D.3 Motion in electromagnetic fields: IB Physics HL exam-style questions
Motion in electromagnetic fields applies the field ideas of D.2 to moving charges. In a uniform electric field a charged particle accelerates along the field lines; in a uniform magnetic field the force qvB sin θ is perpendicular to the velocity, so the speed stays constant and the path is a circle.
Questions use these ideas in velocity selectors, mass spectrometers and cyclotrons, and to find the specific charge of a particle. For currents, F = BIL sin θ gives the force on a wire, and two parallel currents attract or repel with a force per unit length that depends on both currents.
52 questions
259 marks
Paper 1A: 31
Paper 1B: 8
Paper 2: 13
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25 practice questions on D.3 Motion in electromagnetic fields
1D-1A-05
Charged particles in magnetic fields·D.3 Motion in electromagnetic fields
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
An electron accelerated from rest through a potential difference V enters a uniform magnetic field at right angles and follows a circular path of radius r. The potential difference is increased to 4V.
What is the new radius?
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Notes
Step 1eV = ½mv², so v ∝ √V: quadrupling V doubles the speed.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2In the field evB = mv²/r, so r = mv/(eB) ∝ v.
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Step 3The radius doubles: 2r.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThe radius depends on the speed, which has changed.
BThis treats r ∝ V1/4.
CCorrect: r ∝ v ∝ √V, and √4 = 2.
DThis assumes r ∝ V, forgetting the square root in v ∝ √V.
Syllabus understandingD.3 — the motion of a charged particle in a uniform magnetic field; the magnitude and direction of the force on a charge moving in a magnetic field as given by F = qvB sin θ; D.2 (HL) — the work done in moving a charge q in an electric field as given by W = qΔVeCommand term: Determine
2D-1A-06
Velocity selector·D.3 Motion in electromagnetic fields
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
In a velocity selector the uniform electric field is directed towards the bottom of the page and the uniform magnetic field is directed into the page. Positive ions moving to the right at speed v pass through undeflected.
Ions of the same type, with the same speed v, are now sent into the selector from the opposite end, so that they move to the left. What happens to these ions as they enter the fields?
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Step 1For ions moving to the right, the electric force (along the field) is towards the bottom of the page, so the magnetic force qv × B must be towards the top of the page to balance it.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Reversing the velocity reverses the magnetic force, v × B, so it now acts towards the bottom of the page; the electric force qE does not depend on the velocity and is unchanged.
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Step 3Both forces now act towards the bottom of the page, giving a resultant force 2qE: the ions are deflected towards the bottom of the page.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis assumes the selector works for any ion with speed v = E/B, whatever its direction. The balance needs the magnetic force to oppose the electric force, and that depends on the direction of the velocity.
BThis reverses the electric force as well as the magnetic force, so the resultant is still towards the top. The electric force on a positive ion is along the field whatever the ion's velocity.
CCorrect: the magnetic force reverses with the velocity and now adds to the unchanged electric force, both towards the bottom of the page.
DThis puts the force along the magnetic field. The magnetic force is perpendicular to both the velocity and the field, so it lies in the plane of the page.
Syllabus understandingD.3 — the motion of a charged particle in perpendicularly orientated uniform electric and magnetic fields; the magnitude and direction of the force on a charge moving in a magnetic field as given by F = qvB sin θ Command term: Deduce
3D-1A-07
Force between parallel wires·D.3 Motion in electromagnetic fields
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
Three long straight parallel wires X, Y and Z are perpendicular to the page and lie in the same plane. X and Y are a distance d apart, and Y and Z are a distance 2d apart. X and Y each carry a current I into the page; Z carries a current 2I out of the page, as shown.
What is the magnitude of the resultant magnetic force per unit length on Y?
Three long parallel wires perpendicular to the page (⊗ current into the page, ⊙ current out of the page).Show mark scheme
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Step 1X and Y carry currents in the same direction, so X attracts Y (towards X): F/L = μ0I·I/(2πd) = μ0I²/(2πd).
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Y and Z carry currents in opposite directions, so Z repels Y — away from Z, which is also towards X: F/L = μ0(2I)(I)/(2π × 2d) = μ0I²/(2πd).
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Step 3Both forces act towards X, so they add: μ0I²/(2πd) + μ0I²/(2πd) = μ0I²/(πd).
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis treats the force from Z as an attraction (towards Z), so that it cancels the force from X. Currents in opposite directions repel.
BThis includes only the force from the nearer wire X, as if the more distant wire Z had a negligible effect. The larger current in Z makes up for its greater distance.
CThis uses I instead of 2I for the current in Z: μ0I²/(2πd) + μ0I²/(4πd) = 3μ0I²/(4πd).
DCorrect: both forces are μ0I²/(2πd) and both act towards X.
Syllabus understandingD.3 — the force per unit length between parallel wires as given by F/L = μ0I1I2/(2πr); the force is attractive when the currents flow in the same direction Command term: Determine
4D-1A-08
Charged particles in magnetic fields·D.3 Motion in electromagnetic fields
Paper 1AMedium1 mark
Multiple choice · 1 mark4 steps to full marksDetermine
A proton and an alpha particle move in circular paths in the same uniform magnetic field.
What is the ratio (period of the alpha particle) / (period of the proton)?
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Notes
Step 1Radius r = mv/(qB) and period T = 2πr/v.
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All 4 steps must be completed — there is no mark for a part-answer.
Step 2Combining: T = 2πm/(qB) — independent of the speed.
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Step 3Alpha: mass 4mp, charge 2e, so m/q = 4/2 = 2.
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Step 4Tα/Tp = (4/2)/(1/1) = 2.
✓ 1
Answer C
Answer: C · 4 stages of work, one mark
Every option, and why
AThis inverts the ratio.
BThe periods would be equal only if the charge-to-mass ratios were equal; the alpha has half the proton's q/m.
CCorrect: T ∝ m/q, and the alpha particle has twice the mass per unit charge.
DThis uses the mass ratio alone and ignores the doubled charge.
Syllabus understandingD.3 — the motion of a charged particle in a uniform magnetic field; the magnitude and direction of the force on a charge moving in a magnetic field as given by F = qvB sin θ; A.2 — circular motion, v = 2πr/TCommand term: Determine
5D-1A-11
Force on a current-carrying conductor·D.3 Motion in electromagnetic fields
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A horizontal metal rod of mass m and length L hangs from two light, flexible, conducting threads in a uniform vertical magnetic field of flux density B. When there is a current I in the rod, the rod moves sideways and comes to rest with the threads at an angle θ to the vertical.
Which expression gives I?
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Step 1The rod is perpendicular to the vertical field, so the magnetic force BIL is horizontal (perpendicular to both the rod and the field).
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The rod is in equilibrium under its weight mg, the tension in the threads and the horizontal force BIL. Resolving: T cos θ = mg and T sin θ = BIL.
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Step 3Dividing: tan θ = BIL/mg, so I = mg tan θ/(BL).
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis sets BIL equal to mg sin θ, the component of the weight perpendicular to the threads; but only the component BIL cos θ of the horizontal force is perpendicular to the threads.
BThis resolves the weight with the wrong trigonometric function: mg cos θ is the component of the weight along the threads.
CThis inverts the ratio of the components (tan θ = mg/BIL), which would make the deflection smaller as the current increases.
DCorrect: tan θ = BIL/mg.
Syllabus understandingD.3 — the magnitude and direction of the force on a current-carrying conductor in a magnetic field as given by F = BIL sin θ (with the equilibrium of forces, A.2) Command term: Deduce
6D-1A-20
Charged particles in electric fields·D.3 Motion in electromagnetic fields
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
An electron with kinetic energy Ek enters the uniform electric field between two parallel plates. It enters midway between the plates, moving parallel to them. The potential difference between the plates is V. The electron is deflected and strikes the positive plate.
What is the kinetic energy of the electron when it strikes the plate?
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Notes
Step 1The electric force is perpendicular to the plates. As the electron is deflected towards the positive plate it moves along the line of the force, so the field does work on it.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Midway between the plates the potential is halfway between the potentials of the plates, so the electron moves through a potential difference of V/2.
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Step 3Gain in kinetic energy = e × V/2, so the final kinetic energy is Ek + eV/2.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: the electron gains eV/2 on top of the kinetic energy it had on entry.
BThis treats the electric force like a magnetic force, always perpendicular to the velocity. The force is perpendicular only to the initial velocity; once the electron moves towards the plate the field does work on it.
CThis is the gain in kinetic energy only; it leaves out the kinetic energy Ek that the electron had when it entered.
DThis uses the whole potential difference V. The electron starts midway between the plates, so it moves through only V/2.
Syllabus understandingD.3 — the motion of a charged particle in a uniform electric field; D.2 (HL) — the work done in moving a charge q in an electric field as given by W = qΔVeCommand term: Deduce
7D-1A-24
Forces between parallel currents·D.3 Motion in electromagnetic fields
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
Two long parallel wires X and Y lie on a smooth horizontal surface and carry currents in the same direction. The current in Y is three times the current in X, and the mass per unit length of Y is half that of X. The wires are released from rest.
What is (initial acceleration of X)/(initial acceleration of Y)?
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Step 1The forces that the wires exert on each other form a Newton's third law pair: F/L = μ0IXIY/(2πr) for both, so the force per unit length is the same on X and on Y.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Acceleration = (force per unit length)/(mass per unit length).
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Step 3Ratio = (mass per unit length of Y)/(mass per unit length of X) = 1/2.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis takes the force on Y to be three times the force on X because Y carries three times the current: (1/1)/(3/½) = 1/6. The forces are equal by Newton's third law.
BCorrect: equal forces, so the accelerations are inversely proportional to the masses per unit length.
CThis takes the force on X to be three times the force on Y because X lies in the field of the larger current: (3/1)/(1/½) = 3/2. The forces are equal by Newton's third law.
DThis inverts the ratio: the lighter wire Y has the larger acceleration.
Syllabus understandingD.3 — the force per unit length between parallel wires as given by F/L = μ0I1I2/(2πr); Newton's third law (A.2) Command term: Deduce
8D-1A-27
Direction of the magnetic force·D.3 Motion in electromagnetic fields
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A rectangular loop of wire PQRS carries a current and lies in the plane of the page in a uniform magnetic field. The field is also in the plane of the page and is parallel to sides PQ and SR, as shown.
