the vertical deflection D of the spot on the screen from the screen's grid to ±0.5 mm. Va is set to ±1 %.
Theory predicts that D = kVan, where n = −1 and k = VplL/(2d). The plates have length l = (25.0 ± 0.5) mm and separation d = (6.0 ± 0.1) mm, and the screen is L = 0.180 m from the centre of the plates. The graph shows lg(D/mm) against lg(Va/V) for four of the six measurements.
Va / V
D / mm
lg(Va/V)
lg(D/mm)
400
22.5
2.602
1.352
500
18.0
2.699
1.255
600
15.0
800
11.0
2.903
1.041
1000
9.0
1200
7.5
3.079
0.875
Graph drawn to scale
(a)
Explain, by considering the motion of the electrons between the plates, why D decreases when Va is increased.
(2)
(b)
Complete the table.
(1)
(c)
Plot the two missing points on the graph and draw the line of best fit.
(2)
(d)
Determine the gradient of your line and state whether it supports the theory.
(2)
(e)
Calculate k in V mm from the theory and deduce whether the result for Va = 600 V is consistent with it.
(3)
(f)
Identify, with a reason, the measurement of D that is least reliable, and suggest a change to the settings that would reduce its percentage uncertainty.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
a larger Va gives the electrons a larger speed along the plates, so they spend less time between them
✓ 1
the force eE «and acceleration» at right angles to the plates is unchanged, so they gain a smaller vertical velocity «and deflection»; D ∝ 1/v² ∝ 1/Va
✓ 1
OWTTE. MP2 needs the unchanged force/acceleration.
Part (b)
lg Va = 2.778 and 3.000 AND lg D = 1.176 and 0.954
✓ 1
All four needed. Accept 3 or 4 d.p.; do not accept 2 d.p.
Part (c)
both points plotted correctly to within half a small square
✓ 1
Allow ECF from (b).
single straight line of best fit passing through all the error bars
✓ 1
Part (d)
gradient = −1.00
✓ 1
Accept −1.04 to −0.96. No unit.
the gradient is equal to −1 within its uncertainty «steepest and least steep lines give about −1.08 to −0.94», so the data support D ∝ 1/Va
✓ 1
Allow ECF from the candidate's gradient. A conclusion with no reference to −1 scores [0].
Part (e)
k = 24.0 × 25.0 × 180/(2 × 6.0) = 9000 V mm
✓ 1
from the data DVa = 15.0 × 600 = 9000 V mm
✓ 1
Accept the graph value 10intercept if the gradient is −1.
percentage uncertainty of k ≈ 6 % and of DVa ≈ 4 %; the two values agree within these uncertainties, so the result is consistent
✓ 1
Award MP3 only for a comparison that refers to an uncertainty. Allow ECF.
Part (f)
D = 7.5 mm at Va = 1200 V: ±0.5 mm is the largest fraction, 6.7 %, of this reading «compared with 2.2 % for 22.5 mm»
✓ 1
increase Vp «so that all deflections are larger, as long as the spot stays on the screen at the smallest Va»
✓ 1
Accept: use only smaller values of Va, provided the spot stays on the screen. Do not accept «repeat readings» alone.
Answers: (b) 2.778, 3.000; 1.176, 0.954 · (d) −1.00 · (e) k = 9000 V mm; DVa = 9000 V mm (the remaining parts are explanations — see the table above)
88D-1B-43Charge in a uniform electric fieldPaper 1BMedium11 marks
In a time-of-flight mass spectrometer, singly charged positive ions start from rest and are accelerated through a potential difference of 2.00 kV by a uniform electric field. They then drift at constant speed along a field-free tube of length 1.20 m to a detector. A timer starts when a voltage pulse releases the ions and stops when they reach the detector. The acceleration region is short, so the time spent in it may be ignored.
To calibrate the instrument, a student records the time of flight t (uncertainty ±0.02 μs) for five ions whose masses m, in unified atomic mass units (u), are taken from a published table. 1 u = 1.66 × 10−27 kg. The graph shows t against √(m/u) for three of the ions.
Ion
m / u
√(m/u)
t / μs
lithium-7
7.016
2.649
5.53
sodium-23
22.990
9.66
potassium-39
38.964
6.242
12.46
rubidium-85
84.912
18.22
caesium-133
132.905
11.528
22.67
Graph drawn to scale
(a)
Show that the speed of a potassium-39 ion in the drift tube is about 1.0 × 105 m s−1.
(1)
(b)
Show that, if the timer is perfect, t is proportional to √m.
(2)
(c)
Complete the table.
(1)
(d)
Plot the two missing points on the graph and draw the line of best fit.
(2)
(e)
Determine the gradient of your line and, by calculation, its intercept on the t-axis.
(2)
(f)
Outline what the intercept suggests about the timing system.
(1)
(g)
An unknown singly charged ion has a time of flight of 15.73 μs. Determine its mass in u and deduce whether it is nickel-58 (57.935 u), copper-63 (62.930 u) or zinc-66 (65.926 u).
t = L/v = L√(m/2qV); L, q and V are the same for every ion, so t ∝ √m
✓ 1
Must see the step from v to t.
Part (c)
4.795 AND 9.215
✓ 1
Both needed. Accept 3 or 4 s.f.
Part (d)
both points plotted correctly to within half a small square
✓ 1
Allow ECF from (c).
single straight line through all the points, which does not pass through the origin
✓ 1
Part (e)
gradient = 1.93 μs
✓ 1
Accept 1.90–1.96 μs. Unit not required.
intercept = 0.41 μs
✓ 1
Accept 0.2–0.6 μs. Must be calculated from t = gradient × √(m/u) + intercept using a point on the line.
Part (f)
there is a constant delay of about 0.4 μs added to every time «a systematic error, e.g. the timer starts early or the detector/electronics respond late»
✓ 1
OWTTE.
Part (g)
√m = (15.73 − intercept)/gradient = 7.933, so m = 62.9 u
✓ 1
Allow ECF from (e). Accept 61.5–64.5 u. Award [0] for MP1 if the intercept is ignored «66.3 u, which would wrongly suggest zinc-66».
closest to 62.930 u and far from the other two masses, so the ion is copper-63
✓ 1
Conclusion must follow from the candidate's mass.
Answers: (a) 9.95 × 104 m s−1 · (c) 4.795, 9.215 · (e) gradient 1.93 μs; intercept 0.41 μs · (g) 62.9 u (copper-63) (the remaining parts are explanations — see the table above)
89D-1B-44The velocity selectorPaper 1BMedium12 marks
A research group studies the aluminium ions in the plume produced when pulses from a laser strike an aluminium target in a vacuum. The ions have a wide range of speeds. Ions of one speed are picked out by a velocity selector: two parallel plates (12.0 ± 0.1) mm apart, with a potential difference Vp between them, in a uniform magnetic field of fixed flux density B at right angles to both the electric field and the beam.
Ions that pass straight through the selector travel along a field-free drift tube of length (0.800 ± 0.002) m to a detector. Their time of flight t along the tube is recorded with an uncertainty of ±0.2 μs. The uncertainty in Vp is negligible. A student analyses the group's data. The graph shows t against 1/Vp for four of the six settings.
Vp / V
t / μs
(1/Vp) / V−1
12.0
40.1
0.0833
15.0
31.9
20.0
24.1
0.0500
25.0
19.1
0.0400
30.0
16.1
40.0
11.9
0.0250
Graph drawn to scale
(a)
Show that t = BdL/Vp, where d is the separation of the plates and L is the length of the drift tube.
(2)
(b)
Complete the table.
(1)
(c)
Plot the two missing points on the graph and draw the line of best fit.
(2)
(d)
Determine the gradient of your line. State its unit.
(2)
(e)
The uncertainty in the gradient is ±2 %. Determine B and its absolute uncertainty.
(3)
(f)
A calibrated magnetic field sensor placed between the plates reads 0.0490 T. Comment on the student's result.
(1)
(g)
Suggest one change to the apparatus that would reduce the percentage uncertainty in the measured times.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
for an ion that passes straight through, qE = qvB with E = Vp/d, so v = Vp/(Bd)
✓ 1
t = L/v = BdL/Vp
✓ 1
Answer given: must see both steps.
Part (b)
0.0667 AND 0.0333
✓ 1
Both needed. Accept 0.067 and 0.033 V−1.
Part (c)
both points plotted correctly to within half a small square
✓ 1
Allow ECF from (b).
single straight line through all the error bars, passing through or very close to the origin
✓ 1
Part (d)
gradient = 480 μs V «= 4.80 × 10−4 V s»
✓ 1
Accept 475–495 μs V. Gradient from a large triangle.
unit μs V OR V s
✓ 1
Accept T m².
Part (e)
B = gradient/(dL) with the gradient in V s
✓ 1
B = 4.80 × 10−4/(0.0120 × 0.800) = 0.0500 T
✓ 1
Allow ECF from (d).
percentage uncertainty = 2 % + 0.8 % + 0.25 % ≈ 3.1 %, so B = (0.0500 ± 0.0015) T
✓ 1
MP3 is for adding the percentage uncertainties AND matching the precision of value and uncertainty. Accept (0.050 ± 0.002) T.
Part (f)
0.0490 T lies within the range 0.0485–0.0515 T, so the result is consistent with the sensor reading
✓ 1
Allow ECF from (e). Conclusion must refer to the uncertainty range.
Part (g)
use a longer drift tube «so the times are longer and ±0.2 μs is a smaller fraction of each time»
✓ 1
Accept: use smaller values of Vp «slower ions, longer times». Do not accept «repeat the readings» alone.
Answers: (b) 0.0667, 0.0333 V−1 · (d) 480 μs V · (e) (0.0500 ± 0.0015) T (the remaining parts are explanations — see the table above)
A student uses a fine-beam tube to determine the charge-to-mass ratio e/m of the electron. Electrons are accelerated from rest through a potential difference V and then move at right angles to a uniform magnetic field produced by a pair of coils. The flux density is (8.0 ± 0.2) × 10−4 T and is kept constant. The electrons follow a circular path, made visible by the gas in the tube.
The student holds a ruler against the front of the glass bulb and, looking from the front, reads the diameter D of the circle to ±0.2 cm. The table gives r = D/2 and r²; the graph shows r² against V with error bars.
V / V
D / cm
r / cm
r² / 10−3 m²
150
10.1
5.05
2.55
180
11.2
5.60
3.14
210
12.0
6.00
3.60
240
12.9
6.45
4.16
270
13.6
6.80
4.62
300
14.4
7.20
5.18
Graph drawn to scale
(a)
Deduce that r² = (2m/(eB²)) V.
(2)
(b)
Show that the absolute uncertainty in r² for V = 300 V is about 1.4 × 10−4 m².
(1)
(c)
Draw the line of best fit for the data.
(1)
(d)
Determine e/m using the gradient of your line.
(2)
(e)
The uncertainty in the gradient is ±4 %. Determine the absolute uncertainty in your value of e/m and compare your result with the accepted value of 1.76 × 1011 C kg−1.
(3)
(f)
The ruler is about 1 cm in front of the plane of the beam. Explain whether this makes the value of e/m obtained too large or too small.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The magnetic force provides the centripetal force: evB = mv²/r, so r = mv/(eB)
✓ 1
eV = ½mv², so v² = 2eV/m; substituting gives r² = m²v²/(e²B²) = 2mV/(eB²)
✓ 1
Must see both equations combined; the answer is given.
Part (b)
Δr/r = 0.1/7.20 and Δ(r²) = 2 × (0.1/7.20) × 5.184 × 10−3 = 1.44 × 10−4 m²
✓ 1
Must see the doubling of the fractional uncertainty in r (Δr = 0.1 cm) OR an answer to 3 s.f.
Part (c)
Single straight line passing through all the error bars and (close to) the origin
Δ(e/m) = 0.09 × 1.80 × 1011 ≈ 0.2 × 1011 C kg−1, so e/m = (1.8 ± 0.2) × 1011 C kg−1
✓ 1
MP2 is for matching the precision of value and uncertainty. Allow ECF from (d).
The accepted value lies within the range 1.6–2.0 × 1011 C kg−1, so the result is consistent with it
✓ 1
Allow ECF.
Part (f)
Seen from the front, the lines of sight to the two sides of the circle converge, so they cross the ruler (nearer the eye) closer together: every diameter/radius is read too small
✓ 1
OWTTE
e/m = 2V/(B²r²) / the gradient is too small, so e/m is too large
✓ 1
Award MP2 only if the direction of the error in r is correct. Consistent with the student's high value.
Answers: (b) 1.44 × 10−4 m² · (d) e/m = 1.80 × 1011 C kg−1 · (e) (1.8 ± 0.2) × 1011 C kg−1(the remaining parts are explanations — see the table above)
91D-1B-08Parallel current-carrying wiresPaper 1BHard12 marks
A student uses a current balance to determine the magnetic constant μ0. A straight horizontal brass rod rests on an insulating block on a top-pan balance of resolution 0.001 g. An identical rod is clamped directly above it, parallel to it. The two rods are connected in series so that the same current I flows through each, in opposite directions. The length of each rod is (0.300 ± 0.002) m and the separation of their centres is (5.0 ± 0.2) mm.
For each current the student records the increase Δm in the balance reading with the current in one direction, then reverses the current and records it again. The table gives the mean of the two values, with an uncertainty of ±0.002 g. The graph shows Δm against I² with error bars.
I / A
I² / A²
Δm / g
4.0
16
0.019
5.0
25
0.032
6.0
36
0.043
7.0
49
0.061
8.0
64
0.077
9.0
81
0.100
10.0
100
0.121
Graph drawn to scale
(a)
Explain why the balance reading increases when there is a current in the rods.
(2)
(b)
Draw the line of best fit and the lines of maximum and minimum gradient for the data.
(2)
(c)
Determine the gradient of the line of best fit and its absolute uncertainty.
(2)
(d)
Determine μ0 and its absolute uncertainty.
(3)
(e)
The accepted value of μ0 is 1.26 × 10−6 T m A−1. Comment on the student's result.
(1)
(f)
The rods are also in the Earth's magnetic field. Explain how reversing the current and averaging the readings removes the effect of the Earth's field on the result.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The currents in the rods are in opposite directions, so the rods repel each other
✓ 1
The force on the lower rod is downwards, so the force on the balance (and its reading) increases
✓ 1
Award MP2 only if repulsion is stated or implied.
Part (b)
Line of best fit through all the error bars, passing through or close to the origin
✓ 1
Steepest and least steep lines that pass through all the error bars
✓ 1
Part (c)
gradient = 1.21 × 10−3 g A−2
✓ 1
Accept 1.18 × 10−3 to 1.25 × 10−3 g A−2. Unit required.
Accept ±1 % to ±5 %. Allow ECF from the candidate's lines.
Part (d)
Force = Δm g and F = μ0I²L/(2πr), so μ0 = 2πr × gradient × g/L (gradient in kg A−2)
✓ 1
μ0 = 2π × 5.0 × 10−3 × 1.21 × 10−6 × 9.81/0.300 = 1.24 × 10−6 T m A−1
✓ 1
Allow ECF from (c).
percentage uncertainty = 2.5 % + 4 % + 0.7 % ≈ 7 %, so μ0 = (1.24 ± 0.09) × 10−6 T m A−1
✓ 1
MP3 is for adding the fractional uncertainties AND matching precision. Accept (1.2 ± 0.1) × 10−6.
Part (e)
1.26 × 10−6 lies within the range 1.15–1.33 × 10−6 T m A−1, so the result is consistent with the accepted value
✓ 1
Allow ECF from (d).
Part (f)
The Earth's field exerts a force BIL on the lower rod that reverses direction when the current is reversed (it is proportional to I)
✓ 1
Reversing the current reverses both currents, so the force between the rods (proportional to I²) is unchanged; the mean of the two readings therefore contains only the force between the rods
✓ 1
OWTTE
Answers: (c) (1.21 ± 0.03) × 10−3 g A−2 · (d) μ0 = (1.24 ± 0.09) × 10−6 T m A−1(the remaining parts are explanations — see the table above)
92D-1B-14The velocity selectorPaper 1BHard12 marks
In a deflection tube, electrons accelerated from rest through a potential difference Va travel from left to right across the page, midway between two horizontal parallel plates (50.0 ± 0.5) mm apart. The upper plate is positive. A pair of coils produces a uniform magnetic field of flux density (1.20 ± 0.02) mT, perpendicular to the page, in the region between the plates.
For each value of Va (uncertainty ±0.05 kV), the student adjusts the potential difference Vp between the plates until the beam passes through undeflected. Judging this gives an uncertainty in Vp of ±20 V. The student tests whether Vp = cVan, where c and n are constants, by plotting lg (Vp / V) against lg (Va / V). Four points have been plotted.
Va / kV
Vp / V
lg (Va / V)
lg (Vp / V)
1.50
1390
3.176
3.143
2.00
1580
3.301
3.199
2.50
1790
3.00
1940
3.477
3.288
3.50
2110
4.00
2240
3.602
3.350
Graph drawn to scale
(a)
State and explain the direction of the magnetic field.
(2)
(b)
Show that the speed of the electrons when Va = 2.00 kV is about 2.6 × 107 m s−1.
(1)
(c)
Complete the table.