Which statement describes the effect of the magnetic forces on the loop?
A current-carrying rectangular loop in a uniform magnetic field in the plane of the page.Show mark scheme
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Step 1PQ and SR are parallel (or antiparallel) to the field, so F = BIL sin 0° = 0 on them.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2QR and SP are perpendicular to the field and carry currents in opposite directions, so they experience equal forces, one into the page and one out of the page: the resultant force is zero.
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Step 3These two forces act along different lines, a distance PQ apart, so they form a couple: the loop turns about an axis through its centre, in the plane of the page and perpendicular to the field (parallel to QR).
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: zero resultant force, but the equal and opposite forces on QR and SP form a couple.
BEqual and opposite forces give a zero resultant force, but they do not act along the same line, so they do produce a turning effect.
CRotation about an axis parallel to the field would need forces on PQ and SR, which are parallel to the field and experience no force.
DThe force on QR and the force on SP are in opposite directions (into and out of the page) because the currents in these sides are opposite, so they cancel.
Syllabus understandingD.3 — the magnitude and direction of the force on a current-carrying conductor in a magnetic field as given by F = BIL sin θCommand term: Deduce
9D-1A-28
Sign of a charge from its path·D.3 Motion in electromagnetic fields
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A charged particle moves in a circle in a uniform magnetic field directed out of the page. At the instant shown it is moving towards the top of the page, and the force on it is towards the centre O.
Which row gives the sign of the charge and the effect on the radius of the path if the speed of the particle is doubled?
The particle at the right-hand side of its circular path is moving towards the top of the page.
Sign of chargeRadius when speed is doubled
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Step 1For a positive charge moving up in a field out of the page, F = qv × B gives (up) × (out) = to the right — away from O.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The actual force is to the left (towards O), so the charge is negative.
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Step 3r = mv/qB ∝ v: doubling the speed doubles the radius (the period 2πm/qB stays the same).
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: the force direction is opposite to that for a positive charge, and r ∝ v.
BA positive charge here would be pushed outwards (to the right), not towards O.
CThe radius depends on the speed: r = mv/qB.
DWrong sign and wrong dependence.
Syllabus understandingD.3 — the magnitude and direction of the force on a charge moving in a magnetic field as given by F = qvB sin θ; the motion of a charged particle in a uniform magnetic field Command term: Deduce
10D-1A-50
Force on a moving charge·D.3 Motion in electromagnetic fields
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksCalculate
In a cloud-chamber demonstration an alpha particle moves at 1.5 × 107 m s−1 through a uniform magnetic field of flux density 0.050 T. Its velocity makes an angle of 40° with the field lines.
What is the magnitude of the magnetic force on the alpha particle?
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Notes
Step 1F = qvB sin θ, where θ is the angle between the velocity and the field.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 3F = 3.20 × 10−19 × 1.5 × 107 × 0.050 × sin 40° = 1.5 × 10−13 N.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis uses q = e instead of 2e: 1.60 × 10−19 × 1.5 × 107 × 0.050 × sin 40° = 7.7 × 10−14 N.
BCorrect: 2e × 1.5 × 107 × 0.050 × sin 40° = 1.54 × 10−13 N.
CThis uses cos 40° instead of sin 40°: 3.20 × 10−19 × 1.5 × 107 × 0.050 × 0.766 = 1.8 × 10−13 N.
DThis omits the angle factor, as if the particle moved perpendicular to the field: 3.20 × 10−19 × 1.5 × 107 × 0.050 = 2.4 × 10−13 N.
Syllabus understandingD.3 — the magnitude and direction of the force on a charge moving in a magnetic field as given by F = qvB sin θCommand term: Calculate
11D-1A-51
Charged particles in electric fields·D.3 Motion in electromagnetic fields
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A proton and an alpha particle are each accelerated from rest through the same potential difference. Each then enters, midway between them and parallel to them, the uniform electric field between the same pair of horizontal deflecting plates.
How do the vertical deflection on leaving the plates and the time spent between the plates for the alpha particle compare with those for the proton?
Vertical deflection of the alpha particleTime between the plates for the alpha particle
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Step 1Acceleration: qV = ½mv², so v = √(2qV/m). The alpha particle has half the proton's q/m, so it enters at 1/√2 of the proton's speed and spends √2 times as long between plates of length L.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Between the plates the vertical acceleration is qE/m (half as large for the alpha particle) and the deflection is y = ½(qE/m)(L/v)².
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Step 3Substituting v² = 2qV/m: y = EL²/(4V) — independent of q and m, so the deflections are equal.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis halves the deflection because the acceleration is halved, but forgets that the time between the plates is √2 times longer, so t² doubles.
BCorrect: half the acceleration for twice the value of t² gives the same deflection; the lower entry speed gives √2 times the time.
CThe deflection is right, but this takes the speed as proportional to q/m instead of √(q/m), giving a time ratio of 2.
DThis doubles the deflection because t² doubles, but forgets that the alpha particle's acceleration is only half the proton's.
Syllabus understandingD.3 — the motion of a charged particle in a uniform electric field; D.2 (HL) — the work done in moving a charge q in an electric field as given by W = qΔVe; A.1 — the behaviour of projectiles Command term: Deduce
12D-1A-52
Specific charge from a circular path·D.3 Motion in electromagnetic fields
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
Positive ions pass undeflected through a velocity selector in which the electric field strength is 3.0 × 104 V m−1 and the magnetic flux density is 0.060 T. They then enter a region where the only field is a uniform magnetic field of flux density 0.20 T, perpendicular to their velocity. They travel through a semicircle and strike a detector 5.2 cm from the point where they entered this region.
What is the charge-to-mass ratio q/m of the ions?
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Step 1Selector: qE = qvB, so v = E/B = 3.0 × 104/0.060 = 5.0 × 105 m s−1.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The detector is one diameter from the entry point: r = 5.2/2 = 2.6 cm.
AThis puts E in place of the speed: 3.0 × 104/(0.20 × 0.026) = 5.8 × 106 C kg−1.
BThis uses the whole distance 5.2 cm as the radius instead of the diameter: 5.0 × 105/(0.20 × 0.052) = 4.8 × 107 C kg−1.
CCorrect: v = 5.0 × 105 m s−1 and r = 0.026 m give 9.6 × 107 C kg−1.
DThis uses the selector field 0.060 T instead of 0.20 T in the deflection region: 5.0 × 105/(0.060 × 0.026) = 3.2 × 108 C kg−1.
Syllabus understandingD.3 — the motion of a charged particle in perpendicularly orientated uniform electric and magnetic fields; the determination of the charge-to-mass ratio from the path in a uniform magnetic field (r = mv/qB) Command term: Determine
13D-1A-53
Force between parallel wires·D.3 Motion in electromagnetic fields
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksCalculate
Two long, straight, parallel busbars in an electrical substation are 0.25 m apart. During a fault each carries a current of 2.0 kA, in opposite directions. Each busbar is held in place by supports placed every 1.5 m along its length.
What is the magnitude of the magnetic force on a 1.5 m section of one busbar?
All 3 steps must be completed — there is no mark for a part-answer.
Step 2F = 3.2 N m−1 × 1.5 m = 4.8 N.
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Step 3The currents are in opposite directions, so the force is repulsive: the supports must hold the busbars apart.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis divides the force per unit length by the length instead of multiplying: 3.2/1.5.
BThis uses μ0I1I2/(4πr), half the correct force per unit length: 1.6 × 1.5.
CThis is the force per unit length (in N m−1); it has not been multiplied by the 1.5 m length.
DCorrect: 3.2 N m−1 × 1.5 m = 4.8 N.
Syllabus understandingD.3 — the force per unit length between parallel wires as given by F/L = μ0I1I2/(2πr) Command term: Calculate
14D-1A-54
Direction of the magnetic force·D.3 Motion in electromagnetic fields
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A long straight wire in the plane of the page carries a current towards the top of the page. An electron to the right of the wire moves parallel to the wire, also towards the top of the page, as shown.
What is the direction of the magnetic force on the electron?
The wire and the electron's velocity both lie in the plane of the page.Show mark scheme
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Step 1Right-hand grip rule: with the thumb along the current (up the page), the fingers curl so that the field to the right of the wire points into the page.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2For a positive charge moving up the page in a field into the page, the left-hand rule gives a force to the left, towards the wire (like parallel currents attracting).
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Step 3The electron is negative, so the force on it is reversed: away from the wire.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: the field at the electron is into the page, and the force on a negative charge is opposite to that on a positive charge moving the same way, so it is away from the wire.
BThis treats the electron as a positive charge, or as a current in the same direction as the current in the wire, which would be attracted. The electron's motion up the page is equivalent to a conventional current down the page.
CThis gives the direction of the magnetic field of the wire at the electron, not the force. The force is perpendicular to the field.
DThe force is perpendicular both to the field (into the page) and to the velocity, so it lies in the plane of the page; it cannot be out of the page.
Syllabus understandingD.3 — the magnitude and direction of the force on a charge moving in a magnetic field (F = qvB sin θ); the direction of the magnetic field determined from the current direction in a straight wire Command term: Deduce
15D-1A-55
Charged particles in magnetic fields·D.3 Motion in electromagnetic fields
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A proton (mass m, charge +e), a deuteron (mass 2m, charge +e) and an alpha particle (mass 4m, charge +2e) all have the same kinetic energy. Each moves in a circle in the same uniform magnetic field, perpendicular to its velocity.
Which gives the correct relationship between the radii rp, rd and rα of their paths?
Show mark scheme
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Notes
Step 1r = mv/(qB) and the momentum is mv = √(2mEk), so r = √(2mEk)/(qB) ∝ √m/q at the same Ek and B.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Deuteron: √2/1 = √2 times the proton value.
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Step 3Alpha particle: √4/2 = 1, the same as the proton. So rp = rα < rd.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: r ∝ √m/q gives 1 : √2 : 1 for the proton, deuteron and alpha particle.
BThis uses r ∝ √m but ignores the double charge of the alpha particle: 1 : √2 : 2.
CThis uses r ∝ m/q, which holds for equal speeds, not equal kinetic energies: 1 : 2 : 2.
DThis inverts the mass dependence (r ∝ 1/(√mq)): 1 : 1/√2 : 1/4.
Syllabus understandingD.3 — the motion of a charged particle in a uniform magnetic field (r = mv/qB); the kinetic energy of a charged particle stays constant in a magnetic field Command term: Deduce
16D-1A-66
Charged particles in electric fields·D.3 Motion in electromagnetic fields
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
An electron (mass m, charge −e) enters a region of uniform electric field of strength E. On entry its speed is v and it is moving in the same direction as the electric field lines.
What distance does the electron travel in the field before it comes momentarily to rest?
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Step 1The force on the electron, eE, is opposite to the field and so opposite to its velocity: it decelerates with a = eE/m.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 20 = v² − 2as gives s = v²/(2a) = mv²/(2eE).
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Step 3Alternatively, the work done against the electric force equals the initial kinetic energy: eEs = ½mv².
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis is v/a, the time taken to stop, not the distance.
BThis omits the ½ in the kinetic energy (or the 2 in v² = 2as).
CThis inverts the specific charge, using a = mE/e.
DCorrect: eEs = ½mv².
Syllabus understandingD.3 — the motion of a charged particle in a uniform electric field; D.2 — the electric field strength as given by E = F/q; A.1 — the equations of motion for uniform acceleration Command term: Determine
17D-1A-67
Charged particles in electric fields·D.3 Motion in electromagnetic fields
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A proton is released from rest next to the positive plate of a pair of parallel plates in a vacuum. The potential difference between the plates is V and their separation is d. The proton reaches the negative plate after a time t.
The separation of the plates is doubled and the potential difference is unchanged. What is the time now taken by the proton to cross from one plate to the other?
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Notes
Step 1E = V/d, so doubling d halves the field strength and therefore halves the acceleration a = eV/(md).
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2From rest, d = ½at², so t = √(2d/a) = d√(2m/(eV)) ∝ d.
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Step 3Doubling d doubles the time: 2t. (The final speed is unchanged because eV is unchanged; the average speed is the same over twice the distance.)
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis uses the unchanged final speed but ignores the doubled distance.
BThis doubles the distance but keeps the acceleration unchanged, forgetting that E = V/d halves.
CCorrect: the time is proportional to the separation when the potential difference is fixed.
DThis finds that t² becomes four times larger but then forgets to take the square root.
Syllabus understandingD.3 — the motion of a charged particle in a uniform electric field; D.2 — the uniform electric field strength between parallel plates E = V/d; A.1 — the equations of motion for uniform acceleration Command term: Deduce
18D-1A-68
Charged particles in magnetic fields·D.3 Motion in electromagnetic fields
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A proton enters a region of uniform magnetic field, moving at right angles to the field and to the boundary of the region, as shown. The region has parallel boundaries a distance w apart. Inside the region the proton moves along a circular arc of radius 2w.
What is the angle θ through which the proton has been deflected when it leaves the region?
Not to scale. The magnetic field (×) is directed into the page.Show mark scheme
Marking point
Mark
Notes
Step 1On entry the velocity is perpendicular to the boundary, and the magnetic force is perpendicular to the velocity, so the centre of the circular path lies on the entry boundary.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The proton leaves when it has moved a horizontal distance w, after turning through an angle θ about the centre with sin θ = w/r = w/(2w) = 0.5.
—
Step 3The velocity turns through the same angle as the radius: θ = 30°.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis uses tan θ = w/r = 0.5, giving 27°.
BCorrect: sin θ = w/(2w), so θ = 30°.
CThis uses cos θ = w/r = 0.5, giving 60°.
DThis assumes the proton turns through a quarter of a circle before leaving, which needs a radius no larger than w.
Syllabus understandingD.3 — the motion of a charged particle in a uniform magnetic field; the magnitude and direction of the force on a charge moving in a magnetic field F = qvB sin θCommand term: Determine
19D-1A-97
Paths in uniform fields·D.3 Motion in electromagnetic fields
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksIdentify
A proton travelling at constant velocity enters, in turn, two separate regions of space.
In region X there is only a uniform electric field, directed at right angles to the velocity with which the proton enters. In region Y there is only a uniform magnetic field, directed parallel to the velocity with which the proton enters.
Which row describes the path of the proton in each region?
Path in region XPath in region Y
Show mark scheme
Marking point
Mark
Notes
Step 1Region X: the electric force qE has constant magnitude and a constant direction, at right angles to the initial velocity, like the weight of a horizontally launched projectile: the path is a parabola.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Region Y: the angle between the velocity and the field is 0, so F = qvB sin 0 = 0.
—
Step 3With no force, the proton continues in a straight line at constant speed.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: a constant force at right angles to the initial velocity gives a parabola; a velocity parallel to the magnetic field gives no force.
BThis assumes that a magnetic field always deflects a moving charge. The force depends on sin θ, and θ = 0 when the motion is along the field.
CThis treats the electric force as if it stayed at right angles to the velocity, like a magnetic force. The electric force keeps a fixed direction, so the velocity turns towards it and the path is a parabola.
DThis makes both errors: it treats the electric force like a magnetic force, and it ignores the sin θ factor in the magnetic force.
Syllabus understandingD.3 — the motion of a charged particle in a uniform electric field; the motion of a charged particle in a uniform magnetic field; the magnitude and direction of the force on a charge moving in a magnetic field as given by F = qvB sin θ; A.1 — the behaviour of projectiles Command term: Identify
20D-1A-98
Charged particles in electric fields·D.3 Motion in electromagnetic fields
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
Two large parallel metal plates are held vertically, a few centimetres apart, in a vacuum. There is a uniform horizontal electric field between them. A small positively charged sphere is released from rest midway between the plates.
Which describes the path of the sphere before it strikes a plate?
Show mark scheme
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Mark
Notes
Step 1Two forces act on the sphere: its weight mg (vertical) and the electric force qE (horizontal, towards the negative plate). Both are constant.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The resultant force is therefore constant in magnitude and direction, so the acceleration is constant.
—
Step 3Starting from rest, the velocity is always in the direction of this constant acceleration: the sphere moves along a straight line at an angle to the vertical with tan of the angle = qE/(mg).
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis ignores the electric force, as if it acted only on a moving charge. The electric force qE acts on a charge whether it is moving or not.
BThis uses the projectile picture, which needs an initial velocity at an angle to the force. The sphere starts from rest, so its velocity is always along the constant resultant force.
CCorrect: a constant resultant force on a body released from rest gives straight-line motion along the direction of that force.
DThis ignores the weight of the sphere. The weight acts downwards as well, so the resultant force, and the path, are inclined to the horizontal.
Syllabus understandingD.3 — the motion of a charged particle in a uniform electric field; D.2 — the electric field strength as given by E = F/q; A.2 — Newton's second law and the resultant of forces Command term: Deduce
21D-1A-99
Force between parallel wires·D.3 Motion in electromagnetic fields
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
Two long straight parallel wires P and Q carry steady currents. The magnetic force per unit length on Q is F.
The current in P is doubled, the current in Q is tripled and the separation of the wires is doubled.
What is the new magnetic force per unit length on Q?
Show mark scheme
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Mark
Notes
Step 1F/L = μ0IPIQ/(2πr), so the force per unit length is proportional to IPIQ/r.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The product of the currents increases by 2 × 3 = 6.
—
Step 3Doubling the separation halves the force: 6/2 = 3, so the new force per unit length is 3F.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis uses an inverse-square dependence on the separation, as for the force between point charges: 6/2² = 1.5.
BCorrect: (2 × 3)/2 = 3.
CThis includes the change in the currents but ignores the change in the separation.
DThis multiplies by the change in separation instead of dividing: 6 × 2 = 12.
Syllabus understandingD.3 — the force per unit length between parallel wires as given by F/L = μ0I1I2/(2πr) Command term: Determine
22D-1A-100
Force on a current-carrying conductor·D.3 Motion in electromagnetic fields
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A straight wire carries a current I in a uniform magnetic field. The length of wire in the field is 4.0 cm, and the wire makes an angle of 30° with the field lines. The graph shows how the magnetic force F on the wire varies with I.
What is the magnetic flux density?
Magnetic force on the wire against current (drawn to scale).Show mark scheme
Marking point
Mark
Notes
Step 1Gradient of the graph = F/I = 12.0 × 10−3/5.0 = 2.4 × 10−3 N A−1.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2F = BIL sin θ, so the gradient = BL sin 30°.
—
Step 3B = 2.4 × 10−3/(0.040 × 0.50) = 0.12 T.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis multiplies by sin 30° instead of dividing by it: (2.4 × 10−3/0.040) × 0.50 = 0.030 T.
BThis leaves out the angle, as if the wire were perpendicular to the field: 2.4 × 10−3/0.040 = 0.060 T.
CThis uses cos 30°, the component of the field along the wire, instead of sin 30°: 2.4 × 10−3/(0.040 × 0.866) = 0.069 T.
DCorrect: only the component of the field perpendicular to the wire, B sin 30°, exerts a force, so B = gradient/(L sin 30°) = 0.12 T.
Syllabus understandingD.3 — the magnitude and direction of the force on a current-carrying conductor in a magnetic field as given by F = BIL sin θCommand term: Determine
23D-1A-101
Charged particles in magnetic fields·D.3 Motion in electromagnetic fields
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A charged particle moves in a circle in a uniform magnetic field, at right angles to the field. No other forces act.
Which statements are correct?
I. The speed of the particle is constant.
II. The acceleration of the particle is constant.
III. The magnetic force does no work on the particle.
Show mark scheme
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Mark
Notes
Step 1The magnetic force is always perpendicular to the velocity, so it has no component along the direction of motion: it does no work (III is correct).
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2With no work done, the kinetic energy and therefore the speed stay constant (I is correct).
—
Step 3The acceleration has constant magnitude v²/r but always points towards the centre, so its direction changes continuously: the acceleration is not constant (II is wrong).
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis omits III and treats the acceleration as constant because its magnitude is constant. Acceleration is a vector; its direction rotates with the particle.
BCorrect: the force does no work, so the speed is constant, but the centripetal acceleration changes direction all the time.
CThis assumes that an acceleration must change the speed. A force at right angles to the velocity changes only the direction of motion.
DThis accepts II by treating the constant magnitude of the acceleration as a constant vector.