(2)
(d)
Plot the two missing points on the graph and draw the line of best fit for the data.
(1)
(e)
Determine the gradient of the line and outline what it shows about the speed of the electrons.
(2)
(f)
Determine, using the data for Va = 2.00 kV, the charge-to-mass ratio e/m of the electron and its absolute uncertainty.
(3)
(g)
Compare your answer to (f) with the accepted value of 1.76 × 1011 C kg−1.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The electric force on the (negative) electrons is upwards, towards the positive plate, so the magnetic force must be downwards
✓ 1
Into the page
✓ 1
Award MP2 only with a correct reason (left-hand rule with conventional current to the left, or F = qv × B with q negative).
Part (b)
v = E/B = Vp/(dB) = 1580/(0.0500 × 1.20 × 10−3) = 2.63 × 107 m s−1
✓ 1
Must see eE = evB (or v = E/B) with full substitution OR an answer to at least 3 s.f.
Part (c)
lg (Va / V) = 3.398 AND 3.544
✓ 1
Accept 3 or 4 d.p.
lg (Vp / V) = 3.253 AND 3.324
✓ 1
Accept 3 or 4 d.p.; do not accept 2 d.p.
Part (d)
Both points plotted correctly AND a single straight line of best fit
✓ 1
Allow ECF from (c).
Part (e)
gradient = 0.49
✓ 1
Accept 0.47–0.53. No unit.
v = Vp/(dB) ∝ Vp, so v ∝ Va0.5: v² ∝ Va, consistent with ½mv² = eVa
93D-1B-26Circular motion in a magnetic fieldPaper 1BHard12 marks
In a cyclotron, protons move in circular paths at right angles to a uniform magnetic field of flux density B. They are given a push twice in every revolution by an alternating potential difference, whose frequency f must equal the number of revolutions per second of the protons.
A student analyses the operating log of a small cyclotron. For each setting of B (measured to ±0.02 T), the log records the frequency f that gave the largest beam current (±0.2 MHz). The graph shows the data with error bars.
B / T
f / MHz
0.40
6.2
0.60
9.1
0.80
12.1
1.00
15.4
1.20
18.2
1.40
21.4
Graph drawn to scale
(a)
Show that the time for one revolution of a particle of mass m and charge q is 2πm/(qB).
(2)
(b)
Outline why the magnetic field does not change the kinetic energy of a proton.
(1)
(c)
Draw the line of best fit and the lines of maximum and minimum gradient for the data.
(2)
(d)
Determine the gradient of the line of best fit and its absolute uncertainty.
(2)
(e)
Determine the charge-to-mass ratio of the proton and its absolute uncertainty. Compare your result with the value obtained from the data booklet.
(3)
(f)
The protons leave the cyclotron when the radius of their path is 0.150 m. Calculate, in MeV, their kinetic energy when B = 1.40 T.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
qvB = mv²/r, so r = mv/(qB)
✓ 1
T = 2πr/v = 2πm/(qB)
✓ 1
Must see the elimination of v (or r).
Part (b)
The magnetic force is always perpendicular to the velocity, so it does no work on the proton (speed constant)
✓ 1
OWTTE
Part (c)
Line of best fit through all the error bars, passing through or close to the origin
✓ 1
Steepest and shallowest lines that still pass through all the error bars
✓ 1
Part (d)
gradient = 15.2 MHz T−1
✓ 1
Accept 14.8–15.6 MHz T−1.
uncertainty = ½(max − min) ≈ ±1 MHz T−1
✓ 1
Accept ±0.6 to ±1.4 MHz T−1.
Part (e)
f = qB/(2πm) so q/m = 2π × gradient = 2π × 15.2 × 106 = 9.55 × 107 C kg−1
94D-1B-40Parallel current-carrying wiresPaper 1BHard12 marks
A student investigates how the force between two parallel conductors depends on their separation. Two straight, parallel copper rods, each of diameter 4.00 mm, are connected in series so that the same current I = (25.0 ± 0.2) A flows in opposite directions in them. The lower rod rests on a force sensor; the upper rod is clamped directly above it. The length of each rod is (0.250 ± 0.002) m.
The student measures the gap s between the facing surfaces of the rods with vernier calipers. The force sensor measures the magnetic force F on the lower rod with an uncertainty of ±0.03 mN. The theory is F = μ0I²L/(2πr), where r is the separation of the centres of the rods. The graph shows F against 1/r for four of the six measurements.
s / mm
F / mN
r / mm
(1/r) / m−1
1.00
6.27
5.00
200
2.00
5.20
4.00
3.92
8.00
125
6.00
3.10
10.00
100
10.00
2.24
16.00
1.56
20.00
50
Graph drawn to scale
(a)
Outline why the magnetic force on the lower rod is directed downwards.
(1)
(b)
The student first plotted F against 1/s. Explain why this graph would not be a straight line through the origin.
(2)
(c)
Complete the table.
(1)
(d)
Plot the two missing points on the graph and draw the line of best fit.
(2)
(e)
Determine the gradient of your line. State its unit.
(2)
(f)
The uncertainty in the gradient is ±3 %. Determine μ0 and its absolute uncertainty.
(3)
(g)
The accepted value of μ0 is 1.26 × 10−6 T m A−1. Comment on the student's result.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
the currents in the rods are in opposite directions, so the rods repel and the lower rod is pushed down, away from the upper rod
✓ 1
Accept an argument using the field of the upper rod at the lower rod and F = IL × B.
Part (b)
r is the distance between the centres, so r = s + 4.00 mm «one radius of each rod»
✓ 1
F ∝ 1/(s + 4.00 mm), not 1/s, so F is not proportional to 1/s «the graph is a curve whose gradient decreases» OR F × s is not constant «it rises from 6.3 to 25 mN mm»
✓ 1
OWTTE. MP2 needs the link to proportionality.
Part (c)
r = 6.00 mm and 14.00 mm AND 1/r = 167 m−1 and 71.4 m−1
✓ 1
All four values needed. Accept 167 and 71 m−1.
Part (d)
both points plotted correctly to within half a small square
✓ 1
Allow ECF from (c).
single straight line through all the error bars and passing through or very close to the origin
✓ 1
Part (e)
gradient = 0.0313 mN m «= 3.13 × 10−5 N m»
✓ 1
Accept 0.0303–0.0322 mN m. Gradient from a large triangle.
unit mN m OR N m
✓ 1
Accept 10−3 N m.
Part (f)
μ0 = 2π × gradient/(I²L)
✓ 1
μ0 = 2π × 3.13 × 10−5/(25.0² × 0.250) = 1.26 × 10−6 T m A−1
✓ 1
Allow ECF from (e). Gradient must be in N m.
percentage uncertainty = 3 % + 2 × 0.8 % + 0.8 % = 5.4 %, so μ0 = (1.26 ± 0.07) × 10−6 T m A−1
✓ 1
MP3 is for doubling the uncertainty in I AND matching the precision of value and uncertainty. Accept (1.26 ± 0.07) × 10−6 or (1.3 ± 0.1) × 10−6.
Part (g)
the accepted value lies within the range 1.19–1.33 × 10−6 T m A−1, so the result is consistent with it
✓ 1
Allow ECF from (f).
Answers: (c) r = 6.00 mm, 14.00 mm; 1/r = 167 m−1, 71.4 m−1 · (e) 0.0313 mN m · (f) (1.26 ± 0.07) × 10−6 T m A−1(the remaining parts are explanations — see the table above)
95D-2-12Parallel current-carrying wiresPaper 2Easy10 marks
In an aluminium smelter, very large direct currents are carried by long, straight, parallel metal bars called bus bars. Two bus bars P and Q are 0.12 m apart. P carries a current of 3.0 kA into the page and Q carries a current of 3.0 kA out of the page, as shown in the cross-section.
Diagram NOT to scale
(a)
(i)
Draw, on the diagram, three magnetic field lines around P due to the current in P only. Show the direction of the field.
(2)
(ii)
State the direction of the magnetic field of P at the position of Q.
(1)
(iii)
Hence, deduce whether the bus bars attract or repel each other.
(2)
(b)
The bus bars are clamped to supports every 2.5 m. Calculate the magnetic force on a 2.5 m section of Q.
(3)
(c)
During a fault, the current in both bus bars rises briefly to 40 kA.
(i)
Calculate the factor by which the force per unit length increases.
(1)
(ii)
Suggest why the clamps must be designed for this fault and not only for the normal current.
(1)
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Notes
Part (a)(i)
concentric circles centred on P, further apart with increasing distance from P
✓ 1
At least three lines.
arrows showing a clockwise direction
✓ 1
Right-hand grip rule: thumb into the page, fingers curl clockwise.
Part (a)(ii)
towards the bottom of the page «downwards»
✓ 1
Allow ECF from (a)(i).
Part (a)(iii)
the force on Q is to the right, away from P «Fleming's left-hand rule: current out of the page, field down the page»
✓ 1
Allow ECF from (a)(ii).
so the bus bars repel
✓ 1
Accept a conclusion based on «currents in opposite directions repel».
the force becomes about 180 times larger «about 6.7 kN on each section», so clamps designed only for the normal force could break and the bars could be pushed apart
✓ 1
OWTTE
Answers: (b) F/L = 15 N m−1; F = 38 N · (c)(i) 178 (the remaining parts are explanations — see the table above)
96D-2-34Force on a current-carrying wirePaper 2Easy8 marks
A rectangular coil PQRS of one turn lies in the plane of the page between the poles of a magnet. The uniform magnetic field of flux density 0.050 T is directed from left to right, in the plane of the page. Sides PQ and RS are 0.060 m long and perpendicular to the field. Sides QR and SP are 0.040 m long and parallel to the field. The current in the coil is 2.5 A in the direction P → Q → R → S.
Diagram NOT accurately drawn
(a)
Calculate the magnitude of the magnetic force on side PQ.
(2)
(b)
Annotate the diagram, using the symbols ⊙ (out of the page) and ⊗ (into the page), to show the direction of the magnetic force on side PQ and on side RS.
(2)
(c)
Explain why there is no magnetic force on side QR.
(1)
(d)
State the resultant magnetic force on the whole coil.
(1)
(e)
The coil is turned about a vertical axis until side QR makes an angle of 30° with the field.
Calculate the magnitude of the magnetic force on side QR in this position.
(2)
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Notes
Part (a)
F = BIL = 0.050 × 2.5 × 0.060
✓ 1
F = 7.5 × 10−3 N
✓ 1
Award [2] for CNA.
Part (b)
⊗ «into the page» on PQ
✓ 1
⊙ «out of the page» on RS
✓ 1
Award [1] max if both directions are reversed.
Part (c)
the current in QR is parallel to the field, so θ = 0 and F = BIL sin θ = 0
✓ 1
OWTTE
Part (d)
zero «the forces on PQ and RS are equal and opposite, and there is no force on QR or SP»
✓ 1
Part (e)
F = BIL sin θ = 0.050 × 2.5 × 0.040 × sin 30°
✓ 1
F = 2.5 × 10−3 N
✓ 1
Award [2] for CNA.
Answers: (a) 7.5 × 10−3 N · (d) zero · (e) 2.5 × 10−3 N (the remaining parts are explanations — see the table above)
97D-2-71Parallel current-carrying wiresPaper 2Easy12 marks
A home solar installation stores energy in a 48 V lithium battery bank of emf 51.2 V and internal resistance 8.0 mΩ. It is protected by a fast direct-current circuit breaker. Inside the breaker, the current flows along a fixed straight conductor F and then back, in the opposite direction, along a straight moving contact arm A that is parallel to F. A and F are parallel over a length of 40 mm and their centres are 4.0 mm apart. A spring holds the contacts closed with a force of 20 N. The figure shows a cross-section. μ0 = 4π × 10−7 T m A−1.
Diagram NOT to scale
(a)
(i)
Draw, on the figure, the magnetic field line due to the current in F that passes through A. Show the direction of the field.
(1)
(ii)
Hence deduce the direction of the magnetic force on A.
(2)
(b)
The normal current in the breaker is 32 A. Calculate the magnetic force on A.
(2)
(c)
A fault connects the terminals of the battery bank through cables and the breaker of total resistance 0.50 mΩ.
(i)
Show that the current during the fault is about 6 kA.
(1)
(ii)
Determine whether the magnetic force forces the contacts open.
(2)
(iii)
Calculate the smallest current for which the magnetic force can open the contacts.
(2)
(d)
Suggest why this breaker opens more quickly for a larger fault current.
(1)
(e)
Explain why the breaker would not work if the currents in A and F were in the same direction.
(1)
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Notes
Part (a)(i)
circle centred on F passing through A, with an arrow showing a clockwise direction «to the right at A»
✓ 1
Right-hand grip rule with the current into the page.
Part (a)(ii)
use of F = IL × B OR Fleming's left-hand rule with the current in A out of the page and the field to the right
✓ 1
upwards, away from F «the conductors repel»
✓ 1
Allow ECF from (a)(i). Award [1] max for «away from F, because opposite currents repel» with no use of the field.
Allow ECF from (c)(i). Accept 72–73 N; 72 N if 6.0 kA is used.
this is greater than the 20 N spring force, so the contacts are forced open
✓ 1
Conclusion must follow from a comparison with 20 N.
Part (c)(iii)
20 = 2 × 10−7 × I² × 0.040/4.0 × 10−3
✓ 1
I = 3.2 × 103 A
✓ 1
Accept 3160 A. Award [2] for CNA.
Part (d)
the force is proportional to I², so a larger current gives a much larger resultant force and acceleration of the arm «it reaches the open position sooner»
✓ 1
OWTTE. Must refer to the larger force or acceleration.
Part (e)
currents in the same direction attract, so A would be pulled towards F and the contacts pressed more tightly together
✓ 1
OWTTE.
Answers: (b) 2.0 × 10−3 N · (c)(i) 6020 A · (c)(ii) 73 N · (c)(iii) 3.2 × 103 A (the remaining parts are explanations — see the table above)
98D-2-79Force on a current-carrying wirePaper 2Easy11 marks
In a precision laboratory balance, the weighing pan is attached to a horizontal circular coil of 250 turns, each of radius 12 mm. The coil sits in the narrow gap of a permanent magnet, where the magnetic field is horizontal and directed radially outwards, with a flux density of 0.50 T, so the field is perpendicular to the wire at every point. A control circuit adjusts the current I in the coil so that the pan always returns to the same height; the display shows a mass calculated from I.
Diagram NOT to scale
(a)
Show that the length of wire in the magnetic field is about 19 m.
(1)
(b)
State and explain whether the current in the coil, seen from above, must be clockwise or anticlockwise to give an upward force on the coil.
(2)
(c)
Calculate the current needed to support a load of mass 50.000 g.
(2)
(d)
The current is measured with a resolution of 1 μA. Determine the smallest change of mass that the balance can detect.
(2)
(e)
The resistance of the coil is 40 Ω. Calculate the power dissipated in the coil when the 50.000 g load is supported.
(1)
(f)
The magnet warms slightly and its flux density falls by 0.2 %. State and explain the effect on the mass displayed for the 50.000 g load.
(2)
(g)
Suggest why the balance must be calibrated with a standard mass at the place where it is used.
(1)
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Notes
Part (a)
L = 250 × 2π × 0.012 = 18.8 «m»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (b)
clockwise
✓ 1
F = IL × B: on the right-hand side of the coil in the diagram the field points to the right and a clockwise current is towards the bottom of the page, giving a force out of the page «upwards»
✓ 1
Accept any correct application of Fleming's left-hand rule. MP2 only scores if MP1 scores.
A materials-testing laboratory uses a beam of slow positrons. The positrons are emitted in the β⁺ decay of sodium-22 (2211Na), slowed to negligible speed in a thin solid film and then accelerated from rest through a potential difference of 5.00 kV. The beam then passes through a velocity selector: two horizontal parallel plates 12.0 mm apart with a uniform magnetic field of flux density 2.50 mT directed out of the page between them. Mass of a positron = 9.11 × 10−31 kg; charge of a positron = +1.60 × 10−19 C.
Diagram NOT to scale
(a)
Write the equation for the β⁺ decay of sodium-22. The nucleus formed is an isotope of neon (Ne).
(2)
(b)
(i)
State the direction of the electric field that accelerates the positrons, relative to their direction of motion.
(1)
(ii)
Show that the speed of the positrons after acceleration is about 4.2 × 107 m s−1.
(1)
(c)
(i)
Calculate the potential difference between the plates for which the positrons pass straight through the selector.
(2)
(ii)
Deduce which plate must be positive.
(2)
(iii)
Some positrons enter the selector with a speed slightly less than the speed in (b)(ii). Sketch, on the figure, the path of one of these positrons between the plates.
(1)
(d)
The beam also contains electrons moving with the same velocity. Explain why these electrons also pass straight through the selector.
(2)
(e)
After the selector, the beam enters a region containing only a uniform magnetic field of 2.50 mT, at right angles to the beam. Calculate the radius of the path of the positrons and outline how this field separates the positrons from the electrons.
(2)
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Notes
Part (a)
2210Ne AND 0+1e «β⁺» as products
✓ 1
Nucleon and proton numbers needed for Ne.
electron neutrino ν «00ν» as a product
✓ 1
Do not accept an antineutrino.
Part (b)(i)
in the same direction as the motion «from higher to lower potential»
E = vB = 4.19 × 107 × 2.50 × 10−3 = 1.05 × 105 V m−1
✓ 1
V = Ed = 1.05 × 105 × 0.0120 = 1260 V
✓ 1
Award [2] for CNA. Accept 1.2–1.3 kV. Allow ECF from (b)(ii).
Part (c)(ii)
the magnetic force on a positron is towards the bottom of the page «F = qv × B with q positive, v to the right and B out of the page»
✓ 1
Accept a correct use of Fleming's left-hand rule.
so the electric force, and hence the electric field, must be towards the top of the page: the lower plate is positive
✓ 1
Allow ECF from MP1.
Part (c)(iii)
path curving upwards, towards the upper plate, starting along the original direction of motion «the electric force is now larger than the magnetic force»
✓ 1
Do not accept a path that bends sharply or curves downwards. Allow ECF from (c)(ii).
Part (d)
for an electron both the electric force and the magnetic force are reversed «the magnetic force is up and the electric force is down»
✓ 1
the forces are still equal because eE = evB does not depend on the sign of the charge: only particles with v = E/B pass
✓ 1
OWTTE.
Part (e)
r = mv/(eB) = 9.11 × 10−31 × 4.19 × 107/(1.60 × 10−19 × 2.50 × 10−3) = 0.095 m
✓ 1
Allow ECF from (b)(ii).
the electrons curve in the opposite direction «same radius», so the two beams move apart
✓ 1
Answers: (b)(ii) 4.19 × 107 m s−1 · (c)(i) 1260 V · (e) 0.095 m (the remaining parts are explanations — see the table above)
100IBP2S-032-01-11Force on a current-carrying wirePaper 2Easy5 marks
A straight copper lead of length 0.25 m inside an MRI scanner carries a current of 3.0 A. The uniform magnetic field in the scanner has a magnetic flux density of 1.5 T.
(a)
The lead is perpendicular to the field. Calculate the magnitude of the force on the lead.
(2)
(b)
Outline why the force decreases if the lead is turned so that it is no longer perpendicular to the field.
(1)
(c)
The lead is moved so that it makes an angle of 40° with the field. Calculate the new force on the lead.
(2)
Show mark scheme
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Notes
Part (a)
Uses F = BIL = 1.5 × 3.0 × 0.25
✓ 1
F = 1.1 N (accept 1.13 N)
✓ 1
Part (b)
Only the component of the field perpendicular to the current (or of the current perpendicular to the field) produces a force, so F = BIL sin θ decreases as θ falls below 90°
✓ 1
Part (c)
F = BIL sin 40° = 1.125 × sin 40°
✓ 1
F = 0.72 N (accept 0.71 N from using 1.1 N; allow error carried forward from the first part)
✓ 1
Common error: Using cos 40° here gives 0.86 N. θ is measured between the wire and the field, so use sin θ.
101IBP2S-032-01-12Force on a current-carrying wirePaper 2Easy5 marks
A current balance is used to measure the field between the poles of a magnet in an electric-guitar pickup test rig. A horizontal wire of length 0.050 m is perpendicular to the field. The vertical force F on the wire is recorded for different currents I.
Data Row 1, I / A: 1.0, 2.0, 3.0, 4.0, 5.0 Row 2, F / mN: 2.4, 4.9, 7.2, 9.7, 12.1
Data from the question
(a)
Outline why the theory predicts that F/I is constant in this experiment.
(1)
(b)
Determine, using the data, the magnetic flux density B.
(3)
(c)
State one possible cause of a systematic error in this experiment.
(1)
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Notes
Part (a)
F = BIL; B and L are fixed, so F/I = BL is constant
✓ 1
Part (b)
Finds the gradient of F against I (or the mean of F/I) in N A−1, e.g. best fit ≈ 2.42×10−3 N A−1; end points give 9.7×10−3/4.0 = 2.43×10−3 N A−1
✓ 1
Uses B = gradient / L = 2.42×10−3/0.050
✓ 1
B = 0.048 T (accept 0.047–0.049 T)
✓ 1
Part (c)
Any one of: balance not zeroed before the current flows / wire not exactly perpendicular to the field / field not uniform along the whole length of the wire / ammeter with a zero error
✓ 1
Exam tip: Use a best-fit line or widely separated points rather than a single pair of readings, and do not round the gradient before dividing by L. Convert mN to N first.
102IBP2S-032-00-12Force on a moving chargePaper 2Easy7 marks
In the outer magnetosphere, a solar-wind proton moves at 4.0 × 105 m s−1 perpendicular to a uniform magnetic field of strength 5.0 × 10−8 T. Mass of proton = 1.67 × 10−27 kg; mass of electron = 9.11 × 10−31 kg; elementary charge = 1.60 × 10−19 C.
(a)
Calculate the magnitude of the magnetic force on the proton.
(1)
(b)
Explain why the proton moves in a circular path.
(2)
(c)
Calculate the radius of the path.
(2)
(d)
An electron moves with the same speed in the same field. State two ways in which its motion differs from that of the proton.
(2)
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Notes
Part (a)
F = qvB = 1.60 × 10−19 × 4.0 × 105 × 5.0 × 10−8 = 3.2 × 10−21 N
✓ 1
Part (b)
The force is always perpendicular to the velocity, so it does no work. The speed is constant and only the direction changes .
✓ 1
The force has constant magnitude and always acts perpendicular to the motion, towards the centre, so it provides the centripetal force for uniform circular motion .
✓ 1
Part (c)
qvB = mv²/r, so r = mv/(qB) .
✓ 1
r = 1.67 × 10−27 × 4.0 × 105/(1.60 × 10−19 × 5.0 × 10−8) = 8.4 × 104 m . Allow ECF from (a) if r = mv²/F is used.
✓ 1
Part (d)
Any two of the following points:
—
• the radius is much smaller, because the electron's mass is about 1/1800 of the proton's (r ≈ 46 m);
—
• it circulates in the opposite sense, because its charge is negative;
—
• its period is much shorter, or its frequency is much higher, because T = 2πm/(qB).
—
First valid difference .
✓ 1
Second valid difference .
✓ 1
Common error: In r = mv/(qB), use the mass of the particle actually named in the question. Mixing up the proton and electron masses changes the answer by a factor of about 1800.
103IBP2S-032-03-11Parallel current-carrying wiresPaper 2Easy5 marks
A pair of jump-start leads connects a flat car battery to a working one. While the engine is cranking, each lead carries a current of 120 A, and the currents in the two leads are in opposite directions. Over a length of 2.5 m, the leads lie straight and parallel with their centres 1.5 cm apart. μ0 = 4π × 10−7 T m A−1
(a)
State and explain whether the leads attract or repel each other.
(2)
(b)
Calculate the magnetic force per unit length between the leads.
(2)
(c)
Calculate the total magnetic force on one lead over the parallel section.
(1)
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Notes
Part (a)
Each lead lies in the magnetic field produced by the current in the other lead, so each experiences a magnetic force (F = BIL) . Because the currents are in opposite directions, the right-hand rule gives a force on each lead directed away from the other, so the leads repel . Accept: antiparallel currents repel, with this stated as the reason.
104IBP2S-032-03-12Force on a moving chargePaper 2Easy6 marks
In an ion-implantation beamline, a bending magnet turns a beam of protons through 90° so that it can be directed at a silicon wafer. Inside the magnet, the protons move at 4.0 × 106 m s−1 perpendicular to a uniform magnetic field of flux density 0.25 T. Proton mass = 1.67 × 10−27 kg; proton charge = 1.60 × 10−19 C
(a)
Explain why the magnetic field does not change the kinetic energy of the protons.
(2)
(b)
Show that the radius of the protons' path is about 0.17 m.
(2)
(c)
Calculate the time a proton takes to travel through the 90° bend.
(2)
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Notes
Part (a)
The magnetic force is always perpendicular to the proton's velocity . So there is no component of force along the displacement, no work is done, and the speed (and kinetic energy) stays constant .
✓ 2
Part (b)
qvB = mv²/r, so r = mv/(qB) . r = (1.67 × 10−27 × 4.0 × 106)/(1.60 × 10−19 × 0.25) = 0.167 m .
✓ 2
Part (c)
Period T = 2πr/v (or 2πm/(qB)) = 2π × 0.167/(4.0 × 106) = 2.6 × 10−7 s . Time for a quarter circle = T/4 = 6.6 × 10−8 s; accept 6.5–6.6 × 10−8 s . Allow error carried forward from the radius found in the previous part. Alternative method: quarter-circle arc length (πr/2)/v earns both marks.
✓ 2
Exam tip: In a 'show that' question, give your answer to at least one more significant figure than the value quoted, to show you have actually done the calculation.
105IBP2S-032-02-11Paths in electric and magnetic fieldsPaper 2Easy5 marks
In a proton-therapy facility, a beam of protons travelling at 2.0 × 107 m s−1 enters a steering magnet. The magnet provides a uniform magnetic field of 1.2 T perpendicular to the beam. Proton mass = 1.67 × 10−27 kg; proton charge = 1.60 × 10−19 C.
(a)
Explain why the protons move along a circular arc in the field.
(2)
(b)
Calculate the radius of the arc.
(2)
(c)
The field is directed vertically upwards and the protons enter moving horizontally due north. State the direction of the magnetic force on the protons as they enter.
(1)
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Notes
Part (a)
The magnetic force is always perpendicular to the proton's velocity, so it does no work and the speed stays constant . The force therefore has constant magnitude qvB and always points towards a fixed centre, providing the centripetal force for circular motion .
✓ 2
Part (b)
qvB = mv²/r, so r = mv/(qB) . r = (1.67 × 10−27 × 2.0 × 107)/(1.60 × 10−19 × 1.2) = 0.17 m (0.174 m) .
✓ 2
Part (c)
East .
✓ 1
Exam tip: Start from qvB = mv²/r rather than quoting the radius formula. Writing this equation earns the method mark.
106IBP2S-032-02-12Force on a current-carrying wirePaper 2Easy4 marks
A school demonstration of an electromagnetic launcher uses a metal rod of length 0.080 m and mass 0.020 kg. The rod rests across two horizontal, frictionless rails. A current of 15 A flows through the rod. A uniform magnetic field of 0.25 T acts vertically, perpendicular to the rod.
Diagram not to scale. View from above; rod rests across the rails.
(a)
Calculate the magnitude of the magnetic force on the rod.
(2)
(b)
State one change that would reverse the direction of the force on the rod.
(1)
(c)
Calculate the initial acceleration of the rod.
(1)
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Notes
Part (a)
F = BIL . F = 0.25 × 15 × 0.080 = 0.30 N .
✓ 2
Part (b)
Reverse the direction of the current OR reverse the direction of the magnetic field .
✓ 1
Part (c)
a = F/m = 0.30/0.020 = 15 m s−2 . ECF from (a).
✓ 1
Common error: Reversing both the current and the field leaves the force unchanged; reverse only one of them.
107IBP2S-032-04-11Motion in an electric fieldPaper 2Easy5 marks
In a scanning electron microscope used to inspect medical implants, electrons are accelerated from rest through a potential difference of 5.0 kV. The accelerating region is a uniform electric field across a gap of 2.0 cm. Mass of electron = 9.11×10−31 kg.
(a)
State the kinetic energy gained by an electron, in eV.
(1)
(b)
Calculate the speed of an electron as it leaves the accelerating region.
(2)
(c)
Calculate the time an electron takes to cross the 2.0 cm gap.
(2)
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Notes
Part (a)
5.0×103 eV (5.0 keV)
✓ 1
Part (b)
Uses ½mv2 = eV, so v = √(2eV/m)
✓ 1
v = √(2 × 1.60×10−19 × 5.0×103 / 9.11×10−31) = 4.2×107 m s−1
✓ 1
Part (c)
The acceleration is uniform from rest, so the average speed is v/2 (or uses a = v2/2d ≈ 4.4×1016 m s−2)
✓ 1
t = 2d/v = 0.040 / 4.2×107 = 9.5×10−10 s (ECF from (b))
✓ 1
Exam tip: In a uniform field starting from rest, the average speed is half the final speed. Using this avoids having to find the acceleration first.
108IBP2S-032-04-12The velocity selectorPaper 2Easy5 marks
A solar-wind analyser on a space probe uses a velocity selector to measure the speed of incoming protons. The electric field is produced by two parallel plates 4.0 mm apart with a potential difference of 120 V between them. A perpendicular uniform magnetic field of flux density 0.050 T acts in the same region.
(a)
Calculate the electric field strength between the plates.
(1)
(b)
Show that the speed of the protons that pass through undeflected is about 6×105 m s−1.
(2)
(c)
Singly charged oxygen ions (O+) in the solar wind enter the selector with the same velocity as the undeflected protons. State and explain whether these ions are also undeflected.
(2)
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Notes
Part (a)
E = V/d = 120 / 4.0×10−3 = 3.0×104 V m−1
✓ 1
Part (b)
For no deflection, qE = qvB, so v = E/B
✓ 1
v = 3.0×104 / 0.050 = 6.0×105 m s−1
✓ 1
Part (c)
Yes, they are undeflected
✓ 1
Both forces are proportional to q, so the charge cancels in qE = qvB. Mass does not appear in the force balance, so only the speed matters
✓ 1
Common error: Do not argue that heavier ions deflect less inside the selector. When the forces balance there is no resultant force, so the mass is irrelevant.
109D-2-09Force on a current-carrying wirePaper 2Medium10 marks
A model electromagnetic launcher has two horizontal parallel metal rails 0.080 m apart. A metal rod rests across the rails and is fixed to a light trolley. The total mass of the rod and trolley is 0.150 kg. A uniform vertical magnetic field of flux density 0.25 T acts over the whole length of the rails.
A supply drives a constant current of 12 A through the rails and the rod. The diagram shows the arrangement seen from above.
Diagram NOT to scale
(a)
(i)
Draw an arrow on the diagram to show the direction of the magnetic force on the rod.
(1)
(ii)
Calculate the magnitude of the magnetic force on the rod.
(1)
(b)
A constant frictional force of 0.030 N opposes the motion. The rod starts from rest and travels 1.20 m along the rails.
(i)
Calculate the acceleration of the rod and trolley.
(2)
(ii)
Show that the speed of the rod at the end of the rails is about 1.8 m s−1.
(1)
(c)
The potential difference across the terminals of the supply is 3.0 V throughout the launch.
(i)
Determine the efficiency with which energy from the supply is transferred to kinetic energy of the rod and trolley.
(4)
(ii)
Suggest what happens to most of the energy from the supply.
(1)
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Notes
Part (a)(i)
arrow along the rails, to the right «away from the supply»
✓ 1
Fleming's left-hand rule: field upwards «out of the page», current up the page, force to the right.
Part (a)(ii)
F = BIL = 0.25 × 12 × 0.080 = 0.24 N
✓ 1
Part (b)(i)
resultant force = 0.24 − 0.030 = 0.21 N
✓ 1
Allow ECF from (a)(ii).
a = 0.21/0.150 = 1.4 m s−2
✓ 1
Award [1 max] for 1.6 m s−2 «friction ignored».
Part (b)(ii)
v = √(2 × 1.4 × 1.20) = 1.83 m s−1
✓ 1
Must see full substitution OR answer to at least 3 s.f. ALTERNATIVE: 0.21 × 1.20 = ½ × 0.150 × v².
Part (c)(i)
time of launch t = v/a = 1.83/1.4 = 1.31 s
✓ 1
ALTERNATIVE: t from s = ½at².
energy from the supply = VIt = 3.0 × 12 × 1.31 = 47 J
✓ 1
Ek = ½ × 0.150 × 1.83² = 0.25 J
✓ 1
Allow ECF from (b)(ii).
efficiency = 0.25/47 = 0.0053 «0.53 %»
✓ 1
Allow ECF. Accept 0.5–0.6 %.
Part (c)(ii)
it becomes internal energy «heating» of the rails, the rod and the connections because of their resistance
✓ 1
Do not accept «friction» alone «only 0.036 J».
Answers: (a)(ii) 0.24 N · (b)(i) 1.4 m s−2 · (b)(ii) 1.83 m s−1 · (c)(i) 0.53 % (the remaining parts are explanations — see the table above)
110D-2-10Charge-to-mass ratioPaper 2Medium14 marks
In an ion-beam laboratory, positive ions from an ion source are accelerated from rest through a potential difference of 46.7 kV. They then travel along an evacuated beam line, as shown. Two detectors, D₁ and D₂, 2.40 m apart, record each ion as it passes. The time between the two signals is 1.60 μs.
At X the ions enter a bending magnet, a region of uniform magnetic field perpendicular to the page. Inside the magnet the ions follow a quarter circle of radius 0.250 m and leave towards the target.
Diagram NOT to scale
(a)
(i)
Calculate the speed of the ions.
(1)
(ii)
Show that the charge-to-mass ratio q/m of the ions is about 2.4 × 107 C kg−1.
(2)
(b)
(i)
Draw, on the diagram, an arrow to show the direction of the magnetic force on an ion at X.
(1)
(ii)
State the direction of the magnetic field in the bending magnet.
(1)
(iii)
Explain why the path of an ion inside the magnet is an arc of a circle.