Syllabus understandingD.3 — the motion of a charged particle in a uniform magnetic field; Guidance: the kinetic energy of a charged particle stays constant in a magnetic field; A.2 — circular motion and centripetal acceleration Command term: Deduce
24D-1A-102
Charged particles in electric fields·D.3 Motion in electromagnetic fields
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
An alpha particle is released from rest in a uniform electric field in a vacuum. The graph shows how its kinetic energy Ek varies with the distance x it has moved.
What is the electric field strength?
Kinetic energy of the alpha particle against distance moved (drawn to scale).Show mark scheme
Marking point
Mark
Notes
Step 1The gain in kinetic energy equals the work done by the electric force: ΔEk = qEx, so the gradient of the graph is the force qE.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Gradient = 6.0 keV/4.0 mm = 1.5 keV mm−1 = 1.5 × 106 eV m−1, so the force is (1.5 × 106 V m−1) × e.
—
Step 3The alpha particle has charge 2e, so E = 1.5 × 106/2 = 7.5 × 105 V m−1.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis does not convert millimetres to metres: (1.5 × 103 eV per mm)/2 = 750, given in V m−1.
BCorrect: qE = 2eE = 1.5 × 106 eV m−1, so E = 7.5 × 105 V m−1.
CThis uses a charge of e, as for a proton, instead of 2e.
DThis multiplies by 2 instead of dividing by the charge number of 2.
Syllabus understandingD.3 — the motion of a charged particle in a uniform electric field; D.2 — the electric field strength as given by E = F/q; D.2 (HL) — the work done in moving a charge q in an electric field as given by W = q∆Ve; A.3 — the work done by a force Command term: Determine
25D-1A-103
Velocity selector·D.3 Motion in electromagnetic fields
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
Positive ions travelling at speed v pass undeflected through a region of uniform electric field E and uniform magnetic field B. The two fields are perpendicular to each other and to the velocity of the ions.
Which changes would still allow these ions to pass undeflected?
I. The electric field strength and the magnetic flux density are both doubled.
II. The speed of the ions and the electric field strength are both doubled.
III. The direction of the magnetic field is reversed and nothing else is changed.
Show mark scheme
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Mark
Notes
Step 1Undeflected motion needs equal and opposite forces: qE = qvB, so v = E/B.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2I: E/B is unchanged, so the same ions still pass. II: 2E/B = 2v, the new speed of the ions, so they pass.
—
Step 3III: reversing B reverses the magnetic force, which then acts in the same direction as the electric force, so the ions are deflected.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: the condition v = E/B still holds in I and II; in III the two forces no longer oppose each other.
BThis considers only the magnitudes of the forces (accepting III), and rejects II by assuming the selector passes only the speed it was first set for, even though E was changed in proportion.
CThis uses v = EB instead of E/B, which rejects I (the product is four times larger) and accepts II; it also considers only the magnitudes of the forces in III.
DThis considers only the magnitudes of the forces, so it accepts III; the direction of the magnetic force matters as well.
Syllabus understandingD.3 — the motion of a charged particle in perpendicularly orientated uniform electric and magnetic fields; the magnitude and direction of the force on a charge moving in a magnetic field as given by F = qvB sin θCommand term: Deduce
26D-1A-104
Force on a curved conductor·D.3 Motion in electromagnetic fields
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A wire is bent into a semicircle of radius R and placed in a uniform magnetic field of flux density B directed into the page, as shown. There is a current I in the wire.
What is the magnitude of the resultant magnetic force on the semicircular part of the wire?
A semicircular wire of radius R carrying a current I in a uniform magnetic field directed into the page (not to scale).Show mark scheme
Marking point
Mark
Notes
Step 1Every short element of the arc is perpendicular to the field and has a force of magnitude BIΔl, in the plane of the page and perpendicular to the element (along the radius).
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2By symmetry, the components of these forces parallel to the diameter XY cancel in pairs; the components perpendicular to XY all point the same way and add.
—
Step 3Adding the perpendicular components is equivalent to replacing the arc by the straight line XY carrying the same current: F = BI(2R) = 2BIR.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis assumes the forces on the curved wire cancel because they point in many different directions. Only the components parallel to the diameter cancel; the components perpendicular to it add.
BThis uses the radius R as the effective length of the wire. The effective length is the straight-line distance between the ends of the wire, the diameter 2R.
CCorrect: the resultant force is the same as on a straight wire from X to Y, BI × 2R.
DThis adds the magnitudes of the forces on all the elements, BI × (arc length πR), ignoring the fact that they act in different directions.
Syllabus understandingD.3 — the magnitude and direction of the force on a current-carrying conductor in a magnetic field as given by F = BIL sin θCommand term: Determine
27D-1A-105
Motion across a field boundary·D.3 Motion in electromagnetic fields
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A boundary separates two regions of uniform magnetic field, both directed into the page. The flux density is B above the boundary and 2B below it. A positive particle at point P on the boundary moves at right angles to the boundary into the upper region, as shown. In the upper region its path has radius r.
The particle passes into the lower region and then returns to the upper region. What is the distance from P to the point where the particle returns to the upper region?
Two regions of uniform magnetic field directed into the page, separated by a straight boundary (not to scale).Show mark scheme
Marking point
Mark
Notes
Step 1Above the boundary the particle moves at constant speed through a semicircle of radius r and crosses the boundary a distance 2r from P, moving downwards.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Below the boundary r = mv/(q × 2B) = r/2. The force on the downward-moving particle now points back towards P, so it moves through a semicircle of diameter r back towards P.
—
Step 3It reaches the boundary a distance 2r − r = r from P (the path drifts along the boundary by r in each cycle).
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis assumes the particle returns to its starting point, as it would if the field were the same on both sides of the boundary.
BCorrect: the semicircles have diameters 2r and r, and they curve in opposite senses relative to the boundary, so the displacement is 2r − r = r.
CThis subtracts the radius of the lower semicircle (r/2) instead of its diameter: 2r − 0.5r.
DThis assumes the lower semicircle continues in the same direction along the boundary, so that the diameters add: 2r + r.
Syllabus understandingD.3 — the motion of a charged particle in a uniform magnetic field; the magnitude and direction of the force on a charge moving in a magnetic field as given by F = qvB sin θ; Guidance: the kinetic energy of a charged particle stays constant in a magnetic field Command term: Deduce
28D-1A-106
Work and momentum in a magnetic field·D.3 Motion in electromagnetic fields
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A particle of mass m and charge q moves at speed v at right angles to a uniform magnetic field and completes one quarter of a circle.
Which row gives the work done on the particle by the magnetic force and the magnitude of the change in momentum of the particle during this quarter circle?
Work done by the magnetic forceMagnitude of the change in momentum
Show mark scheme
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Mark
Notes
Step 1The magnetic force is always perpendicular to the velocity, so it does no work: the work done is zero and the speed stays v.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2After a quarter circle the momentum, still of magnitude mv, points at 90° to its original direction.
—
Step 3The change in momentum is the vector difference of two perpendicular vectors of magnitude mv: |Δp| = √((mv)² + (mv)²) = √2 mv.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis assumes that, because the speed is constant, the momentum is constant. Momentum is a vector, and its direction has changed by 90°.
BThis adds the magnitudes of the initial and final momenta, as for a reversal of direction (a half circle); after a quarter circle the momenta are perpendicular.
CThis takes the work done as force × arc length: qvB × (πr/2) with r = mv/(qB) gives πmv²/2. The force is perpendicular to the displacement at every instant, so the work done is zero.
DCorrect: no work is done, and the vector change in momentum has magnitude √2 mv.
Syllabus understandingD.3 — the motion of a charged particle in a uniform magnetic field; Guidance: the kinetic energy of a charged particle stays constant in a magnetic field; A.2 — change in momentum (impulse); A.3 — work done by a force Command term: Deduce
29D-1A-107
Charged particles in electric fields·D.3 Motion in electromagnetic fields
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksCalculate
An electron is between two parallel metal plates in a vacuum. The plates are 2.0 cm apart and the potential difference between them is 150 V.
What is the magnitude of the acceleration of the electron?
Show mark scheme
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Notes
Step 1E = V/d = 150/0.020 = 7.5 × 103 V m−1.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2F = eE = 1.60 × 10−19 × 7.5 × 103 = 1.2 × 10−15 N.
AThis uses the mass of a proton, 1.67 × 10−27 kg, instead of the mass of the electron.
BThis uses the separation in centimetres (2.0) instead of metres when calculating E = V/d.
CThis uses the potential difference in place of the field strength, eV/me, leaving out the division by the separation.
DCorrect: a = eV/(med) = 1.60 × 10−19 × 150/(9.11 × 10−31 × 0.020) = 1.3 × 1015 m s−2.
Syllabus understandingD.3 — the motion of a charged particle in a uniform electric field; D.2 — the electric field strength between parallel plates as given by E = V/d; A.2 — Newton's second law Command term: Calculate
30D-1A-108
Specific charge from a circular path·D.3 Motion in electromagnetic fields
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
Ions of one type move in circular paths at right angles to a uniform magnetic field. The graph shows how the number of revolutions per second f of the ions varies with the magnetic flux density B.
What is the charge-to-mass ratio of the ions?
Number of revolutions per second of the ions against magnetic flux density (drawn to scale).Show mark scheme
Marking point
Mark
Notes
Step 1qvB = mv²/r gives r = mv/(qB), so the period is T = 2πr/v = 2πm/(qB), independent of the speed.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2f = 1/T = (q/m) × B/(2π): the gradient of the graph is (q/m)/(2π).
AThis divides the gradient by 2π instead of multiplying by it.
BThis takes the gradient itself as q/m, leaving out the factor 2π that links the frequency to the angular speed qB/m.
CThis takes the time for one revolution as πm/(qB), the time for a semicircle, so that q/m = π × gradient.
DCorrect: q/m = 2π × gradient = 4.8 × 107 C kg−1 (these could be alpha particles).
Syllabus understandingD.3 — the motion of a charged particle in a uniform magnetic field; the magnitude and direction of the force on a charge moving in a magnetic field as given by F = qvB sin θ; Guidance: the determination of the charge-to-mass ratio for a charged particle by investigating its path in a uniform magnetic field; A.2 — circular motion Command term: Determine
31D-1A-109
Deflection between parallel plates·D.3 Motion in electromagnetic fields
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
Electrons enter the uniform electric field between two parallel plates, moving parallel to the plates with speed v. They leave the plates moving at an angle θ to their original direction, without striking a plate.
The potential difference between the plates is halved and the speed with which the electrons enter is halved. The electrons still leave without striking a plate.
By what factor does tan θ change?
Show mark scheme
Marking point
Mark
Notes
Step 1Between plates of length L and separation d, the time is L/v and the acceleration across the plates is eV/(md).
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Velocity component across the plates on leaving = eVL/(mdv), so tan θ = (eVL/(mdv))/v = eVL/(mdv²).
—
Step 3tan θ ∝ V/v²: (1/2)/(1/2)² = 2.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis halves tan θ because the field is halved, ignoring the change in speed.
BThis takes tan θ ∝ V/v, so the two changes cancel. Halving v doubles the time between the plates, and so the sideways velocity, and also halves the forward velocity, so tan θ ∝ 1/v².
CCorrect: halving V halves the acceleration, while halving v doubles the time between the plates (doubling the sideways velocity) and halves the forward velocity: 0.5 × 2 × 2 = 2.
DThis includes the effect of halving the speed (× 4) but ignores the halving of the potential difference.
Syllabus understandingD.3 — the motion of a charged particle in a uniform electric field; D.2 — E = V/d between parallel plates; A.1 — the behaviour of projectiles Command term: Deduce
32D-1B-04
Force on a current-carrying conductor·D.3 Motion in electromagnetic fields
Paper 1BMedium7 marks
Data-based question6 steps to full marksDetermine
A U-shaped permanent magnet rests on a digital top-pan balance, with its uniform horizontal field between the poles. A straight horizontal copper wire is held between the poles, perpendicular to the field, by a clamp that does not touch the magnet or the balance. The length of wire in the field is 5.00 cm ± 0.10 cm. The balance was set to zero with the magnet on the pan, but the student did not re-zero it before taking readings.
The student records the balance reading m for six values of the current I in the wire. The readings increase as the current increases. Each reading has an uncertainty of ±0.01 g; the uncertainty in I is negligible. The data are shown in the table and in the graph.
I / A
1.0
2.0
3.0
4.0
5.0
6.0
m / g
0.43
0.84
1.25
1.66
2.06
2.47
Graph drawn to scale. The error bars (±0.01 g) are smaller than the points.
(a)
The force on the wire.
(i)
Deduce the direction of the magnetic force on the wire.
(1)
(ii)
Draw the line of best fit and determine its gradient, including its unit.
(2)
(b)
The magnetic field.
(i)
Determine the magnetic flux density B between the poles.
(2)
(ii)
The uncertainty in the gradient is ±2 %. Determine the absolute uncertainty in B.
(1)
(c)
(i)
The magnet is turned on the pan about a vertical axis so that the field, still horizontal, makes an angle of 30° with the wire. Predict the gradient of the new graph of m against I.
(1)
Show mark scheme
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Notes
Part (a)(i)
The reading increases, so the wire exerts a downward force on the magnet; by Newton's third law the magnet exerts an equal and opposite force on the wire: upwards
✓ 1
Both the downward force on the magnet and the Newton's third law link are needed.
Part (a)(ii)
A single straight line of best fit through (or very close to) all six points
✓ 1
The line need not pass through the origin.
Gradient = 0.408 g A−1 (4.08 × 10−4 kg A−1)
✓ 1
Accept 0.400–0.415 g A−1. Unit required for this mark.
Part (b)(i)
Magnetic force BIL = (change in reading) × g, so the gradient Δm/ΔI = BL/g and B = g × gradient/L
✓ 1
Award for relating the gradient to B. A value from a single reading divided by its current (e.g. 2.47 g/6.0 A, ignoring the offset) scores this mark only.
B = 9.81 × 4.08 × 10−4/0.0500 = 0.080 T
✓ 1
Allow ECF from (a)(ii). Accept 0.078–0.082 T. The conversion from g to kg is required.
Allow ECF from (b)(i). Accept 0.0032 T. Omitting the uncertainty in the length (0.002 T) scores 0.
Part (c)(i)
Only the component of the field perpendicular to the wire contributes: F = BIL sin 30°, so the gradient halves: 0.408 × sin 30° = 0.20 g A−1
✓ 1
Allow ECF from (a)(ii). Accept 0.20–0.21 g A−1. Using cos 30° (0.35 g A−1) scores 0.
Answers: (a)(i) upwards · (a)(ii) 0.408 g A−1 · (b)(i) 0.080 T · (b)(ii) ±0.003 T · (c)(i) 0.20 g A−1(the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the magnitude and direction of the force on a current-carrying conductor in a magnetic field as given by F = BIL sin θ; A.2 — Newton's third law of motion; Tools 3 — gradient of a linear graph with its unit, propagating percentage uncertainties Command term: Determine
33D-1B-10
Specific charge of the electron·D.3 Motion in electromagnetic fields
Paper 1BEasy7 marks
Data-based question6 steps to full marksDetermine
In a fine-beam tube, electrons are accelerated from rest through a potential difference V and then enter a uniform magnetic field of flux density B = 1.00 × 10−3 T at right angles to their velocity. They move in a circle whose radius r is read from a scale behind the beam. Theory predicts that r² = (2m/(eB²))V, where e/m is the specific charge of the electron.
B is known to ±2 %, and the gradient of the graph has an uncertainty of ±3 %. The graph shows four of the five data points.
V / V
100
200
300
400
500
r / cm
3.4
4.8
5.8
6.7
7.5
r² / cm²
11.6
23.0
44.9
56.3
Graph drawn to scale.
(a)
The graph.
(i)
Complete the table.
(1)
(ii)
Plot the missing point, draw the line of best fit and determine its gradient in m² V−1.
(2)
(b)
The specific charge.
(i)
Determine the specific charge e/m of the electron.
(2)
(ii)
Determine the absolute uncertainty in your value of e/m, and comment on whether the accepted value, 1.76 × 1011 C kg−1, agrees with it.
(2)
Show mark scheme
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Mark
Notes
Part (a)(i)
5.8² = 33.6 cm²
✓ 1
Part (a)(ii)
Point plotted correctly and a straight line of best fit through (or very close to) the origin
✓ 1
Allow ECF from (a)(i).
Gradient = 0.113 cm² V−1 = 1.13 × 10−5 m² V−1
✓ 1
Accept 1.09–1.16 × 10⁻⁵ m² V⁻¹. The conversion from cm² to m² is required for this mark.
Allow ECF from (a)(ii). Accept 1.72–1.84 × 10¹¹ C kg⁻¹.
Part (b)(ii)
e/m ∝ B−2 × (gradient)−1, so the percentage uncertainty is 2 × 2 % + 3 % = 7 %
✓ 1
Adding 2 % + 3 % = 5 % scores 0 for this mark.
Δ(e/m) = 0.07 × 1.77 × 1011 ≈ 0.1 × 1011 C kg−1: (1.8 ± 0.1) × 1011 C kg−1, a range that includes 1.76 × 1011, so the result agrees with the accepted value
✓ 1
Allow ECF from (b)(i). The comparison must use the candidate's own range.
Answers: (a)(i) 33.6 cm² · (a)(ii) 1.13 × 10−5 m² V−1 · (b)(i) 1.8 × 1011 C kg−1 · (b)(ii) ±0.1 × 1011 C kg−1(the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the motion of a charged particle in a uniform magnetic field; the determination of the charge to mass ratio for a charged particle by investigating its path in a uniform magnetic field; Tools 3 — linearised graph, gradient with unit conversion, propagating uncertainties through a power Command term: Determine
34D-1B-12
Charged particles in electric fields·D.3 Motion in electromagnetic fields
Paper 1BMedium7 marks
Data-based question6 steps to full marksDetermine
In an electron deflection tube, electrons accelerated from rest through a potential difference Va enter, midway between them, the uniform electric field between two horizontal deflecting plates. The plates are 4.0 cm long and 1.0 cm apart and the potential difference between them is fixed at 40.0 V. After leaving the plates the electrons travel a further 18.0 cm to a fluorescent screen. The vertical deflection y of the spot on the screen is read from a scale ruled on the screen; each reading has an uncertainty of ±1 mm.
Theory predicts that y = kVd/Va, where Vd is the potential difference between the plates and k depends only on the dimensions of the tube. For this tube the dimensions give k = 0.40 m. The graph shows five of the six data points.
Va / V
y / mm
(1/Va) / 10−3 V−1
400
41.8
2.50
500
34.3
2.00
600
28.4
800
22.2
1.25
1000
17.9
1.00
1200
15.4
0.83
Graph drawn to scale. Error bars show ±1 mm.
(a)
The graph.
(i)
Complete the table.
(1)
(ii)
Plot the missing point and draw the line of best fit. Determine the gradient of the line, including its unit.
(2)
(b)
Interpreting the line.
(i)
The uncertainty in the gradient is ±5 %. Determine k from your gradient and discuss whether your value agrees with the value given by the dimensions of the tube.
(2)
(ii)
Determine the intercept of the line on the y axis. Suggest a cause of this intercept and explain why it does not affect your value of k.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
1/600 = 1.67 × 10−3 V−1
✓ 1
Accept 1.667.
Part (a)(ii)
Point plotted at (1.67, 28.4) and a single straight line of best fit drawn through the error bars; the line does not pass through the origin
✓ 1
Allow ECF from (a)(i).
Gradient = 15.9 mm per 10−3 V−1 = 15.9 V m
✓ 1
Accept 15.4–16.4 V m. The unit (V m) is required for this mark.
Part (b)(i)
k = gradient/Vd = 15.9/40.0 = 0.40 m
✓ 1
Allow ECF from (a)(ii). Accept 0.385–0.41 m.
±5 % gives k = (0.40 ± 0.02) m, a range that includes 0.40 m, so the result agrees with the value from the dimensions of the tube
✓ 1
The comparison must use the candidate's own value and uncertainty (ECF).
Part (b)(ii)
Line extrapolated to 1/Va = 0: intercept ≈ +2 mm
✓ 1
Accept 1.5–2.7 mm. Allow ECF from the candidate's line.
The zero of the scale is not at the position of the undeflected spot (the spot is about 2 mm off zero when Vd = 0): a systematic error that adds the same amount to every reading, so it shifts the line without changing its gradient, and k comes from the gradient
✓ 1
Both the cause and the reason that the gradient is unchanged are needed. Do not accept "random error" or "parallax" on its own.