(2)
(iv)
Determine the magnetic flux density in the bending magnet.
(2)
(c)
The ion source can produce helium ions He⁺ (charge +e, mass 6.64 × 10−27 kg) and carbon ions C³⁺ (charge +3e, mass 1.99 × 10−26 kg).
(i)
Determine whether the beam could consist of He⁺ ions, of C³⁺ ions, or of either.
(2)
(ii)
Explain why no arrangement of electric and magnetic fields can separate these two kinds of ion if they enter it with the same velocity.
(2)
(iii)
Suggest a measurement that would identify the ions in the beam.
Must see full substitution OR answer to at least 3 s.f.
Part (b)(i)
arrow at X towards the top of the page «towards the centre of the arc»
✓ 1
Arrow must be perpendicular to the beam at X.
Part (b)(ii)
into the page
✓ 1
F = qv × B with q > 0, v to the right and F towards the top of the page. Allow ECF from (b)(i).
Part (b)(iii)
the magnetic force is always perpendicular to the velocity, so it does no work and the speed «and so the force qvB» is constant
✓ 1
a force of constant magnitude always perpendicular to the velocity is a centripetal force, giving uniform circular motion
✓ 1
OWTTE
Part (b)(iv)
qvB = mv²/r, so B = v/((q/m)r)
✓ 1
B = 1.50 × 106/(2.41 × 107 × 0.250) = 0.249 T
✓ 1
Accept 0.24–0.25 T. Award [2] for CNA. Allow ECF from (a).
Part (c)(i)
q/m for He⁺ = 1.60 × 10−19/6.64 × 10−27 = 2.41 × 107 C kg−1 AND q/m for C³⁺ = 4.80 × 10−19/1.99 × 10−26 = 2.41 × 107 C kg−1
✓ 1
Both values needed.
both equal the measured q/m, so the beam could be either «the data cannot decide»
✓ 1
MP2 only scores if MP1 scores. Do not accept a choice of one ion.
Part (c)(ii)
the electric force qE and the magnetic force qvB are both proportional to q, so the acceleration a = F/m depends only on q/m and the velocity
✓ 1
ions with the same q/m and the same velocity have the same acceleration at every point, so they follow identical paths
✓ 1
OWTTE
Part (c)(iii)
measure the kinetic energy of the ions, e.g. with a detector that stops them: a C³⁺ ion has three times the kinetic energy «140 keV, not 46.7 keV» of a He⁺ ion
✓ 1
Accept measuring the momentum or the mass of the ions, e.g. from a collision. Do not accept another E- or B-field measurement.
Answers: (a)(i) 1.50 × 106 m s−1 · (a)(ii) 2.41 × 107 C kg−1 · (b)(iv) 0.249 T · (c)(i) 2.41 × 107 C kg−1 for both ions (the remaining parts are explanations — see the table above)
111D-2-13Charge in a uniform electric fieldPaper 2Medium13 marks
A space probe is driven by an ion engine. Xenon ions, each of mass 2.18 × 10−25 kg and charge +1.60 × 10−19 C, start almost at rest at grid P. A uniform electric field accelerates them to grid Q, which is 1.20 mm from P. The potential difference between the grids is 1100 V.
The ions leave the engine as a beam. The current in the beam is 1.40 A.
Diagram NOT to scale
(a)
(i)
Show that the speed of an ion at Q is about 4.0 × 104 m s−1.
(1)
(ii)
Determine the time an ion takes to travel from P to Q.
(2)
(b)
(i)
Show that about 9 × 1018 ions leave the engine each second.
(1)
(ii)
Determine the force exerted on the probe by the engine.
(3)
(iii)
The probe has a mass of 650 kg. Determine whether the engine could increase the speed of the probe by 1.0 km s−1 in 30 days. Assume that the mass of the probe does not change.
(3)
(c)
Electrons are injected into the ion beam after it passes grid Q. Explain why this is necessary.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Ek = qV = 1.60 × 10−19 × 1100 = 1.76 × 10−16 J, so v = √(2 × 1.76 × 10−16/2.18 × 10−25) = 4.02 × 104 m s−1
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (a)(ii)
the field is uniform, so the acceleration is uniform and the mean speed is v/2 «= 2.0 × 104 m s−1»
✓ 1
ALTERNATIVE: E = V/d = 9.2 × 105 V m−1, a = qE/m = 6.7 × 1011 m s−2, then t = v/a.
t = 1.20 × 10−3/2.0 × 104 = 6.0 × 10−8 s
✓ 1
Award [2] for CNA. Award [1 max] for 3.0 × 10−8 s «t = d/v».
Part (b)(i)
n = I/e = 1.40/1.60 × 10−19 = 8.75 × 1018 s−1
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (b)(ii)
momentum of one ion = mv = 2.18 × 10−25 × 4.02 × 104 = 8.76 × 10−21 kg m s−1
✓ 1
force = rate of change of momentum of the ions = n × mv «by Newton's third law, equal to the force on the probe»
✓ 1
F = 8.75 × 1018 × 8.76 × 10−21 = 0.077 N
✓ 1
Allow ECF. Accept 0.076–0.077 N.
Part (b)(iii)
a = F/m = 0.077/650 = 1.18 × 10−4 m s−2
✓ 1
Allow ECF from (b)(ii).
in 30 days Δv = 1.18 × 10−4 × 30 × 86 400 = 3.1 × 102 m s−1
✓ 1
ALTERNATIVE: time needed = 1000/a = 8.5 × 106 s ≈ 98 days.
this is less than 1.0 km s−1, so it could not
✓ 1
The conclusion must be consistent with the candidate's value.
Part (c)
positive charge continually leaves the probe in the beam
✓ 1
by conservation of charge the probe would build up an equal negative charge
✓ 1
the negatively charged probe would attract the positive ions back, reducing the force «the injected electrons neutralize the beam, so the probe stays uncharged»
✓ 1
OWTTE
Answers: (a)(i) 4.02 × 104 m s−1 · (a)(ii) 6.0 × 10−8 s · (b)(ii) 0.077 N · (b)(iii) 3.1 × 102 m s−1(the remaining parts are explanations — see the table above)
112D-2-30Charge in a uniform electric fieldPaper 2Medium13 marks
In an ion-energy analyser, two parallel wire-mesh grids G1 and G2 are 12.0 mm apart in a vacuum. Protons, each with a kinetic energy of 450 eV, pass through G1 moving at right angles to the grids, towards G2 (Figure 1). A potential difference V is applied between the grids so that G2 is positive relative to G1. The field between the grids is uniform.
e = 1.60 × 10−19 C; mass of a proton = 1.67 × 10−27 kg.
Figure 1Figure 2 — axes for (c)(ii)
(a)
State the direction of the electric field between the grids.
(1)
(b)
V = 300 V.
(i)
Calculate the magnitude of the deceleration of a proton between the grids.
(2)
(ii)
Determine whether the protons reach G2 and, if they do, state their kinetic energy there in eV.
(2)
(c)
V is increased to 600 V.
(i)
Determine the distance from G1 at which a proton momentarily stops.
(2)
(ii)
Sketch, on Figure 2, a graph to show how the velocity v of a proton varies with time t from the moment it passes through G1, with speed v0, until it returns to G1. Take the direction from G1 to G2 as positive.
(3)
(iii)
State the kinetic energy, in eV, of the proton when it returns to G1.
(1)
(d)
A beam contains protons with a range of kinetic energies.
Suggest how the analyser could be used to find the number of protons with kinetic energies between 400 eV and 500 eV.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
from G2 towards G1 «opposite to the direction of motion of the protons»
✓ 1
Part (b)(i)
E = 300 / 0.0120 = 2.5 × 104 V m−1
✓ 1
a = eE/m = 2.4 × 1012 m s−2
✓ 1
Award [2] for CNA.
Part (b)(ii)
work done against the field across the whole gap = 300 eV, which is less than 450 eV, so the protons reach G2
✓ 1
ALTERNATIVE: stopping distance = (450/300) × 12.0 mm = 18 mm > 12.0 mm.
kinetic energy at G2 = 450 − 300 = 150 eV
✓ 1
Part (c)(i)
450 eV = e × E × s with E = 600/0.0120 = 5.0 × 104 V m−1 OR s/12.0 mm = 450/600
✓ 1
s = 9.0 mm
✓ 1
Award [2] for CNA.
Part (c)(ii)
straight line starting at v0 with a constant negative gradient
✓ 1
line continues through v = 0 into negative values
✓ 1
line ends at −v0 at twice the time at which v = 0
✓ 1
Do not award MP3 if the graph is curved or the final speed differs from v0.
Part (c)(iii)
450 eV «the field does equal positive work on the return journey»
✓ 1
Part (d)
a detector beyond G2 counts the protons that pass through; only protons with kinetic energy «in eV» greater than V reach it
✓ 1
OWTTE
measure the count «rate» at V = 400 V and at V = 500 V; the difference is the number with energies between 400 eV and 500 eV
✓ 1
Answers: (b)(i) 2.4 × 1012 m s−2 · (b)(ii) yes; 150 eV · (c)(i) 9.0 mm · (c)(iii) 450 eV (the remaining parts are explanations — see the table above)
113D-2-31Force on a current-carrying wirePaper 2Medium10 marks
In a moving-coil loudspeaker, a coil of 60 turns, each of radius 12 mm, sits in the narrow gap of a magnet. The magnetic field in the gap is radial, with flux density 0.80 T, so the field is perpendicular to the wire at every point. The coil is fixed to a cone; together they have a mass of 8.0 g. The suspension of the cone behaves as a spring of spring constant 1.6 × 103 N m−1.
The diagram shows a cross-section through the axis of the coil. In the coil, the current is out of the page at the top and into the page at the bottom; the field points away from the axis.
Diagram NOT accurately drawn
(a)
Annotate the diagram with arrows to show the direction of the magnetic force on the coil at the top and at the bottom of the cross-section.
(2)
(b)
Show that the total length of wire in the coil is about 4.5 m.
(1)
(c)
There is a steady current of 0.50 A in the coil.
(i)
Calculate the magnitude of the magnetic force on the coil.
(1)
(ii)
Calculate the displacement of the cone from its equilibrium position.
(1)
(d)
Show that the natural frequency of oscillation of the cone is about 71 Hz.
(1)
(e)
The coil now carries a signal current of constant amplitude whose frequency is slowly increased from 20 Hz to 200 Hz.
(i)
Describe and explain how the amplitude of vibration of the cone changes.
(3)
(ii)
Suggest why loudspeaker designers make the motion of the cone heavily damped.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
arrow on the top wire pointing to the left, along the axis «towards the cone»
✓ 1
Use of F = IL × B or Fleming's left-hand rule.
arrow on the bottom wire also pointing to the left
114D-2-33Parallel current-carrying wiresPaper 2Medium12 marks
Three long, straight, parallel cables X, Y and Z lie in a row on a horizontal cable tray. The diagram shows a cross-section. X carries a current of 200 A and Y carries 150 A, both out of the page; Z carries 300 A into the page. The distance between X and Y is 0.040 m and between Y and Z is 0.060 m.
μ0 = 4π × 10−7 T m A−1.
Diagram NOT accurately drawn
(a)
State and explain the direction of the magnetic force on Y due to X.
(2)
(b)
(i)
Draw, on the diagram, the magnetic field line due to the current in Z that passes through Y. Show its direction.
(1)
(ii)
Hence deduce the direction of the magnetic force on Y due to Z.
(1)
(c)
Determine the magnitude and direction of the resultant magnetic force per unit length on Y.
(3)
(d)
Calculate the magnitude of the resultant magnetic force per unit length on X and state its direction.
(3)
(e)
Determine whether there would be a resultant magnetic force on Y if the current in Z were reversed.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
towards X «to the left»
✓ 1
the currents in X and Y are in the same direction, and parallel currents in the same direction attract
✓ 1
MP2 only scores if MP1 scores.
Part (b)(i)
circle centred on Z passing through Y, with a clockwise arrow «up the page at Y»
✓ 1
Part (b)(ii)
to the left, away from Z «F = IL × B with the current in Y out of the page and B up the page»
✓ 1
Allow ECF from (b)(i).
Part (c)
force per unit length due to X = 2 × 10−7 × 200 × 150 / 0.040 = 0.150 N m−1
✓ 1
Accept μ0/2π = 2 × 10−7 T m A−1.
force per unit length due to Z = 2 × 10−7 × 300 × 150 / 0.060 = 0.150 N m−1
✓ 1
resultant = 0.30 N m−1 towards X «to the left»
✓ 1
Both forces act in the same direction. Allow ECF from (a) and (b)(ii).
Part (d)
due to Y: 0.150 N m−1 towards Y «attraction»
✓ 1
due to Z: 2 × 10−7 × 200 × 300 / 0.100 = 0.120 N m−1 away from Z «repulsion»
✓ 1
Distance X to Z = 0.100 m.
resultant = 0.030 N m−1 towards Y «to the right»
✓ 1
Award [3] for CNA with direction.
Part (e)
with Z reversed, the force on Y due to Z is 0.150 N m−1 towards Z «attraction», opposite to the 0.150 N m−1 towards X
✓ 1
the two forces cancel, so there would be no resultant force on Y
✓ 1
Conclusion must follow from the comparison. Allow ECF from (c).
Answers: (c) 0.30 N m−1 towards X · (d) 0.030 N m−1 towards Y · (e) no resultant force (the remaining parts are explanations — see the table above)
115D-2-36Force on a moving chargePaper 2Medium8 marks
Close to the equator the Earth's magnetic field is horizontal and directed towards the north, with a flux density of 3.0 × 10−5 T. A cosmic-ray proton moves vertically downwards towards the surface at a speed of 2.0 × 107 m s−1. In the diagram, up the page is vertically upwards, the right of the page is north and east is out of the page.
e = 1.60 × 10−19 C; mass of a proton = 1.67 × 10−27 kg.
Diagram NOT accurately drawn
(a)
Calculate the magnitude of the magnetic force on the proton.
(1)
(b)
Determine the direction of this force.
(2)
(c)
Show that the radius of curvature of the path of the proton is about 7 km.
(1)
(d)
Calculate the angle, in radians, through which the velocity of the proton turns while it travels 1.0 km along its path. Assume that the field is uniform.
(1)
(e)
Near the Earth's magnetic poles the field is almost vertical.
Explain why protons arriving vertically there are hardly deflected by the field.
(2)
(f)
State the direction of the magnetic force on an electron moving vertically downwards at the equator.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
F = evB = 1.60 × 10−19 × 2.0 × 107 × 3.0 × 10−5 = 9.6 × 10−17 N
✓ 1
sin θ = 1 since v ⊥ B.
Part (b)
use of F = qv × B OR Fleming's left-hand rule with the current downwards and the field towards the north
✓ 1
towards the east «out of the page»
✓ 1
Award [1] max for “east” with no supporting reasoning.
116D-2-37Motion in an electric fieldPaper 2Medium15 marks
An electron is accelerated from rest through a potential difference of 2500 V. It then enters, along the centre line, the region between two horizontal parallel plates that are 40 mm long and 20 mm apart. The potential difference between the plates is 240 V and the upper plate is positive. Ignore edge effects.
e = 1.60 × 10−19 C; me = 9.11 × 10−31 kg; g = 9.81 m s−2.
Diagram NOT accurately drawn
(a)
Calculate the speed of the electron as it enters the region between the plates.
(2)
(b)
(i)
Calculate the electric field strength between the plates.
(1)
(ii)
Show that the acceleration of the electron between the plates is about 2.1 × 1015 m s−2.
(1)
(iii)
Determine the deflection of the electron, at right angles to the plates, as it leaves the region between them.
(3)
(c)
Sketch, on the diagram, the path of the electron between the plates and after it has left them.
(2)
(d)
Explain why the path of the electron between the plates is a parabola.
(2)
(e)
Determine whether the electron would still leave the region between the plates if the potential difference between the plates were increased to 2000 V.
(3)
(f)
Outline why the gravitational force on the electron can be ignored.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
eV = ½mv², so v = √(2 × 1.60 × 10−19 × 2500 / 9.11 × 10−31)
✓ 1
v = 2.96 × 107 m s−1
✓ 1
Award [2] for CNA.
Part (b)(i)
E = 240 / 0.020 = 1.2 × 104 V m−1
✓ 1
Part (b)(ii)
a = eE / m = 1.60 × 10−19 × 1.2 × 104 / 9.11 × 10−31 = 2.11 × 1015 «m s−2»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (b)(iii)
t = 0.040 / 2.96 × 107 = 1.35 × 10−9 s «horizontal velocity constant»
✓ 1
Allow ECF from (a).
y = ½at²
✓ 1
y = 1.9 × 10−3 m «1.9 mm»
✓ 1
Accept 1.9–2.0 mm. Award [3] for CNA.
Part (c)
between the plates the path curves upwards «towards the positive plate» with increasing gradient
✓ 1
after the plates the path is a straight line in the direction of motion at the exit «tangential»
✓ 1
Do not award MP2 if the path keeps curving or bends sharply at the exit.