Answers: (a)(i) 1.67 × 10−3 V−1 · (a)(ii) 15.9 V m · (b)(i) 0.40 m · (b)(ii) ≈ +2 mm (the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the motion of a charged particle in a uniform electric field; D.2 — the uniform electric field strength between parallel plates as given by E = V/d; A.1 — the behaviour of projectiles; Tools 3 — linearising a relationship, gradient with its unit, interpreting an intercept as a systematic error; Inquiry 3 — comparing a result with the accepted scientific context Command term: Determine
35D-1B-18
Force between parallel wires·D.3 Motion in electromagnetic fields
Paper 1BEasy7 marks
Data-based question5 steps to full marksDetermine
Two straight aluminium rods of diameter 2.0 mm are held horizontal, one vertically above the other, with a distance r = 10.0 mm between their centres. They overlap for a length of 0.400 m and are connected in series so that they carry the same current I in opposite directions. The lower rod rests on insulating supports on a top-pan balance, which is set to zero when there is no current.
For each current the increase in the balance reading Δm is recorded, and the magnetic force on the lower rod is found from F = Δm × g. The graph shows four of the five points.
I / A
I² / A²
Δm / g
F / mN
10
100
0.080
0.78
15
225
0.181
1.78
20
400
0.327
3.21
25
625
0.507
4.97
30
900
0.734
Not to scale. The overlap length of the rods is 0.400 m and r is the distance between their centres.Graph drawn to scale.
(a)
The graph.
(i)
Complete the table.
(1)
(ii)
Plot the missing point and, using the full range of the data, determine the gradient of the graph, including its unit.
(2)
(b)
The permeability of free space.
(i)
Determine a value for μ0.
(2)
(ii)
Another student measured r as the gap between the facing surfaces of the rods, 8.0 mm. Determine the value of μ0 this student would obtain from the same data, and state the type of error this introduces.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
0.734 × 10−3 × 9.81 = 7.20 mN
✓ 1
Accept 7.2.
Part (a)(ii)
Point plotted correctly and a straight line of best fit through (or very close to) the origin, using all five points
✓ 1
Allow ECF from (a)(i).
Gradient = 8.0 × 10−6 N A−2
✓ 1
Accept 7.8–8.2 × 10⁻⁶. Unit required. A gradient from a single point or from the first two points only scores [1 max] for the part.
Part (b)(i)
F = μ0I²L/(2πr), so the gradient = μ0L/(2πr) and μ0 = 2π × 0.0100 × gradient/0.400
✓ 1
μ0 = 1.26 × 10−6 T m A−1
✓ 1
Allow ECF from (a)(ii). Accept 1.22–1.29 × 10⁻⁶; accept N A⁻².
Part (b)(ii)
μ0 = 1.26 × 10−6 × 8.0/10.0 = 1.0 × 10−6 T m A−1
✓ 1
Allow ECF from (b)(i).
A systematic error: the same underestimated r is used for every reading, so μ0 is always about 20 % too small and repeating the readings does not reduce it
✓ 1
Answers: (a)(i) 7.20 mN · (a)(ii) 8.0 × 10−6 N A−2 · (b)(i) 1.26 × 10−6 T m A−1 · (b)(ii) 1.0 × 10−6 T m A−1(the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the force per unit length between parallel wires as given by F/L = μ0I1I2/(2πr); Tools 3 — gradient with its unit using the full range of the data; Inquiry 3 — systematic error Command term: Determine
36D-1B-23
Velocity selector·D.3 Motion in electromagnetic fields
Paper 1BHard8 marks
Data-based question6 steps to full marksDetermine
Electrons are accelerated from rest through a potential difference Va. They then pass between two horizontal parallel plates 5.0 mm apart, where a uniform magnetic field of flux density 1.20 mT acts perpendicular to both the electric field between the plates and the velocity of the electrons. For each value of Va the potential difference Vp between the plates is adjusted until the beam passes through undeflected.
The flux density is known to ±2 % and the plate separation to ±3 %. The graph shows Vp² against Va.
Va / kV
Vp / V
Vp² / 10³ V²
0.50
81
6.6
1.00
111
12.3
1.50
139
19.3
2.00
161
25.9
2.50
177
31.3
3.00
197
38.8
Graph drawn to scale.
(a)
The relationship.
(i)
Show that Vp² = 2(e/m)d²B²Va, where d is the separation of the plates.
(2)
(ii)
Draw the line of best fit and determine its gradient, including its unit.
(2)
(b)
The specific charge of the electron.
(i)
Determine e/m.
(1)
(ii)
The uncertainty in the gradient is ±3 %. Determine the absolute uncertainty in your value of e/m and discuss whether it agrees with the accepted value of 1.76 × 1011 C kg−1.
(2)
(iii)
Identify the measurement whose uncertainty contributes most to the uncertainty in e/m.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
eVa = ½mv², so v² = 2(e/m)Va
✓ 1
Undeflected: eVp/d = evB (electric force balances magnetic force), so v = Vp/(Bd); substituting gives Vp² = 2(e/m)d²B²Va
✓ 1
The use of E = Vp/d must be seen.
Part (a)(ii)
A single straight line through (or very close to) the origin and the points
✓ 1
Gradient = 12.8 × 10³ V² kV−1 = 12.8 V
✓ 1
Accept 12.4–13.2 V. The unit V (or 10³ V² kV⁻¹) is required for this mark.
Allow ECF from (a)(ii). Accept 1.72–1.83 × 10¹¹ C kg⁻¹.
Part (b)(ii)
e/m ∝ gradient × d−2 × B−2, so the percentage uncertainty is 3 % + 2 × 3 % + 2 × 2 % = 13 %
✓ 1
Adding 3 % + 3 % + 2 % = 8 % scores 0 for this mark.
Δ(e/m) = 0.13 × 1.78 × 1011 ≈ 0.2 × 1011 C kg−1; the range (1.8 ± 0.2) × 1011 C kg−1 includes 1.76 × 1011 C kg−1, so the result agrees with the accepted value
✓ 1
Allow ECF from (b)(i). Accept 0.23 × 10¹¹ C kg⁻¹. The conclusion must be consistent with the candidate's own range.
Part (b)(iii)
The plate separation d: because d appears squared, its 3 % uncertainty contributes 6 %, more than B (4 %) or the gradient (3 %)
✓ 1
Allow ECF from the candidate's propagation in (b)(ii). The comparison of contributions is required; "the plate separation" alone scores 0.
Answers: (a)(ii) 12.8 V · (b)(i) 1.78 × 1011 C kg−1 · (b)(ii) ±0.2 × 1011 C kg−1 · (b)(iii) the plate separation (the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the motion of a charged particle in perpendicularly orientated uniform electric and magnetic fields; the determination of the charge-to-mass ratio for a charged particle; D.2 — E = V/d between parallel plates; D.2 (HL) — W = qΔVe; Tools 3 — linearising a relationship, gradient with its unit, propagating uncertainties through powers Command term: Determine
37D-1B-30
Time-of-flight mass spectrometer·D.3 Motion in electromagnetic fields
Paper 1BEasy7 marks
Data-based question5 steps to full marksDetermine
In a time-of-flight mass spectrometer, singly charged positive ions are produced at rest and accelerated through a potential difference V by a uniform electric field in a very short gap. They then drift at constant speed along a field-free tube of length D = 1.200 m to a detector. An electronic timer measures the flight time t from the moment the ions are produced; the time spent in the accelerating gap is negligible.
The student records t for five types of ion of known mass m, measured in unified atomic mass units u. Theory predicts that t = D√(m/(2eV)), where m is the mass of the ion in kg.
ion
m / u
√(m / u)
t / μs
Li+
7
2.65
5.47
Na+
23
4.80
9.66
K+
39
12.41
Rb+
85
9.22
18.19
Cs+
133
11.53
22.68
Flight time t against √(m / u), with the line of best fit (drawn to scale).
(a)
(i)
Complete the table for the K+ ion.
(1)
(ii)
Determine the gradient and the intercept on the t axis of the line of best fit.
(2)
(b)
(i)
Use the gradient to determine the accelerating potential difference V.
(2)
(ii)
An unknown singly charged ion has a flight time of 14.8 μs. Determine its mass in u.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
√39 = 6.24
✓ 1
Accept 6.2 or 6.245.
Part (a)(ii)
Gradient from a large triangle = 1.94 μs
✓ 1
Accept 1.90–1.98 μs (the unit is μs, since √(m / u) has no unit).
Intercept = 0.34 μs
✓ 1
Accept 0.2–0.5 μs.
Part (b)(i)
Gradient = D√(u/(2eV)), so V = D²u/(2e × gradient²)
✓ 1
The gradient must be converted to seconds.
V = 1.200² × 1.66 × 10−27/(2 × 1.60 × 10−19 × (1.937 × 10−6)²) = 1990 V
✓ 1
Allow ECF from (a)(ii). Accept 1.9–2.1 kV.
Part (b)(ii)
The intercept is a constant delay added to every time, so it is subtracted: √(m / u) = (14.8 − 0.34)/1.94 = 7.47
✓ 1
Allow ECF from (a)(ii). Reading the line on the graph at t = 14.8 μs is equivalent.
m = 56 u
✓ 1
Accept 54–57 u. Ignoring the intercept (58 u): [1 max] for the part.
Answers: (a)(i) 6.24 · (a)(ii) 1.94 μs; 0.34 μs · (b)(i) 1.99 kV · (b)(ii) 56 u (the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the motion of a charged particle in a uniform electric field; D.2 (HL) — the work done in moving a charge q in an electric field as given by W = q∆Ve; A.1 — motion at constant velocity; Tools 3 — linearising a relationship, gradient and intercept of a linear graph, a systematic error shown by an intercept Command term: Determine
38D-1B-31
Force between parallel conductors·D.3 Motion in electromagnetic fields
Paper 1BMedium7 marks
Data-based question5 steps to full marksDetermine
Two long, straight, horizontal copper rods are connected in series so that each carries a current of 20.0 A, in opposite directions. The lower rod is clamped to a vertical micrometer stage, which sets the separation r between the centres of the rods. The upper rod, of length 0.300 m, hangs from a digital force sensor by an insulating thread; the sensor is set to zero with no current, so that it measures the magnetic force F on the upper rod. For each separation F is recorded. Each value of F has an uncertainty of ±0.05 mN.