Part (d)
there is no force parallel to the plates, so the component of velocity parallel to the plates is constant
✓ 1
the force at right angles to the plates is constant «uniform field», so the displacement at right angles is proportional to t² and hence to (distance along the plates)², as for a horizontally launched projectile
✓ 1
OWTTE
Part (e)
y ∝ potential difference between the plates «a ∝ E ∝ V and the time between the plates is unchanged»
✓ 1
ALTERNATIVE: full recalculation with E = 1.0 × 105 V m−1.
y = 1.92 × 2000/240 = 16 mm
✓ 1
Allow ECF from (b)(iii).
this is greater than 10 mm «half the separation», so the electron hits the positive plate and does not leave
✓ 1
Conclusion must follow from the comparison with 10 mm.
Part (f)
the weight «8.9 × 10−30 N» is negligible compared with the electric force «1.9 × 10−15 N»
✓ 1
Must see a comparison. Do not accept “the electron has negligible mass” alone.
Answers: (a) 2.96 × 107 m s−1 · (b)(i) 1.2 × 104 V m−1 · (b)(iii) 1.9 mm · (e) 16 mm; no (the remaining parts are explanations — see the table above)
117D-2-48Force on a current-carrying wirePaper 2Medium13 marks
A high-voltage direct-current (HVDC) link transmits 2.0 GW of power along a straight overhead line 800 km long that runs north–south. The line has two horizontal conductors, A and B, 12 m apart. The potential difference between A and B is 1.0 MV. The current flows north in A and returns south in B, as shown in the plan view. Each conductor has a resistance of 0.010 Ω per kilometre and a mass of 7.6 kg per metre.
At the line, the Earth's magnetic field has a flux density of 5.0 × 10−5 T and is directed towards the north at 60° below the horizontal. μ0 = 4π × 10−7 T m A−1.
Diagram NOT to scale
(a)
Show that the current in each conductor is 2000 A.
(1)
(b)
Calculate the percentage of the transmitted power that is dissipated in the two conductors.
(2)
(c)
(i)
Calculate the magnitude of the force per unit length on A due to the Earth's magnetic field.
(2)
(ii)
Draw, on the diagram, arrows to show the direction of the force due to the Earth's magnetic field on A and on B.
(2)
(d)
(i)
State and explain whether A and B attract or repel each other.
(2)
(ii)
Calculate the magnitude of the force per unit length between A and B.
(1)
(e)
Determine whether the magnetic forces on the conductors need to be considered when the sag of the line between its towers is calculated.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
I = P/V = 2.0 × 109 / 1.0 × 106 «= 2000 A»
✓ 1
Part (b)
R = 800 × 0.010 = 8.0 Ω per conductor, so power lost = 2 × 2000² × 8.0 = 6.4 × 107 W
✓ 1
percentage = 3.2 %
✓ 1
Award [1] max for 1.6 % «one conductor only».
Part (c)(i)
F/L = BI sin θ with θ = 60° «only the vertical component of the field acts»
✓ 1
F/L = 5.0 × 10−5 × 2000 × sin 60° = 8.7 × 10−2 N m−1
✓ 1
Award [1] max for 0.10 N m−1 or 0.050 N m−1.
Part (c)(ii)
horizontal arrow on A pointing west «to the left»
✓ 1
Fleming's left-hand rule with the downward component of the field.
horizontal arrow on B pointing east «to the right»
✓ 1
Part (d)(i)
the currents are in opposite directions
✓ 1
so the conductors repel «the field of one at the other gives a force away from it»
total magnetic force per metre on A = 8.7 × 10−2 + 6.7 × 10−2 = 0.15 N m−1 «both forces act westwards on A»
✓ 1
Allow ECF from (c)(i) and (d)(ii).
the magnetic forces are about 0.2 % «less than 1 %» of the weight «and horizontal», so they need not be considered
✓ 1
MP3 only scores if MP1 and MP2 score.
Answers: (b) 3.2 % · (c)(i) 8.7 × 10−2 N m−1 · (d)(ii) 6.7 × 10−2 N m−1 · (e) under 1 % of the weight — no (the remaining parts are explanations — see the table above)
118D-2-49Charge in a uniform electric fieldPaper 2Medium13 marks
A photomultiplier detects very faint flashes of light in a medical scanner. Light releases an electron from electrode 0. The electron is accelerated from rest across a uniform electric field to electrode 1 (a "dynode"), 4.0 mm away, which is at a potential difference of +100 V relative to electrode 0. When an electron strikes a dynode it releases 4 electrons, which are accelerated in the same way to the next dynode. There are 10 dynodes. The tube is evacuated.
e = 1.60 × 10−19 C; me = 9.11 × 10−31 kg.
Diagram NOT to scale
(a)
(i)
State, in eV, the kinetic energy of an electron when it reaches dynode 1.
(1)
(ii)
Show that the speed of the electron when it reaches dynode 1 is about 5.9 × 106 m s−1.
(1)
(iii)
Determine the time taken by the electron to travel from electrode 0 to dynode 1.
(2)
(b)
The electrons that leave the tenth dynode reach a collector in a pulse that lasts 3.0 ns.
Calculate the average current in this pulse when a single electron is released from electrode 0.
(3)
(c)
A stray magnetic field acts into the page, as shown.
(i)
Sketch, on the diagram, the path of the electron from electrode 0 towards dynode 1.
(2)
(ii)
Determine whether the operation of the tube is affected by the Earth's magnetic field of 5.0 × 10−5 T and by the 4.0 mT field near a small permanent magnet. Use the speed in (a)(ii) for your estimate.
(3)
(d)
Suggest why photomultipliers of this kind cannot be used inside the very strong magnetic field of an MRI scanner.
I = Δq/Δt = 1.7 × 10−13 / 3.0 × 10−9 = 5.6 × 10−5 A
✓ 1
Allow ECF from MP1 and MP2.
Part (c)(i)
path starts at right angles to electrode 0 «along the electric field direction»
✓ 1
path curves towards the bottom of the page
✓ 1
Do not award MP2 for a circular arc curving upwards.
Part (c)(ii)
r = mv/eB: Earth's field gives r = 0.67 m
✓ 1
Accept 0.67–0.68 m.
4.0 mT gives r = 8.4 × 10−3 m «8.4 mm»
✓ 1
the Earth's field: radius much greater than 4.0 mm, so negligible effect; the 4.0 mT field: radius comparable with the gap, so the electrons are deflected noticeably and some miss the next dynode «the tube is affected»
✓ 1
Both conclusions required for MP3.
Part (d)
in a field of several tesla the radius of the path would be far smaller than the gap, so electrons would curl round and never reach the dynodes
✓ 1
OWTTE
Answers: (a)(i) 100 eV · (a)(ii) 5.93 × 106 m s−1 · (a)(iii) 1.35 × 10−9 s · (b) 5.6 × 10−5 A · (c)(ii) 0.67 m and 8.4 mm (the remaining parts are explanations — see the table above)
119D-2-74Circular motion in a magnetic fieldPaper 2Medium13 marks
In an electron-beam evaporator used to coat telescope mirrors with aluminium, electrons leave a hot filament with negligible speed and are accelerated through a potential difference of 5.00 kV by the uniform electric field between the filament and an anode 8.0 mm away. The beam leaves the gun moving vertically upwards. At P it enters a region of uniform magnetic field at right angles to the page, where it turns through 180° on a semicircular path of radius 30 mm and then travels vertically down into a crucible of aluminium beside the gun. The beam current is 0.20 A. Ignore relativistic effects. e = 1.60 × 10−19 C; me = 9.11 × 10−31 kg.
Diagram NOT to scale
(a)
(i)
Calculate the electric field strength between the filament and the anode.
(1)
(ii)
Show that the speed of the electrons leaving the gun is about 4.2 × 107 m s−1.
(1)
(iii)
Determine the time an electron takes to travel from the filament to the anode.
(2)
(b)
(i)
Draw, on the figure, an arrow to show the direction of the magnetic force on an electron at P.
(1)
(ii)
State the direction of the magnetic field.
(1)
(iii)
Calculate the magnetic flux density.
(2)
(iv)
Outline why the electrons strike the aluminium with the same speed as they leave the gun.
(1)
(c)
(i)
Calculate the power delivered by the beam.
(1)
(ii)
60 % of this power is used to evaporate aluminium that is already at its boiling point. Specific latent heat of vaporization of aluminium = 1.09 × 107 J kg−1. Determine the mass of aluminium evaporated in one minute.
(2)
(d)
Suggest why the gun is placed beside the crucible, with the beam bent through 180°, instead of pointing directly at the aluminium from above.
uniform acceleration from rest, so mean speed = v/2
✓ 1
ALTERNATIVE: a = eE/me = 1.10 × 1017 m s−2 and t = v/a.
t = 2 × 8.0 × 10−3/4.2 × 107 = 3.8 × 10−10 s
✓ 1
Award [2] for CNA. Award [1] max for 1.9 × 10−10 s «d/v».
Part (b)(i)
arrow at P pointing to the right «towards the centre of the semicircle»
✓ 1
Must be perpendicular to the beam at P.
Part (b)(ii)
into the page
✓ 1
F = qv × B with q negative, v up the page and F to the right. Allow ECF from (b)(i).
Part (b)(iii)
evB = mev²/r, so B = mev/(er)
✓ 1
B = 9.11 × 10−31 × 4.19 × 107/(1.60 × 10−19 × 0.030) = 8.0 × 10−3 T
✓ 1
Allow ECF from (a)(ii). Award [2] for CNA.
Part (b)(iv)
the magnetic force is perpendicular to the velocity, so it does no work and the kinetic energy «speed» is unchanged
✓ 1
OWTTE.
Part (c)(i)
P = IV = 0.20 × 5.00 × 103 = 1000 W
✓ 1
Part (c)(ii)
energy in one minute = 0.60 × 1000 × 60 = 3.6 × 104 J
✓ 1
m = Q/L = 3.6 × 104/1.09 × 107 = 3.3 × 10−3 kg
✓ 1
Allow ECF from (c)(i). Award [2] for CNA.
Part (d)
the hot aluminium vapour rises from the crucible; a gun beside and below the crucible is out of the stream of vapour, so the filament is not coated or damaged
✓ 1
OWTTE.
Answers: (a)(i) 6.25 × 105 V m−1 · (a)(iii) 3.8 × 10−10 s · (b)(iii) 8.0 × 10−3 T · (c)(i) 1000 W · (c)(ii) 3.3 × 10−3 kg (the remaining parts are explanations — see the table above)
120D-2-77Circular motion in a magnetic fieldPaper 2Medium12 marks
Uranium can be enriched in uranium-235 by electromagnetic separation. Singly charged positive uranium ions, charge +1.60 × 10−19 C, start from rest and are accelerated through a potential difference of 35.0 kV. They enter a uniform magnetic field at S, moving at right angles to the field, travel through a semicircle and are collected on the boundary. Mass of a uranium-235 ion = 3.90 × 10−25 kg; mass of a uranium-238 ion = 3.95 × 10−25 kg.
Diagram NOT to scale
(a)
Show that the speed of a uranium-235 ion entering the field is about 1.7 × 105 m s−1.
(1)
(b)
The radius of the path of the uranium-235 ions is 1.20 m. Calculate the magnetic flux density.
(2)
(c)
Draw, on the figure, the paths of the uranium-235 ions and of the uranium-238 ions in the field. Label each path.
(2)
(d)
Show that the distance between the points where the two kinds of ion reach the boundary is about 15 mm.
(2)
(e)
The ion beam carries a current of 0.10 A. 0.72 % of the ions are uranium-235 ions.
(i)
Calculate the number of uranium-235 ions collected per second.
(2)
(ii)
Determine whether one separator could collect 1.0 kg of uranium-235 in a year.
(2)
(f)
Outline why most nuclear fission reactors need fuel that is enriched in uranium-235.
mass per year = 4.5 × 1015 × 3.90 × 10−25 × 3.15 × 107 = 55 g «0.055 kg»
✓ 1
Allow ECF from (e)(i).
this is much less than 1.0 kg, so one separator could not
✓ 1
Conclusion must follow from the candidate's value.
Part (f)
uranium-235 is the isotope that undergoes neutron-induced fission; natural uranium has too little of it to sustain a chain reaction «in reactors moderated by ordinary water»
✓ 1
OWTTE.
Answers: (b) 0.344 T · (d) 15 mm · (e)(i) 4.5 × 1015 s−1 · (e)(ii) 55 g (the remaining parts are explanations — see the table above)
121D-2-78Force on a moving chargePaper 2Medium13 marks
In a scanning electron microscope, electrons are accelerated from rest through a potential difference of 2.00 kV and travel down the evacuated column. Scanning coils produce a uniform magnetic field, directed out of the page, over a 15 mm length of the beam. The beam is moved across the sample by changing the field. The centre of the coils is 40 mm above the sample. Ignore the Earth's magnetic field. e = 1.60 × 10−19 C; me = 9.11 × 10−31 kg.
Diagram NOT to scale
(a)
Calculate the speed of the electrons.
(2)
(b)
The flux density in the coils is 0.10 mT.
(i)
Show that the radius of the path of an electron in the coils is about 1.5 m.
(1)
(ii)
Outline why the speed of the electrons is not changed by the coils.
(1)
(iii)
Show that the beam turns through an angle of about 0.010 rad in the coils.
(1)
(iv)
Estimate the displacement of the spot on the sample.
(2)
(c)
Deduce the direction in which the spot on the sample is displaced.
(2)
(d)
The accelerating potential difference is increased to 4.50 kV. Determine the factor by which the flux density in the coils must change to give the same displacement of the spot.
(2)
(e)
Samples that are electrical insulators are coated with a very thin layer of metal that is earthed. Explain why.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
eV = ½mev²
✓ 1
v = √(2 × 1.60 × 10−19 × 2.00 × 103/9.11 × 10−31) = 2.65 × 107 m s−1
122IBP2S-032-01-13Charge-to-mass ratioPaper 2Medium8 marks
In an anti-doping laboratory, a magnetic-sector mass spectrometer is used to identify a substance in an athlete's sample. Singly charged positive ions start from rest. They are accelerated through a potential difference of 2500 V and then enter a uniform magnetic field of 0.40 T at right angles. The ions travel in a semicircle of radius 0.15 m. Use e = 1.60 ×10−19 C and 1 u = 1.66 ×10−27 kg.
Diagram not to scale. The magnetic field fills the shaded region.
(a)
Show that the speed of an ion entering the magnetic field is \(v=\sqrt{\frac{2qV}{m}}\).
(2)
(b)
Show that the charge-to-mass ratio of the ion is \(\frac{q}{m}=\frac{2V}{B^2r^2}\).
(2)
(c)
Calculate the charge-to-mass ratio of the ion.
(2)
(d)
Determine the mass of the ion in atomic mass units, u.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Work done by the electric field equals the gain in kinetic energy: qV = ½mv2
✓ 1
Rearranges to v = √(2qV/m)
✓ 1
Part (b)
The magnetic force provides the centripetal force: qvB = mv2/r, so v = qBr/m
123IBP2S-032-01-14Parallel current-carrying wiresPaper 2Medium7 marks
In a data centre, two long, straight, parallel copper busbars carry current to the servers. The busbars are 0.12 m apart, and each section is 2.0 m long. Under normal operation, each busbar carries 1500 A, and the currents are in opposite directions. μ0 = 4π ×10−7 T m A−1
(a)
State and explain whether the busbars attract or repel each other.
(2)
(b)
Calculate the magnitude of the force on one 2.0 m section of busbar during normal operation.
(2)
(c)
During a short-circuit fault, the current in each busbar briefly rises to 25 kA. Determine the force on the same section during the fault.
(2)
(d)
Outline why the busbars must be rigidly clamped.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Each busbar lies in the magnetic field of the other (direction from the right-hand grip rule), so it experiences a force F = BIL
✓ 1
Applying the motor rule (or Fleming's left-hand rule) shows that for opposite currents the force on each busbar points away from the other, so they repel
The force is proportional to I2, so it scales by (25000/1500)2 ≈ 278, or substitute directly into F = μ0I2L/(2πr)
✓ 1
F ≈ 2.1 ×103 N
✓ 1
Allow error carried forward from the previous part.
—
Part (d)
The large forces during a fault could bend the busbars or move them apart and damage the supports, so the clamps must withstand forces of the order of 103 N
✓ 1
Common error: Both currents increase during the fault, so the force scales with the square of the current ratio, not the ratio itself.
124IBP2S-032-01-16Force on a current-carrying wirePaper 2Medium5 marks
In a science museum demonstration, a horizontal metal rod of mass 12 g and length 0.20 m hangs from two light, flexible conducting leads. The rod lies at right angles to a uniform horizontal magnetic field of 0.25 T. When there is a current in the rod, the magnetic force on it is vertically upward.
(a)
State the rule used to predict the direction of the magnetic force on the rod from the directions of the current and the field.
(1)
(b)
Calculate the current for which the tension in the leads becomes zero.
(2)
(c)
The current is suddenly increased to 3.0 A. Determine the initial upward acceleration of the rod.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A right-hand rule for F = IL × B (or Fleming's left-hand rule)
✓ 1
Part (b)
At zero tension, the magnetic force balances the weight: BIL = mg
✓ 1
I = 0.012 × 9.8 / (0.25 × 0.20) = 2.4 A
✓ 1
Part (c)
Resultant force = BIL − mg = 0.25 × 3.0 × 0.20 − 0.012 × 9.8 = 0.15 − 0.1176 = 0.0324 N
✓ 1
a = 0.0324 / 0.012 = 2.7 m s−2
✓ 1
Allow use of g = 9.81 m s−2. Allow ECF for an incorrect value of the weight.
—
Common error: Convert 12 g to 0.012 kg. The weight still acts, so the acceleration comes from the resultant force, not from BIL alone.
125IBP2S-032-00-13Charge-to-mass ratioPaper 2Medium7 marks
A student analyses data from an isotope separator. Singly charged positive ions start from rest, are accelerated through a potential difference V, and then enter a uniform magnetic field of strength B = 0.150 T at right angles to it. The radius r of each ion's path is measured. V / V: 500, 1000, 1500, 2000 r / cm: 5.63, 7.97, 9.76, 11.27
Data from the question
(a)
Show that \(r^2 = \frac{2m}{qB^2}V\), where m is the mass and q the charge of an ion.
(2)
(b)
Using the data, determine the charge-to-mass ratio q/m of the ions.
(3)
(c)
Determine the mass of one ion and suggest which element it is.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Energy gained: qV = ½mv², so v² = 2qV/m .
✓ 1
Magnetic force provides the centripetal force: r = mv/(qB), so r² = m²v²/(q²B²) = (2m/(qB²))V .
✓ 1
Part (b)
Calculate r²/V, or the gradient of a graph of r² against V, e.g. r² = 1.270 × 10−2 m² at V = 2000 V .
✓ 1
Gradient ≈ 6.35 × 10−6 m² V−1, with the data consistent with a constant value .
m = e/(q/m) = 1.60 × 10−19/1.40 × 107 = 1.1 × 10−26 kg . ECF from (b).
✓ 1
This is about 6.9 u, which suggests a lithium-7 ion .
✓ 1
Exam tip: When the stem asks you to use the data, use r² against V. A single pair of values earns less credit than a gradient or a check of several rows.
126IBP2S-032-00-16The velocity selectorPaper 2Medium9 marks
An electromagnetic blood-flow meter measures blood flow in an artery during surgery. A uniform magnetic field is applied perpendicular to the artery. Two electrodes on opposite sides of the artery, along a line perpendicular to both the blood flow and the field, measure a potential difference. Blood contains positive and negative ions that move with the blood.
(a)
Explain why positive and negative ions are deflected towards opposite sides of the artery.
(2)
(b)
Explain why the potential difference between the electrodes quickly reaches a steady value.
(2)
(c)
Show that the steady potential difference is V = vBd, where v is the blood speed and d is the internal diameter of the artery.
(2)
(d)
The artery has an internal diameter of 4.0 mm, the field strength is 0.30 T, and the measured potential difference is 0.48 mV. Calculate the blood speed.
(2)
(e)
Calculate the volume flow rate of blood through the artery.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Each moving ion experiences a magnetic force qvB, perpendicular to both its velocity and the field .
✓ 1
The force direction depends on the sign of the charge, so positive and negative ions are pushed towards opposite electrodes .
✓ 1
Part (b)
The separated charges produce an electric field across the artery, which exerts a force on the ions opposing the magnetic force .
✓ 1
Charge builds up until qE = qvB. The net sideways force is then zero, so no further separation occurs .
127IBP2S-032-03-13Parallel current-carrying wiresPaper 2Medium7 marks
A student uses a current balance to measure the force between two long, parallel wires. The same current I flows through both wires in the same direction, and the centres of the wires are 2.0 cm apart. The force per unit length F/L on one wire is measured for several currents.
I / A: 2.0, 4.0, 6.0, 8.0 (F/L) / ×10−4 N m−1: 0.40, 1.6, 3.6, 6.4
Data from the question
(a)
State whether the force between the wires is attractive or repulsive.
(1)
(b)
Show that the data are consistent with F/L being proportional to I2.
(2)
(c)
Determine a value for μ0 from the data.
(3)
(d)
Outline why the separation of the wires is measured between their centres rather than between their surfaces.
(1)
Show mark scheme
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Notes
Part (a)
Attractive, because the currents are in the same direction .
✓ 1
Part (b)
Calculates (F/L)/I2 for all four rows, e.g. 0.40×10−4/4.0 = 1.6×10−4/16 = 3.6×10−4/36 = 6.4×10−4/64 = 1.0×10−5 N m−1 A−2. Equivalent evidence also earns this mark: I2 scales by 4, 9 and 16 while F/L scales by the same factors, or F/L against I2 gives a straight line through the origin . States that the ratio is constant, so F/L ∝ I2 .
✓ 2
Part (c)
Uses F/L = μ0I2/(2πr), so (F/L)/I2 = μ0/(2πr) . Substitutes μ0 = 2π × 0.020 × 1.0×10−5, using r in metres and the constant from (b), a mean of the rows or a gradient of F/L against I2 . μ0 = 1.26×10−6 T m A−1 (or N A−2); accept 1.3×10−6. Allow error carried forward from (b) .
✓ 3
Part (d)
B = μ0I/(2πr) uses the distance from the wire's axis. By symmetry, the field outside a wire of circular cross-section is the same as if all the current flowed along its central axis, so the centre-to-centre distance is the correct r .
✓ 1
Exam tip: When data should show proportionality, calculating the ratio for every row, or plotting a straight line through the origin, gives the most convincing evidence.
128IBP2S-032-03-14Motion in an electric fieldPaper 2Medium7 marks
In a continuous inkjet printer, charged ink droplets travel horizontally at 20 m s−1 between two horizontal deflecting plates. The plates are 1.0 cm long and 1.0 mm apart, with a potential difference of 1500 V between them. Each droplet has a mass of 1.2 × 10−10 kg and a charge of 1.5 × 10−13 C. Assume the field between the plates is uniform.
Diagram not to scale. The plates are horizontal.
(a)
Calculate the electric field strength between the plates.
(1)
(b)
Show that the acceleration of a droplet due to the electric field is about 1.9 × 103 m s−2.
(2)
(c)
Calculate the time a droplet spends between the plates.
(1)
(d)
Determine the vertical deflection of a droplet as it leaves the plates.
(2)
(e)
Explain why the effect of gravity on the droplet can be ignored in this calculation.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
E = V/d = 1500/(1.0 × 10−3) = 1.5 × 106 V m−1 .
✓ 1
Part (b)
F = qE = 1.5 × 10−13 × 1.5 × 106 = 2.25 × 10−7 N . a = F/m = 2.25 × 10−7/(1.2 × 10−10) = 1.875 × 103 ≈ 1.9 × 103 m s−2 .
✓ 2
Part (c)
The horizontal velocity is constant, so t = 0.010/20 = 5.0 × 10−4 s .
✓ 1
Part (d)
Uses y = ½at² with the vertical acceleration from (b) and the time from (c) . y = 0.5 × 1.875 × 103 × (5.0 × 10−4)2 = 2.3 × 10−4 m . Allow error carried forward from (b) and (c).
✓ 2
Part (e)
Either: g = 9.8 m s−2 is very small compared with the electric acceleration of about 1.9 × 103 m s−2. Or: the weight, 1.2 × 10−9 N, is very small compared with the electric force of 2.25 × 10−7 N. Either comparison (about 0.5%) shows the gravitational deflection is negligible .
✓ 1
Common error: Do not apply the acceleration to the horizontal motion. The field acts only perpendicular to the plates, so the horizontal speed stays at 20 m s⁻¹.
129IBP2S-032-02-14Parallel current-carrying wiresPaper 2Medium9 marks
Two long, straight, parallel copper busbars in a data centre are 0.050 m apart. They carry equal currents I in opposite directions. An engineer measures the force per unit length F/L on one busbar for several currents.
I / A: 200, 400, 600, 800 F/L / N m−1: 0.16, 0.64, 1.44, 2.56
The force per unit length between parallel wires is \(\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi r}\).
Data from the question
(a)
State whether the busbars attract or repel each other.
(1)
(b)
Explain how a force on one busbar arises from the current in the other.
(2)
(c)
Show that the data are consistent with F/L being proportional to I2.
(2)
(d)
Determine a value for μ0 from the data.
(2)
(e)
During a short-circuit fault, the current in each busbar rises briefly to 2.0 × 104 A. Estimate the force on a 2.0 m length of busbar.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Repel .
✓ 1
Part (b)
The current in one busbar produces a magnetic field at the other busbar, perpendicular to it . The current in the second busbar is perpendicular to this field, so it experiences a force F = BIL, directed away from the first busbar .
✓ 2
Part (c)
Calculates (F/L)/I² for every row, e.g. 0.16/200², 0.64/400², 1.44/600² and 2.56/800² . All four give 4.0 × 10−6 N m−1 A−2; the ratio is constant, so F/L ∝ I² . An equivalent ratio argument also earns both marks: doubling I from 200 A to 400 A multiplies F/L by 4, tripling it to 600 A multiplies F/L by 9, and quadrupling it to 800 A multiplies F/L by 16.
✓ 2
Part (d)
μ0 = 2πr(F/L)/I² . μ0 = 2π × 0.050 × 4.0 × 10−6 = 1.3 × 10−6 T m A−1 . ECF from (c).
✓ 2
Part (e)
F/L = 4.0 × 10−6 × (2.0 × 104)² = 1.6 × 103 N m−1 . F = 1.6 × 103 × 2.0 = 3.2 × 103 N . ECF from (c) or (d).
✓ 2
Exam tip: To test for proportionality to I², calculate (F/L)/I² for every row. Checking only that F/L increases with I is not enough.
130IBP2S-032-02-15The velocity selectorPaper 2Medium8 marks
A sports anti-doping laboratory uses a mass spectrometer. Singly charged positive ions pass through a velocity selector, which has an electric field of 4.0 × 104 V m−1 perpendicular to a magnetic field of 0.20 T. The ions then enter a second region, which contains only a uniform magnetic field of 0.50 T perpendicular to their velocity. One ion species follows a semicircle of radius 0.12 m. Elementary charge = 1.60 × 10−19 C.
Diagram not to scale. The magnetic field fills the shaded region.
(a)
Explain why only ions with one particular speed leave the velocity selector undeflected.
(3)
(b)
Calculate the speed of the ions leaving the selector.
(1)
(c)
Determine the mass of this ion species.
(2)
(d)
A heavier isotope of the same singly charged ion also passes through the selector. State and explain whether its semicircle has a larger or smaller radius.
(2)
Show mark scheme
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Mark
Notes
Part (a)
The electric force qE and the magnetic force qvB act in opposite directions . The ions are undeflected only when qE = qvB, i.e. v = E/B . For faster ions the magnetic force is larger, and for slower ions it is smaller, so these ions are deflected .
✓ 3
Part (b)
v = E/B = 4.0 × 104/0.20 = 2.0 × 105 m s−1 .
✓ 1
Part (c)
m = qBr/v . m = 1.60 × 10−19 × 0.50 × 0.12/2.0 × 105 = 4.8 × 10−26 kg . ECF from (b).
✓ 2
Part (d)
Larger radius . It has the same speed and charge in the same field, and r = mv/(qB) ∝ m .
✓ 2
Common error: Using the selector field (0.20 T) instead of the deflection field (0.50 T) when calculating the mass.
131IBP2S-032-04-14Charge in a uniform electric fieldPaper 2Medium8 marks
A gridded ion thruster on a deep-space probe ionises xenon atoms. The singly charged Xe+ ions are accelerated from rest through a potential difference of 1.2 kV. The ion beam carries a current of 1.5 A. Mass of a xenon ion = 131 u, with 1 u = 1.66×10−27 kg.
(a)
Calculate the kinetic energy of one ion leaving the thruster, in J.
(1)
(b)
Calculate the speed of the ions.
(2)
(c)
Determine the thrust on the probe.
(3)
(d)
The thruster also emits electrons into the beam. Outline why this is necessary.
(2)
Show mark scheme
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Notes
Part (a)
Ek = qV = 1.60×10−19 × 1200 = 1.9×10−16 J
✓ 1
Part (b)
m = 131 × 1.66×10−27 = 2.17×10−25 kg
✓ 1
v = √(2Ek/m) = 4.2×104 m s−1 (ECF from (a))
✓ 1
Part (c)
Number of ions emitted per second = I/e = 1.5/1.60×10−19 = 9.4×1018 s−1
✓ 1
Thrust = rate of change of momentum = (ions per second) × mv
✓ 1
Thrust = 9.4×1018 × 2.17×10−25 × 4.2×104 ≈ 0.086 N (ECF from (b))
✓ 1
Part (d)
The beam carries positive charge away, so without the electrons the probe would become increasingly negatively charged
✓ 1
A negatively charged probe would attract the positive ions back, reducing the thrust; emitting electrons keeps the probe neutral
✓ 1
Exam tip: Thrust is rate of change of momentum. Find the number of ions per second from I/e, then multiply by the momentum of each ion.
132IBP2S-032-04-15Force on a moving chargePaper 2Medium6 marks
An accelerator mass spectrometer used for radiocarbon dating of archaeological samples contains a velocity selector. Singly charged positive carbon-14 ions travel horizontally due east into the selector. The uniform electric field in the selector is vertically downward, with strength 2.4×104 V m−1. A uniform magnetic field acts in the same region. The selector is designed to pass ions of speed 8.0×104 m s−1 undeflected.
(a)
State the direction of the electric force on an ion.
(1)
(b)
Determine the magnitude and direction of the magnetic flux density required.
(3)
(c)
A carbon-14 ion enters the selector with a speed of 1.0×105 m s−1. Determine the magnitude and direction of the initial resultant force on this ion.
(2)
Show mark scheme
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Mark
Notes
Part (a)
Vertically downward, in the direction of the field, because the ion is positive
✓ 1
Part (b)
qE = qvB, so B = E/v
✓ 1
B = 2.4×104 / 8.0×104 = 0.30 T
✓ 1
The magnetic force must be upward; with the velocity due east, the field is horizontal and points due north
✓ 1
Part (c)
FB = evB = 1.60×10−19 × 1.0×105 × 0.30 = 4.8×10−15 N upward and FE = eE = 3.84×10−15 N downward (ECF from (b))
✓ 1
Resultant = 9.6×10−16 N vertically upward
✓ 1
Common error: The electric force does not change with speed. Only the magnetic force increases, so a faster ion feels a resultant force in the direction of the magnetic force.
133D-2-08Circular motion in a magnetic fieldPaper 2Hard15 marks
Polonium-210 (21084Po) decays by emitting alpha particles, all with a kinetic energy of 5.30 MeV.
In an evacuated chamber, a narrow beam of these alpha particles moving to the right enters a large region of uniform magnetic field of flux density 1.50 T, directed into the page.
Mass of an alpha particle = 6.64 × 10−27 kg. Charge of an alpha particle = +3.20 × 10−19 C.
Diagram NOT to scale
(a)
(i)
State the nuclear equation for this decay. The nucleus formed is an isotope of lead (Pb).
(2)
(ii)
Suggest why the alpha particles must travel in a vacuum in this experiment.
(1)
(b)
(i)
Show that the speed of the alpha particles is about 1.6 × 107 m s−1.
(2)
(ii)
Draw, on the diagram, the path of the alpha particles in the magnetic field.
(2)
(iii)
Calculate the radius of the path.
(2)
(c)
Explain why the kinetic energy of the alpha particles does not change while they are in the magnetic field.
(2)
(d)
A student suggests that protons with the same kinetic energy, 5.30 MeV, could be distinguished from these alpha particles by the radius of their path in the same field. Determine whether the student is correct.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
42α «or 42He» as a product
✓ 1
Nucleon and proton numbers both needed.
20682Pb as the other product
✓ 1
Nucleon and proton numbers both needed.
Part (a)(ii)
alpha particles are strongly ionizing, so in air they would lose their energy within a few centimetres «and not complete the path»
✓ 1
OWTTE
Part (b)(i)
Ek = 5.30 × 106 × 1.60 × 10−19 = 8.48 × 10−13 J
✓ 1
v = √(2 × 8.48 × 10−13/6.64 × 10−27) = 1.60 × 107 m s−1
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (b)(ii)
path curves towards the top of the page as soon as the particles enter the field
✓ 1
Fleming's left-hand rule: field into the page, positive particles moving to the right, force up the page.
path is an arc of a circle «constant curvature»: a semicircle leaving the field through its left-hand edge, moving to the left
In an ion implanter used in making computer chips, singly charged boron ions, each of charge +1.60 × 10−19 C, are accelerated from rest through a potential difference of 20.0 kV. They then enter a velocity filter: two horizontal parallel plates 20.0 mm apart, with a uniform magnetic field of flux density 0.0500 T directed into the page between them.
The filter is set so that boron-11 ions, of mass 1.83 × 10−26 kg, pass straight through. Boron-10 ions have a mass of 1.66 × 10−26 kg.
Diagram NOT to scale
(a)
(i)
Draw and label, on the ion shown in the diagram, the electric force and the magnetic force on a boron-11 ion that passes straight through the filter.
(2)
(ii)
Deduce which plate is positive.
(1)
(b)
(i)
Show that the speed of a boron-11 ion entering the filter is about 5.9 × 105 m s−1.
(1)
(ii)
Calculate the potential difference between the plates.