The student suspects that F = krn, where k and n are constants, and plots a graph of ln(F / mN) against ln(r / mm).
r / mm
F / mN
ln(r / mm)
ln(F / mN)
4.0
5.96
1.39
1.79
6.0
3.94
1.79
1.37
8.0
3.02
2.08
12.0
1.97
2.48
0.68
16.0
1.50
2.77
0.41
24.0
0.99
3.18
−0.01
ln(F / mN) against ln(r / mm), with the line of best fit extended to ln(r / mm) = 0 (drawn to scale).
(a)
(i)
Complete the table.
(1)
(ii)
Plot the missing point. Determine the gradient of the line and state what it shows about the relationship between F and r.
(2)
(b)
(i)
Use the intercept of the line on the vertical axis to determine a value for μ0.
(2)
(ii)
Determine the absolute uncertainty in ln(F / mN) for r = 24.0 mm, and explain why this point carries less weight than the point for r = 4.0 mm.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
ln 3.02 = 1.11
✓ 1
Accept 1.105.
Part (a)(ii)
Point plotted at (2.08, 1.11); gradient = −1.00
✓ 1
Allow ECF from (a)(i) for the point. Accept −0.95 to −1.05. No unit.
n = gradient ≈ −1, so F ∝ 1/r (inversely proportional), as predicted by F/L = μ0I1I2/(2πr)
✓ 1
The conclusion must agree with their gradient.
Part (b)(i)
Intercept = 3.18, so Fr = e3.18 = 24.0 mN mm = 2.4 × 10−5 N m
✓ 1
Accept an intercept of 3.10–3.25. The extrapolation to ln(r / mm) = 0 is required.
μ0 = 2π × Fr/(I²L) = 2π × 2.4 × 10−5/(20.0² × 0.300) = 1.26 × 10−6 T m A−1
✓ 1
Allow ECF from their intercept. Accept 1.2–1.35 × 10−6 T m A−1 (or N A−2).
Part (b)(ii)
Δ(ln F) = ΔF/F = 0.05/0.99 = ±0.05
✓ 1
Accept ±0.05.
The same ±0.05 mN is a larger fraction of the smaller force at large r (for r = 4.0 mm, Δ(ln F) = ±0.008), so its error bar is about six times longer
✓ 1
A comparison of the fractional uncertainties is required.
Answers: (a)(i) 1.11 · (a)(ii) −1.00; F ∝ r−1 · (b)(i) 1.26 × 10−6 T m A−1 · (b)(ii) ±0.05 (the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the force per unit length between parallel wires as given by F/L = μ0I1I2/(2πr); Tools 3 — linearising a power law with logarithms, gradient and intercept, extrapolation, uncertainty of a logarithm Command term: Determine
39D-1B-32
Force on a conductor and friction·D.3 Motion in electromagnetic fields
Paper 1BMedium7 marks
Data-based question5 steps to full marksDetermine
Two straight, parallel, horizontal metal rails are fixed 6.0 cm apart in a uniform vertical magnetic field of flux density B. A flat aluminium bar of length 10.0 cm and mass 8.0 g lies across the rails, at right angles to them. When a switch is closed there is a current I in the bar and the bar slides along the rails from rest. A light gate records the time t for the bar to slide 0.300 m. The student calculates the acceleration of the bar from a = 2s/t².
The current is kept constant during each run. Friction between the bar and the rails is constant. The magnetic fields of the currents in the rails can be ignored.
I / A
t / s
a / m s−2
4.0
0.722
1.15
5.0
0.542
2.04
6.0
0.456
2.89
7.0
0.395
8.0
0.356
4.73
9.0
0.327
5.61
Acceleration of the bar against current, with the line of best fit (drawn to scale).
(a)
(i)
Calculate the missing value of a.
(1)
(ii)
Determine the gradient of the line of best fit, including its unit, and the intercept on the I axis.
(2)
(b)
(i)
Determine B.
(2)
(ii)
Determine the coefficient of dynamic friction between the bar and the rails.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
a = 2 × 0.300/0.395² = 3.85 m s−2
✓ 1
Accept 3.8–3.9 m s−2.
Part (a)(ii)
Gradient = 0.90 m s−2 A−1
✓ 1
Accept 0.86–0.94. The unit is required for this mark.
Intercept on the I axis = 2.7 A
✓ 1
Accept 2.5–2.9 A.
Part (b)(i)
BIL − Ff = ma, so a = (BL/m)I − Ff/m: gradient = BL/m, with L = 0.060 m, the separation of the rails
✓ 1
Using the length of the bar, 0.100 m: [0] for this mark.
B = gradient × m/L = 0.90 × 0.0080/0.060 = 0.119 T
✓ 1
Allow ECF from (a)(ii). Accept 0.115–0.125 T.
Part (b)(ii)
At the intercept the magnetic force just balances friction: Ff = BIL0 = 0.119 × 2.73 × 0.060 = 2.0 × 10−2 N
✓ 1
Allow ECF from (a)(ii) and (b)(i). ALT: Ff/m = −(intercept on the a axis) = 2.44 m s−2.
μd = Ff/(mg) = 2.0 × 10−2/(0.0080 × 9.81) = 0.25
✓ 1
Accept 0.23–0.27.
Answers: (a)(i) 3.85 m s−2 · (a)(ii) 0.90 m s−2 A−1; 2.7 A · (b)(i) 0.119 T · (b)(ii) 0.25 (the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the magnitude and direction of the force on a current-carrying conductor in a magnetic field as given by F = BIL sin θ; A.2 — Newton's second law and dynamic friction Ff = μdFN; A.1 — the equations of motion for uniform acceleration; Tools 3 — gradient and intercept of a linear graph with units Command term: Determine
40D-2-03
Mass spectrometer·D.3 Motion in electromagnetic fields
Paper 2Hard10 marks
Short answer & extended response7 steps to full marksDeduce
Singly ionised magnesium ions are accelerated from rest through a potential difference of 2.00 kV. They then enter a region of uniform magnetic field of flux density 0.400 T, directed perpendicular to their velocity, travel through a semicircle and strike a flat detector plate that lies along the line through the point of entry. The beam contains the isotopes 24Mg and 26Mg, with ion masses of 24 u and 26 u.
(a)
Entering the field.
(i)
Show that the speed of the 24Mg ions entering the field is about 1.3 × 105 m s−1.
(1)
(ii)
Explain why the ions move along a circular path at constant speed in the magnetic field.
(2)
(b)
Separating the isotopes.
(i)
Calculate the radius of the path of the 24Mg ions.
(1)
(ii)
Determine the distance between the points where the two isotopes strike the detector.
(3)
(c)
Alternatives.
(i)
A student suggests replacing the magnetic field with a uniform electric field perpendicular to the beam. Deduce, without calculation, that this cannot separate the two isotopes.
(2)
(ii)
The ions enter the magnetic field moving to the right on the page and curve towards the top of the page. State the direction of the magnetic field.
Accept 6.3–6.5 mm. Award [2 max] for 3.2 mm (difference in radii only).
Part (c)(i)
In the electric field the sideways deflection after a distance x along the beam is y = ½(qE/m)(x/v)² = qEx²/(4Ek)
✓ 1
Accept the argument in words: a = qE/m is larger for the lighter ion but it spends less time in the field, and the two effects cancel.
Both ions have the same charge and the same kinetic energy qV, so y is the same at every x: their paths are identical, whatever their mass
✓ 1
Part (c)(ii)
Into the page
✓ 1
From F = qv × B or the left-hand rule for a positive charge.
Answers: (a)(i) 1.27 × 105 m s−1 · (b)(i) 7.9 × 10−2 m · (b)(ii) 6.4 mm · (c)(ii) into the page (the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the motion of a charged particle in a uniform magnetic field and in a uniform electric field; the magnitude and direction of the force F = qvB sin θ; the kinetic energy of a charged particle stays constant in a magnetic field; D.2 (HL) — W = qΔVeCommand term: Deduce
41D-2-05
Specific charge and de Broglie wavelength·D.3 Motion in electromagnetic fields
Paper 2Medium13 marks
Short answer & extended response9 steps to full marksDetermine
In a fine-beam tube, electrons are accelerated from rest through a potential difference of 3.00 kV. They then move perpendicular to a uniform magnetic field of flux density 1.95 mT, and the radius of their circular path is measured as 9.5 cm.
(a)
The beam.
(i)
Show that the speed of the electrons entering the field is about 3.2 × 107 m s−1.
(1)
(ii)
Outline why the kinetic energy of an electron stays constant in the magnetic field.
(1)
(b)
The specific charge.
(i)
Show that e/m = 2V/(B²r²), where V is the accelerating potential difference.
(2)
(ii)
Determine e/m from the measurements and compare it with the value calculated from the data booklet.
(2)
(iii)
The electrons move to the right in the plane of the page and their path curves towards the top of the page. State and explain the direction of the magnetic field.
(2)
(c)
Wave behaviour of the electrons.
(i)
Calculate the de Broglie wavelength of the electrons.
(2)
(ii)
The beam is directed at a thin layer of graphite. The regularly spaced rows of atoms, about 2.1 × 10−10 m apart, act as a diffraction grating for the electrons. Estimate the angle from the straight-through direction at which the first-order maximum of the diffraction pattern occurs.
(1)
(iii)
The accelerating potential difference is increased to 12.0 kV. Deduce the new angle of the first-order maximum.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
v = √(2eV/m) = √(2 × 1.60 × 10−19 × 3000/9.11 × 10−31) = 3.25 × 107 m s−1
✓ 1
Full substitution or an answer to at least 3 s.f. is required.
Part (a)(ii)
The magnetic force is perpendicular to the velocity, so no work is done on the electron
Data booklet: 1.60 × 10−19/9.11 × 10−31 = 1.76 × 1011 C kg−1, so the two agree to within 1 %
✓ 1
The comparison must be quantitative.
Part (b)(iii)
Out of the page
✓ 1
The force on the electrons is towards the top of the page; for a negative charge the conventional current is to the left, and the left-hand rule (or F = qv × B with q < 0) then gives a field out of the page
✓ 1
The reasoning must deal with the negative charge.
Part (c)(i)
p = mv = 9.11 × 10−31 × 3.25 × 107 = 2.96 × 10−23 kg m s−1
d sin θ = λ: sin θ = 2.2 × 10−11/2.1 × 10−10 = 0.11, so θ ≈ 0.11 rad (about 6°)
✓ 1
Allow ECF from (c)(i). Accept 0.10–0.11 rad.