(2)
(c)
Boron-10 ions, accelerated through the same potential difference, also enter the filter. Determine whether they pass straight through. If they do not, state the direction in which they are deflected.
(3)
(d)
A few boron-11 ions lose two electrons instead of one before they are accelerated through the same potential difference. Determine whether these ions pass straight through the filter.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
magnetic force towards the top of the page AND electric force towards the bottom of the page, both labelled
✓ 1
Fleming's left-hand rule: positive ions moving to the right, field into the page, magnetic force up the page.
arrows of equal length
✓ 1
Part (a)(ii)
the upper plate, because the electric force on a positive ion must point down, so the field points from the upper plate to the lower plate
Must see full substitution OR answer to at least 3 s.f.
Part (b)(ii)
qE = qvB, so E = vB = 5.91 × 105 × 0.0500 = 2.96 × 104 V m−1
✓ 1
V = Ed = 2.96 × 104 × 0.0200 = 591 V
✓ 1
Award [2] for CNA. Accept 590–592 V.
Part (c)
boron-10 ions gain the same kinetic energy but have less mass, so they are faster: v = 5.91 × 105 × √(1.83/1.66) = 6.21 × 105 m s−1
✓ 1
the magnetic force qvB increases while the electric force qE is unchanged
✓ 1
so they do not pass straight through; they are deflected towards the top of the page «towards the positive plate»
✓ 1
MP3 only scores if MP2 scores.
Part (d)
the kinetic energy qV doubles, so v = √2 × 5.91 × 105 = 8.36 × 105 m s−1
✓ 1
doubling q doubles both the electric and the magnetic force, so whether they balance depends only on whether v = E/B
✓ 1
v > E/B, so the magnetic force is larger and the ions do not pass straight through «they are deflected towards the top of the page»
✓ 1
MP3 only scores if a larger speed is found or argued.
Answers: (b)(i) 5.91 × 105 m s−1 · (b)(ii) 591 V · (c) 6.21 × 105 m s−1 · (d) 8.36 × 105 m s−1(the remaining parts are explanations — see the table above)
135D-2-32Circular motion in a magnetic fieldPaper 2Hard14 marks
Electrons are accelerated from rest through a potential difference of 2.0 kV. They then enter, at right angles to its boundary, a region of uniform magnetic field of flux density 1.5 mT directed into the page. The field region has two parallel boundaries 50 mm apart and extends far above and below the beam.
e = 1.60 × 10−19 C; me = 9.11 × 10−31 kg.
Diagram NOT accurately drawn
(a)
Calculate the speed of the electrons as they enter the field.
(2)
(b)
Show that the radius of the path of the electrons in the field is about 0.10 m.
(1)
(c)
Draw, on the diagram, an arrow to show the direction of the magnetic force on an electron as it enters the field, and the path of the electron in the field and after it leaves the field.
(3)
(d)
Determine the angle through which the electrons are deflected by the field.
(3)
(e)
Calculate the time that an electron spends in the field.
(2)
(f)
The accelerating potential difference is reduced to 400 V.
Determine whether the electrons now leave the field through the far boundary.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
eV = ½mv², so v = √(2 × 1.60 × 10−19 × 2.0 × 103 / 9.11 × 10−31)
Must see full substitution OR answer to 3 s.f. Allow ECF from (a).
Part (c)
arrow at the entry point pointing down the page «towards the bottom of the page»
✓ 1
Direction from F = qv × B with q negative.
path in the field is a circular arc curving downwards, tangential to the initial direction
✓ 1
path after leaving the field is a straight line, tangential to the arc at the exit point
✓ 1
Do not award MP3 if the path continues to curve outside the field.
Part (d)
the centre of the circular path lies on the entry boundary, a distance r from the entry point
✓ 1
sin θ = w/r = 0.050 / 0.10
✓ 1
Allow ECF from (b).
θ = 30°
✓ 1
Accept 29°–31°. Allow ECF from (b).
Part (e)
arc length = rθ = 0.10 × 0.52 rad = 0.052 m
✓ 1
θ must be in radians. Allow ECF from (b) and (d).
t = 0.052 / 2.65 × 107 = 2.0 × 10−9 s
✓ 1
Accept 1.9–2.0 × 10−9 s. Award [2] for CNA.
Part (f)
v ∝ √V, so r ∝ √V «for the same B»
✓ 1
r = 0.10 × √(400/2000) = 0.045 m
✓ 1
Accept 0.045 m. Award MP2 for a correct radius by direct calculation.
this is less than 50 mm, so the electrons do not reach the far boundary; they turn through a semicircle and leave back through the entry boundary
✓ 1
MP3 requires a comparison with 50 mm.
Answers: (a) 2.65 × 107 m s−1 · (b) 0.101 m · (d) 30° · (e) 2.0 × 10−9 s · (f) r = 0.045 m; no (the remaining parts are explanations — see the table above)
136D-2-35Charge in a uniform electric fieldPaper 2Hard14 marks
In an electron energy analyser, two large horizontal parallel plates are 10.0 mm apart in a vacuum. The potential difference between them is 150 V and the lower plate is positive. Electrons, each with a kinetic energy of 200 eV, enter the region between the plates through a small hole H in the lower plate, moving at 40° to the plate.
e = 1.60 × 10−19 C; me = 9.11 × 10−31 kg.
Diagram NOT accurately drawn
(a)
State the direction of the electric force on an electron between the plates.
(1)
(b)
Calculate the magnitude of the acceleration of an electron between the plates.
(2)
(c)
Show that the speed of the electrons as they enter through H is about 8.4 × 106 m s−1.
(1)
(d)
Determine whether the electrons reach the upper plate.
(4)
(e)
Calculate the distance from H at which the electrons return to the lower plate.
(2)
(f)
Sketch, on the diagram, the path of an electron between the plates.
(2)
(g)
A second small hole is cut in the lower plate at the distance found in (e), and a detector is placed below it.
Suggest how this arrangement selects electrons of one kinetic energy.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
vertically downwards, towards the «positive» lower plate
component of velocity at right angles to the plates = 8.4 × 106 × sin 40° = 5.39 × 106 m s−1
✓ 1
the perpendicular component falls to zero at the greatest height: vy² = 2ah
✓ 1
greatest height = vy²/2a = 5.5 mm
✓ 1
ALTERNATIVE: energy for the perpendicular motion = 200 sin² 40° = 82.6 eV, so height = 82.6 / 1.5 × 104 m = 5.5 mm. Allow ECF.
5.5 mm is less than 10.0 mm, so the electrons do not reach the upper plate
✓ 1
Conclusion must follow from a calculation.
Part (e)
time between the plates = 2vy/a = 4.09 × 10−9 s
✓ 1
Allow ECF from (b) and (d).
distance = v cos 40° × t = 26 mm
✓ 1
Accept 26–27 mm. Award [2] for CNA.
Part (f)
a symmetrical curve «parabola» starting at H at about 40° and returning to the lower plate
✓ 1
highest point about halfway between the plates, clearly below the upper plate
✓ 1
Allow ECF from (d).
Part (g)
for a given angle and field, the landing distance is proportional to the kinetic energy «faster electrons land further from H»
✓ 1
Accept x = 2Ek sin 2θ/eE or a qualitative argument.
so only electrons with kinetic energy 200 eV reach the second hole and the detector; changing the potential difference selects a different energy
✓ 1
OWTTE
Answers: (b) 2.63 × 1015 m s−2 · (c) 8.38 × 106 m s−1 · (d) greatest height 5.5 mm; no · (e) 26 mm (the remaining parts are explanations — see the table above)
137D-2-46Force on a current-carrying wirePaper 2Hard14 marks
In a sodium-cooled nuclear fission reactor, liquid sodium carries thermal energy from the core to a heat exchanger. The sodium is circulated by an electromagnetic pump that has no moving parts.
The diagram shows the pump duct from above. The duct has a rectangular cross-section: it is 0.12 m wide (between the two electrodes) and 0.050 m deep (into the page). Over a 0.25 m length of the duct a uniform magnetic field of flux density 0.40 T acts into the page. A current of 4000 A flows across the sodium from one electrode to the other, as shown.
Diagram NOT to scale
(a)
Draw, on the diagram, an arrow to show the direction of the magnetic force on the sodium between the electrodes.
(1)
(b)
Calculate the magnitude of the magnetic force on the sodium.
(2)
(c)
Show that the pump increases the pressure of the sodium by about 3.2 × 104 Pa.
(1)
(d)
The designers want to double this pressure increase without changing the current or the magnetic flux density.
Suggest, with a reason, one change to the dimensions of the duct.
(3)
(e)
The pump moves 5.0 kg of sodium each second. The temperature of the sodium rises by 120 K as it passes through the core. Specific heat capacity of liquid sodium = 1.30 × 103 J kg−1 K−1; density of liquid sodium = 880 kg m−3.
(i)
Calculate the rate at which the sodium removes thermal energy from the core.
(2)
(ii)
Determine the rate at which the magnetic force does work on the sodium in the duct.
(3)
(f)
Outline why the sodium transfers its energy to the water that drives the turbines through a heat exchanger instead of being used to boil the water directly.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
arrow to the right, along the duct «in the direction of the flow»
✓ 1
Fleming's left-hand rule OR F = IL × B.
Part (b)
the current crosses the duct, so the length of conductor in F = BIL is the electrode separation, 0.12 m
✓ 1
Do not award MP1 for L = 0.25 m «400 N».
F = 0.40 × 4000 × 0.12 = 192 N
✓ 1
Award [2] for CNA.
Part (c)
Δp = F/A = 192 / (0.12 × 0.050) = 3.2 × 104 «Pa»
✓ 1
Allow ECF from (b). Must see full substitution OR answer to 3 s.f.
Part (d)
Δp = BIw/(wh) = BI/h
✓ 1
so the pressure increase does not depend on the width of the duct or on the length of the electrodes
✓ 1
Do not accept «double the width» or «longer electrodes».
halve the depth of the duct «to 0.025 m, in the direction of the field»
✓ 1
MP3 only scores if MP1 scores.
Part (e)(i)
P = (Δm/Δt)cΔT = 5.0 × 1.30 × 103 × 120
✓ 1
P = 7.8 × 105 W
✓ 1
Award [2] for CNA.
Part (e)(ii)
volume flowing per second = 5.0 / 880 = 5.68 × 10−3 m³ s−1
✓ 1
speed of sodium v = 5.68 × 10−3 / (0.12 × 0.050) = 0.947 m s−1
✓ 1
P = Fv = 192 × 0.947 = 182 W
✓ 1
Allow ECF from (b). Accept 180–185 W. Accept Δp × volume flow rate.
Part (f)
the sodium passing through the core becomes radioactive «and reacts violently with water»
✓ 1
the heat exchanger keeps it in a closed loop separate from the water and turbines, so no radioactive material reaches them
✓ 1
OWTTE
Answers: (b) 192 N · (c) 3.2 × 104 Pa · (e)(i) 7.8 × 105 W · (e)(ii) 182 W (the remaining parts are explanations — see the table above)
138D-2-47Paths in electric and magnetic fieldsPaper 2Hard14 marks
A fusion reactor is heated by a beam of fast deuterium atoms. The beam is made by accelerating deuterons (deuterium nuclei, D⁺) from rest through a potential difference of 80.0 kV. The ion source also produces some protons (H⁺), which are accelerated through the same potential difference and must be removed from the beam.
Charge of a deuteron or a proton = +1.60 × 10−19 C; mass of a deuteron = 3.34 × 10−27 kg; mass of a proton = 1.67 × 10−27 kg.
Diagram drawn to scale
(a)
(i)
Show that the kinetic energy of each ion after acceleration is 1.28 × 10−14 J.
(1)
(ii)
Calculate the speed of a deuteron after acceleration.
(1)
(b)
In a first design, the ions pass through an electric deflector: two parallel plates 0.200 m long and 50.0 mm apart, with a potential difference of 4.00 kV between them. The ions enter midway between the plates, moving parallel to them.
(i)
Determine the deflection of a deuteron, at right angles to the plates, as it leaves the region between them.
(3)
(ii)
Deduce whether this electric deflector can separate the protons from the deuterons.
(2)
(c)
In a second design, the ions enter the region of uniform magnetic field of flux density 0.400 T shown in the diagram, moving at right angles to the field.
(i)
Calculate the radius of the path of the deuterons.
(1)
(ii)
Show that (radius of path of deuterons)/(radius of path of protons) = √2.
(2)
(iii)
Sketch, on the diagram, the paths of the deuterons and of the protons in the field.
(2)
(d)
The deuterons are turned into neutral atoms before they enter the reactor, which contains a very strong magnetic field.
time between the plates t = 0.200 / 2.77 × 106 = 7.22 × 10−8 s
✓ 1
Allow ECF from (a)(ii).
y = ½at² = 10.0 mm
✓ 1
Award [3] for CNA. Accept 9.9–10.1 mm.
Part (b)(ii)
y = qEL²/2mv² = qEL²/4Ek, which depends only on the charge and the kinetic energy
✓ 1
ALTERNATIVE: a proton has half the mass «double the acceleration» but spends 1/√2 of the time between the plates, so ½at² is unchanged.
protons and deuterons have the same charge and the same kinetic energy, so they are deflected equally: the deflector cannot separate them
✓ 1
MP2 only scores if MP1 scores.
Part (c)(i)
r = mv/qB = 3.34 × 10−27 × 2.77 × 106 / (1.60 × 10−19 × 0.400) = 0.144 m
✓ 1
Allow ECF from (a)(ii).
Part (c)(ii)
r = mv/qB = √(2mEk)/qB «since mv = √(2mEk)»
✓ 1
same q, Ek and B, so r ∝ √m and rD/rH = √(3.34/1.67) = √2
✓ 1
Accept a numerical ratio from calculated radii: 0.144/0.102 = 1.41.
Part (c)(iii)
both paths are circular arcs starting at P, curving upwards «anticlockwise», that return to the left boundary
✓ 1
Award [0] if the paths curve downwards.
deuteron path leaves the field about 29 cm above P and proton path about 20 cm above P «proton radius smaller by a factor of √2»
✓ 1
Accept any proton arc clearly smaller, with ratio of diameters about 0.7.
Part (d)
charged particles in a strong magnetic field experience a force perpendicular to their velocity and move in small circles «spirals» around the field lines
✓ 1
so ions would be turned back or trapped near the edge; neutral atoms feel no magnetic force and travel in straight lines into the centre of the plasma
✓ 1
OWTTE
Answers: (a)(ii) 2.77 × 106 m s−1 · (b)(i) 10.0 mm · (b)(ii) same deflection — cannot separate · (c)(i) 0.144 m (the remaining parts are explanations — see the table above)
139D-2-72Force on a moving chargePaper 2Hard14 marks
Auroras are produced when electrons from space travel down along the Earth's magnetic field lines into the upper atmosphere near the poles. Near the north magnetic pole the field is vertical, directed downwards, with a flux density of 5.0 × 10−5 T. An electron of kinetic energy 5.0 keV moves in this field with its velocity at 40° to the field. Figure 1 shows the path of the electron from the side; Figure 2 shows it seen from above. e = 1.60 × 10−19 C; me = 9.11 × 10−31 kg.
Diagrams NOT to scale
(a)
Show that the speed of the electron is about 4.2 × 107 m s−1.
(1)
(b)
Calculate the magnitude of the magnetic force on the electron.
(2)
(c)
Explain why the path of the electron is a helix, as shown in Figure 1.
(2)
(d)
(i)
Calculate the radius of the helix.
(2)
(ii)
Show that the time for one turn of the helix is about 7.2 × 10−7 s.
(1)
(iii)
Determine the distance that the electron moves along the field during one turn.
(2)
(e)
Draw, on Figure 2, an arrow to show the direction in which the electron moves round the circle.
(1)
(f)
Collisions with oxygen atoms in the atmosphere cause the emission of green light of wavelength 558 nm.
(i)
Calculate, in eV, the energy of one photon of this light.
(2)
(ii)
Estimate the greatest number of these photons that could result from the kinetic energy of one electron.
Award [0] for MP1 if sin 40° is omitted «3.4 × 10−16 N».
F = 1.60 × 10−19 × 4.19 × 107 × 5.0 × 10−5 × sin 40° = 2.2 × 10−16 N
✓ 1
Award [2] for CNA.
Part (c)
the velocity component along the field experiences no force «F = qvB sin 0° = 0», so it is constant
✓ 1
the component at right angles to the field gives a force of constant size always perpendicular to it, so that motion is circular; the two together give a helix
✓ 1
OWTTE.
Part (d)(i)
r = mev sin 40°/(eB) = 9.11 × 10−31 × 2.69 × 107/(1.60 × 10−19 × 5.0 × 10−5)
✓ 1
Allow ECF from (a).
r = 3.1 m
✓ 1
Award [1] max for 4.8 m «total speed used». Award [2] for CNA.
Allow ECF from (f)(i). The kinetic energy is unchanged by the magnetic field.
Answers: (b) 2.2 × 10−16 N · (d)(i) 3.1 m · (d)(iii) 23 m · (f)(i) 2.23 eV · (f)(ii) 2.2 × 103(the remaining parts are explanations — see the table above)
140D-2-73Charge in a uniform electric fieldPaper 2Hard12 marks
In accelerator mass spectrometry, carbon atoms from a small archaeological sample are counted directly. Negative carbon ions C⁻ leave an ion source with negligible kinetic energy and are accelerated along an evacuated tube from a point at 0 V towards a central terminal at a potential of +2.50 MV. In a thin gas stripper at the terminal, each ion loses four electrons and becomes a C³⁺ ion, with no change in speed. The C³⁺ ions are then accelerated along the rest of the tube back to 0 V.
Mass of a carbon-14 ion = 2.32 × 10−26 kg; e = 1.60 × 10−19 C.
Diagram NOT to scale
(a)
Explain why the ions gain kinetic energy both before and after they reach the terminal.
(2)
(b)
Show that the kinetic energy of a C³⁺ ion leaving the accelerator is 10.0 MeV.
(2)
(c)
Calculate the speed of a carbon-14 ion leaving the accelerator.
(2)
(d)
The C³⁺ ions then enter a uniform magnetic field at right angles to their velocity. Determine the magnetic flux density for which carbon-14 ions follow a circular path of radius 1.10 m.
(2)
(e)
Before the accelerator, the C⁻ ions travel to the right across the page through a magnetic field directed out of the page. State the direction of the magnetic force on the ions.
(1)
(f)
The half-life of carbon-14 is 5730 years.
(i)
The ratio (number of carbon-14 atoms)/(number of carbon-12 atoms) in the sample is one-eighth of the ratio in living wood. Determine the age of the sample.
(2)
(ii)
Suggest why counting carbon-14 atoms is better than measuring the activity of a sample of a few milligrams.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
before the terminal the ions are negative, so the force is towards the higher potential of the terminal «in the direction of motion»
✓ 1
after stripping the ions are positive, so the force is away from the terminal, towards 0 V «still in the direction of motion»; the field does positive work in both halves
✓ 1
OWTTE. Award [1] max for «the field accelerates them» without reference to the change of charge.
Part (b)
first half: Ek = qV = e × 2.50 MV = 2.50 MeV
✓ 1
second half: 3e × 2.50 MV = 7.50 MeV, so total = 10.0 MeV
✓ 1
Must see both stages.
Part (c)
Ek = 10.0 × 106 × 1.60 × 10−19 = 1.60 × 10−12 J
✓ 1
v = √(2 × 1.60 × 10−12/2.32 × 10−26) = 1.17 × 107 m s−1
✓ 1
Award [2] for CNA.
Part (d)
qvB = mv²/r with q = 3e, so B = mv/(3er)
✓ 1
Do not award MP1 for q = e.
B = 2.32 × 10−26 × 1.17 × 107/(4.80 × 10−19 × 1.10) = 0.516 T
✓ 1
Allow ECF from (c). Award [1] max for 1.5 T «q = e».
Part (e)
towards the top of the page
✓ 1
F = qv × B with q negative; a positive ion would be forced down the page.
Part (f)(i)
1/8 = (1/2)³, so three half-lives have passed
✓ 1
age = 3 × 5730 = 1.72 × 104 years
✓ 1
Accept 17 190 years. Award [2] for CNA.
Part (f)(ii)
the half-life is very long, so few nuclei decay that the decay rate «less than one decay per hour for a few milligrams» is hard to distinguish from background; counting atoms uses all the carbon-14 nuclei, not only those that decay
✓ 1
OWTTE. Accept: much smaller samples can be used.
Answers: (c) 1.17 × 107 m s−1 · (d) 0.516 T · (f)(i) 1.72 × 104 years (the remaining parts are explanations — see the table above)
141D-2-76Charge in a uniform electric fieldPaper 2Hard12 marks
A drift-tube linear accelerator accelerates protons for the production of medical isotopes. It consists of a row of hollow metal tubes in a vacuum (Figure 1). Alternate tubes are connected to the two terminals of an alternating supply of frequency 10.0 MHz. Protons enter tube 1 with a kinetic energy of 50 keV. Each time a proton crosses the 10 mm gap between two tubes, the potential difference across that gap is 100 kV and the field in the gap is uniform. Mass of a proton = 1.67 × 10−27 kg; e = 1.60 × 10−19 C.
Figure 1 — diagram NOT to scaleFigure 2 — axes for (e)
(a)
Explain why a proton moves at constant speed inside a tube.
(2)
(b)
Calculate the magnitude of the electric force on a proton in a gap.
(2)
(c)
Show that the speed of a proton in tube 2 is about 5.4 × 106 m s−1.
(1)
(d)
To be accelerated at every gap, a proton must spend half a period of the supply inside each tube.
(i)
Explain why the polarity of the supply must reverse while the protons are inside a tube.
(2)
(ii)
Calculate the length of tube 2.
(1)
(iii)
Determine the length of tube 13.
(2)
(e)
Sketch, on Figure 2, a graph to show how the length of a tube varies with the tube number n for tubes 1 to 13.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
there is no electric field inside a hollow conductor, so there is no «electric» force on the proton
✓ 1
by Newton's first law «with no resultant force» the velocity of the proton is constant
✓ 1
OWTTE.
Part (b)
E = V/d = 100 × 103/0.010 = 1.0 × 107 V m−1
✓ 1
F = eE = 1.6 × 10−12 N
✓ 1
Award [2] for CNA.
Part (c)
Ek = 150 keV, so v = √(2 × 150 × 103 × 1.60 × 10−19/1.67 × 10−27) = 5.36 × 106 «m s−1»
✓ 1
Must see 50 + 100 = 150 keV. Must see full substitution OR answer to 3 s.f.
Part (d)(i)
a proton is accelerated only if the field in the gap is in its direction of motion «the tube it leaves is at the higher potential»
✓ 1
neighbouring gaps have opposite field directions for one polarity, so the polarity must have reversed by the time the proton reaches the next gap; inside a tube the proton is shielded, so the reversal does not affect it
✓ 1
OWTTE.
Part (d)(ii)
L = v × T/2 = v/2f = 5.36 × 106/(2 × 10.0 × 106) = 0.27 m
✓ 1
Allow ECF from (c).
Part (d)(iii)
tube 13: 12 gaps crossed, Ek = 50 + 12 × 100 = 1250 keV, so v = 1.55 × 107 m s−1
✓ 1
Do not award MP1 for 13 gaps «1350 keV».
L = 1.55 × 107/(2 × 10.0 × 106) = 0.77 m
✓ 1
Accept use of L ∝ √Ek: 0.27 × √(1250/150). Award [2] for CNA.
Part (e)
increasing curve starting at about 0.15 m for n = 1 and passing through about 0.77 m at n = 13
✓ 1
Allow ECF from (d).
gradient decreasing with n «L ∝ √Ek, not a straight line»
✓ 1
Do not award MP2 for a straight line or a curve that levels off completely.
Answers: (b) 1.6 × 10−12 N · (d)(ii) 0.27 m · (d)(iii) 0.77 m (the remaining parts are explanations — see the table above)
142IBP2S-032-01-15The velocity selectorPaper 2Hard8 marks
A particle detector on a space station identifies charged particles. The particles first pass through a speed selector, which has a uniform electric field of 6.0 ×104 V m−1 perpendicular to a uniform magnetic field of 0.050 T. The particles that emerge undeflected then enter a region containing only a uniform magnetic field of 0.050 T, perpendicular to their velocity. One type of particle follows a circular arc of radius 0.25 m.
Diagram not to scale. The magnetic field fills the shaded region.
(a)
Explain why only particles with speed E/B pass through the selector undeflected.
(2)
(b)
Calculate the speed of the undeflected particles.
(1)
(c)
Determine the charge-to-mass ratio of the particle.
(2)
(d)
Suggest the identity of the particle.
(1)
(e)
An alpha particle (mass 6.64 ×10−27 kg, charge 3.20 ×10−19 C) also passes through the selector undeflected. Determine the radius of its path in the 0.050 T field.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The electric force qE and the magnetic force qvB act in opposite directions
✓ 1
They balance only when qE = qvB, i.e. v = E/B. Faster or slower particles feel a net force and are deflected
✓ 1
Part (b)
v = 6.0 ×104 / 0.050 = 1.2 ×106 m s−1
✓ 1
Part (c)
q/m = v/(Br)
✓ 1
= 1.2 ×106 / (0.050 × 0.25) = 9.6 ×107 C kg−1
✓ 1
ECF from (b).
—
Part (d)
Proton, since e/mp = 1.60 ×10−19/1.67 ×10−27 ≈ 9.6 ×107 C kg−1
143IBP2S-032-00-15Motion in an electric fieldPaper 2Hard9 marks
In a continuous inkjet printer, a charged ink droplet of mass 1.2 × 10−10 kg and charge −1.5 × 10−13 C enters the region between two deflecting plates. It moves parallel to the plates at 18 m s−1. The plates are 1.2 cm long and produce a uniform field of 8.0 × 105 V m−1 perpendicular to the droplet's initial velocity. After leaving the plates, the droplet travels a further 1.5 cm, measured parallel to its original direction, before reaching the paper. There is no field beyond the plates. Ignore gravity and air resistance.
Diagram not to scale. Plates, droplet path and paper seen from the side.
(a)
Calculate the acceleration of the droplet between the plates.
(2)
(b)
Outline why the path of the droplet between the plates is parabolic.
(2)
(c)
Determine the deflection of the droplet as it leaves the plates.
(2)
(d)
Determine the total deflection of the droplet when it reaches the paper.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
F = qE = 1.5 × 10−13 × 8.0 × 105 = 1.2 × 10−7 N
✓ 1
a = F/m = 1.2 × 10−7/1.2 × 10−10 = 1.0 × 103 m s−2
✓ 1
Part (b)
The force is constant and perpendicular to the initial velocity, so the velocity component parallel to the plates stays constant .
✓ 1
The perpendicular displacement increases as t², while the parallel displacement increases as t, so the path is parabolic .
✓ 1
Part (c)
Time between plates: t = 0.012/18 = 6.7 × 10−4 s
✓ 1
y = ½at² = 0.5 × 1.0 × 103 × (6.67 × 10−4)² = 2.2 × 10−4 m . ECF from (a).
✓ 1
Part (d)
Perpendicular velocity at exit: vy = at = 1.0 × 103 × 6.67 × 10−4 = 0.67 m s−1
✓ 1
Time to reach the paper = 0.015/18 = 8.3 × 10−4 s, giving an extra deflection of 0.67 × 8.3 × 10−4 = 5.6 × 10−4 m
✓ 1
Total deflection = 2.2 × 10−4 + 5.6 × 10−4 = 7.8 × 10−4 m, about 0.78 mm . ECF from (c).
✓ 1
Common error: Beyond the plates there is no force, so the droplet moves in a straight line. Do not keep applying the acceleration over the extra 1.5 cm.
144IBP2S-032-03-15Charge-to-mass ratioPaper 2Hard7 marks
A sports anti-doping laboratory uses a mass spectrometer to compare the magnesium isotopes 24Mg and 26Mg. Singly charged ions first pass through a velocity selector. It has an electric field of 4.0 × 104 V m−1 perpendicular to a magnetic field of 0.080 T. Both fields are perpendicular to the ion beam and arranged so that the electric and magnetic forces on an ion act in opposite directions. The ions that emerge undeflected then enter a separate uniform magnetic field of 0.50 T at right angles to their velocity. They travel through a semicircle before striking a flat detector. 1 u = 1.66 × 10−27 kg; e = 1.60 × 10−19 C. Take the ion masses as 24 u and 26 u.
(a)
Explain why only ions with one particular speed pass through the velocity selector undeflected.
(2)
(b)
Calculate the speed of the ions leaving the selector.
(1)
(c)
Show that the radius of the path of the 24Mg ions in the 0.50 T field is about 0.25 m.
(2)
(d)
Determine the distance between the points where the two isotopes strike the detector.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The electric force qE does not depend on speed, whereas the opposing magnetic force qvB is proportional to speed . The resultant force is zero only when qE = qvB, i.e. v = E/B, so slower or faster ions are deflected .
✓ 2
Part (b)
v = E/B = 4.0 × 104/0.080 = 5.0 × 105 m s−1 .
✓ 1
Part (c)
Use qvB = mv2/r to give r = mv/(qB) .
✓ 1
r = (24 × 1.66 × 10−27 × 5.0 × 105)/(1.60 × 10−19 × 0.50) = 0.249 m ≈ 0.25 m .
✓ 1
Part (d)
r for 26Mg = (26/24) × 0.249 = 0.270 m . Each ion lands one diameter from the entry point, so separation = 2(r26 − r24) = 2 × 0.0208 ≈ 0.042 m. Accept 0.041–0.042 m, and allow error carried forward from earlier parts .
✓ 2
Exam tip: After a semicircle, an ion lands one diameter from its entry point. The separation on the detector is therefore twice the difference in radii. Keep unrounded radii until the final step, because the difference between them is small.
145IBP2S-032-03-16Parallel current-carrying wiresPaper 2Hard8 marks
In a science-museum demonstration, a long, thin horizontal wire floats above a long, fixed horizontal wire that runs parallel to it. The upper wire is held only by the magnetic force, with light vertical guides preventing sideways motion. Both wires carry the same current I. The upper wire has a mass per unit length of 2.0 × 10−3 kg m−1 and floats with its centre 4.0 mm above the centre of the lower wire. μ0 = 4π × 10−7 T m A−1; g = 9.8 m s−2
Diagram not to scale. Cross-section of the two long wires.
(a)
State how the directions of the two currents must be related.
(1)
(b)
Determine the current I.
(3)
(c)
The upper wire is pushed slightly downwards and released. Explain whether its equilibrium is stable for vertical displacements.
(2)
(d)
The current in both wires is reduced to 15 A. Determine the new equilibrium separation of the wires.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The currents must be in opposite directions, so that the force on the upper wire is repulsive (upwards) .
✓ 1
Part (b)
Magnetic force per unit length = weight per unit length: μ0I²/(2πh) = (m/L)g . I² = 2π × 0.0040 × 2.0 × 10−3 × 9.8/(4π × 10−7) = 392 A² . I = 19.8 A ≈ 20 A (either value accepted) .
✓ 3
Part (c)
When h decreases, the repulsive force (∝ 1/h) increases and becomes greater than the weight . The resultant force is upwards, returning the wire towards its equilibrium position (the reverse happens for an upward displacement), so the equilibrium is stable .
✓ 2
Part (d)
h = μ0I²/(2π(m/L)g) . h = (4π × 10−7 × 15²)/(2π × 2.0 × 10−3 × 9.8) = 2.3 × 10−3 m (2.3 mm) .
✓ 2
Exam tip: Equate forces per unit length on both sides. The length of wire then cancels and is not needed.
146IBP2S-032-04-13Charge in a uniform electric fieldPaper 2Hard8 marks
An engineer calibrates the accelerating voltage Va of an electron gun in a test rig. Electrons are accelerated from rest through Va. They then enter, parallel to the plates, a uniform field between two deflecting plates of length L = 5.0 cm and separation d = 4.0 cm. The engineer varies the deflecting p.d. Vp and measures the vertical deflection y of the beam at the end of the plates.
147IBP2S-032-04-16Motion in an electric fieldPaper 2Hard9 marks
An electron detector on a space-weather satellite uses a horizontal metal plate with a small entrance hole. Above the plate there is a uniform electric field of strength 500 V m−1, directed vertically upward. An electron passes up through the hole with speed 3.0×106 m s−1 at 30° above the plate. It then returns to the plate, where it strikes a detector strip. Mass of electron = 9.11×10−31 kg. Elementary charge = 1.60×10−19 C.
Diagram not to scale. Electric field points vertically upward.
(a)
Explain why the gravitational force on the electron can be neglected.
(2)
(b)
Calculate the magnitude of the acceleration of the electron.
(2)
(c)
Determine the maximum height of the electron above the plate.
(2)
(d)
Determine the distance from the entrance hole at which the electron strikes the detector strip.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Weight mg ≈ 9.11×10−31 × 9.8 ≈ 8.9×10−30 N, while the electric force eE = 1.60×10−19 × 500 = 8.0×10−17 N
✓ 1
The weight is about 10−13 of the electric force, so it has no measurable effect on the motion
✓ 1
Part (b)
a = eE/m
✓ 1
a = 8.0×10−17 / 9.11×10−31 = 8.8×1013 m s−2. It is directed downward because the electron's charge is negative, so the force is opposite to the field
✓ 1
Part (c)
Initial vertical velocity component = 3.0×106 sin 30° = 1.5×106 m s−1
✓ 1
h = vy2/(2a) = (1.5×106)2 / (2 × 8.8×1013) = 1.3×10−2 m (ECF from (b))
✓ 1
Part (d)
Time to return to the plate t = 2vy/a = 2 × 1.5×106 / 8.8×1013 = 3.4×10−8 s (ECF from (b))
✓ 1
Horizontal velocity component = 3.0×106 cos 30° = 2.6×106 m s−1, which is constant because there is no horizontal force
✓ 1
Distance = 2.6×106 × 3.4×10−8 = 8.9×10−2 m. Accept answers from 8.8×10−2 m to 8.9×10−2 m, which arise from rounding the intermediate values
✓ 1
Exam tip: This is projectile motion with eE/m in place of g. Resolve the initial velocity and treat each direction separately. Keep unrounded values in your calculator until the final step.
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