Part (c)(iii)
λ = h/p and p ∝ √V, so quadrupling V halves λ
✓ 1
sin θ halves (small angle), so θ ≈ 0.053 rad
✓ 1
Allow ECF from (c)(ii).
Answers: (a)(i) 3.25 × 107 m s−1 · (b)(ii) 1.75 × 1011 C kg−1 · (b)(iii) out of the page · (c)(i) 2.2 × 10−11 m · (c)(ii) ≈ 0.11 rad · (c)(iii) ≈ 0.053 rad (the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the motion of a charged particle in a uniform magnetic field; the determination of the charge-to-mass ratio for a charged particle by investigating its path in a uniform magnetic field; E.2 (HL) — the de Broglie wavelength λ = h/p; diffraction of particles as evidence of the wave nature of matter; C.3 (HL) — the diffraction grating nλ = d sin θ Command term: Determine
42D-2-08
The cyclotron·D.3 Motion in electromagnetic fields
Paper 2Medium12 marks
Short answer & extended response9 steps to full marksDetermine
A cyclotron accelerates protons using a uniform magnetic field of flux density 1.50 T. The protons cross the gap between the two D-shaped electrodes twice in each revolution, and at each crossing they are accelerated through a potential difference of 50.0 kV.
(a)
The motion.
(i)
Explain why the protons move in semicircles inside each electrode.
(1)
(ii)
Show that the time for one revolution is T = 2πm/(eB), independent of the speed of the proton.
(2)
(iii)
Calculate the frequency of the alternating potential difference needed.
(1)
(b)
Reaching 10.0 MeV.
(i)
Determine the number of revolutions needed for a proton starting from rest to reach a kinetic energy of 10.0 MeV, and the time this takes.
(2)
(ii)
Calculate the radius of the path of a proton of kinetic energy 10.0 MeV.
(2)
(c)
The extracted beam, a current of 50 μA of 10.0 MeV protons, strikes a copper target of mass 0.20 kg. Specific heat capacity of copper = 385 J kg−1 K−1.
(i)
Show that the power delivered to the target is about 500 W.
(2)
(ii)
Estimate the initial rate of increase of the temperature of the target.
(1)
(iii)
Suggest why the model in (a)(ii) fails for protons of much higher energy.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
There is no electric field inside the electrodes; the magnetic force is perpendicular to the velocity and of constant magnitude, so it provides a centripetal force at constant speed
Two crossings per revolution: 100 keV per revolution, so 10.0 MeV/100 keV = 100 revolutions
✓ 1
Award [1 max] for 200 revolutions (one crossing per revolution) carried through correctly.
t = 100/2.29 × 107 = 4.4 × 10−6 s
✓ 1
Allow ECF from (a)(iii).
Part (b)(ii)
v = √(2Ek/m) = √(2 × 1.60 × 10−12/1.67 × 10−27) = 4.38 × 107 m s−1
✓ 1
r = mv/(eB) = 0.305 m
✓ 1
Part (c)(i)
Number of protons per second = I/e = 50 × 10−6/1.60 × 10−19 = 3.1 × 1014 s−1
✓ 1
ALT: P = IV with V = 1.0 × 10⁷ V.
P = 3.13 × 1014 × 10.0 × 1.60 × 10−13 J = 500 W
✓ 1
Part (c)(ii)
ΔT/Δt = P/(mc) = 500/(0.20 × 385) = 6.5 K s−1
✓ 1
Allow ECF from (c)(i). Assumes all the beam energy becomes internal energy of the target, with no energy lost at first.
Part (c)(iii)
At 10 MeV the speed is already 0.15c; as the speed approaches c Newtonian mechanics no longer applies, so the period is no longer independent of speed and the protons fall out of step with the alternating potential difference
✓ 1
Do not require any relativistic formula.
Answers: (a)(iii) 2.29 × 107 Hz · (b)(i) 100; 4.4 × 10−6 s · (b)(ii) 0.305 m · (c)(i) 500 W · (c)(ii) 6.5 K s−1(the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the motion of a charged particle in a uniform magnetic field; the magnitude and direction of the force on a charge moving in a magnetic field as given by F = qvB sin θ; (guidance) the kinetic energy of a charged particle stays constant in a magnetic field; D.2 (HL) — W = qΔVe in joules and electronvolts; B.5 — current as the rate of flow of charge; B.1 — Q = mcΔT; A.5 — the limits of Newtonian mechanics at speeds approaching cCommand term: Determine
43D-2-09
Charged particles in electric and magnetic fields·D.3 Motion in electromagnetic fields
Paper 2Medium14 marks
Short answer & extended response8 steps to full marksDetermine
Two horizontal parallel plates, 4.0 cm long and 1.2 cm apart, have a potential difference of 240 V between them; the upper plate is positive. A uniform magnetic field, perpendicular to the electric field, can also be applied in the region between the plates. A narrow beam of electrons, accelerated from rest through 2.00 kV, enters the region midway between the plates, moving horizontally to the right, as shown.
Not to scale. The magnetic field is not shown.
(a)
Both fields applied.
(i)
Show that a charged particle moving perpendicular to both fields passes through undeflected when its speed is v = E/B, whatever its charge and mass.
(2)
(ii)
Determine the magnetic flux density needed for the electron beam to pass through undeflected.
(3)
(iii)
State and explain the direction of the magnetic field.
(2)
(b)
The magnetic field is switched off.
(i)
Determine the vertical deflection of the electrons as they leave the plates.
(3)
(c)
The magnetic field is switched on again.
(i)
The accelerating potential difference is increased to 2.50 kV. Explain in which direction the beam is now deflected.
(2)
(ii)
The electrons are replaced by protons accelerated from rest through 2.00 kV. Deduce whether the protons pass through undeflected.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Undeflected when the electric and magnetic forces are equal and opposite: qE = qvB
✓ 1
q cancels and m does not appear, so v = E/B for any charged particle
✓ 1
Part (a)(ii)
E = V/d = 240/0.012 = 2.0 × 104 V m−1
✓ 1
v = √(2eV/m) = √(2 × 1.60 × 10−19 × 2000/9.11 × 10−31) = 2.65 × 107 m s−1
✓ 1
B = E/v = 2.0 × 104/2.65 × 107 = 7.5 × 10−4 T
✓ 1
Allow ECF from (a)(i).
Part (a)(iii)
The electric force on the (negative) electrons is upwards, towards the positive plate, so the magnetic force must be downwards
✓ 1
Into the page (from F = qv × B with q negative, or the left-hand rule with the conventional current to the left)
✓ 1
Award this mark only with a correct direction for the magnetic force.
Part (b)(i)
Time between the plates t = 0.040/2.65 × 107 = 1.51 × 10−9 s
✓ 1
Allow ECF from the speed in (a)(ii).
a = eE/m = 1.60 × 10−19 × 2.0 × 104/9.11 × 10−31 = 3.51 × 1015 m s−2, constant and vertical
✓ 1
y = ½at² = 4.0 × 10−3 m (4.0 mm), upwards
✓ 1
As for a horizontally launched projectile. Accept 3.9–4.1 mm.
Part (c)(i)
The speed is greater: the electric force eE is unchanged, but the magnetic force evB increases
✓ 1
The resultant force is downwards, so the beam is deflected towards the lower (negative) plate
✓ 1
The direction must be consistent with the candidate's answer to (a)(iii).
Part (c)(ii)
The same potential difference gives the same kinetic energy, but v = √(2eV/m) ∝ 1/√m, so the protons are about 43 times slower and v ≪ E/B
✓ 1
Allow ECF from (a)(i).
The magnetic force is much smaller than the electric force, so the protons are deflected in the direction of the electric force on them (downwards, towards the negative plate)
✓ 1
Answers: (a)(ii) 7.5 × 10−4 T · (a)(iii) into the page · (b)(i) 4.0 mm (the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the motion of a charged particle in perpendicularly orientated uniform electric and magnetic fields; the motion of a charged particle in a uniform electric field; the magnitude and direction of F = qvB sin θ; A.1 — projectile motion with a constant acceleration perpendicular to the initial velocity Command term: Determine
44D-2-27
Force between parallel wires·D.3 Motion in electromagnetic fields
Paper 2Hard13 marks
Short answer & extended response8 steps to full marksExplain
In a demonstration of magnetic levitation, a long straight wire X is fixed horizontally on a bench and carries a current of 50 A into the page. A light, straight aluminium wire Y, parallel to X, carries a current of 5.0 A and is free to move vertically between two glass guides, as shown. Y has a mass per unit length of 5.3 × 10−4 kg m−1 and floats at rest at a distance r vertically above X.
Not to scale. Both wires are long, straight and horizontal, perpendicular to the page.
(a)
The magnetic force.
(i)
State the direction of the magnetic field due to X at the position of Y.
(1)
(ii)
Explain why the current in Y must be in the opposite direction to the current in X for Y to float.
(2)
(b)
Equilibrium.
(i)
Show that r is about 9.6 mm.
(2)
(ii)
The current in Y is increased by 20 %. Determine the new equilibrium separation.
(1)
(iii)
State the magnitude and direction of the magnetic force per unit length on X.
(1)
(c)
Stability.
(i)
Y is pushed a small distance downwards and released. Explain why it moves back towards its original position.
(2)
(ii)
When Y is a small distance x from its equilibrium position, the resultant force per unit length on it is approximately −(λg/r)x, where λ is its mass per unit length. Show that Y performs simple harmonic motion and determine the period.
(2)
(iii)
Explain why Y would not remain above X if the glass guides were removed.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
To the right, horizontally (the field lines around X are clockwise when its current is into the page)
✓ 1
Part (a)(ii)
The force on Y must be upwards, away from X, to balance its weight — a repulsion
✓ 1
Parallel currents in the same direction attract, so the currents must be opposite; check: current out of the page in a field to the right gives an upward force by the left-hand rule
✓ 1
Part (b)(i)
Magnetic force per unit length = weight per unit length: μ0IXIY/(2πr) = λg
✓ 1
r = 4π × 10−7 × 50 × 5.0/(2π × 5.3 × 10−4 × 9.81) = 9.6 × 10−3 m
✓ 1
Must see the substitution or an unrounded 9.62 mm.
Part (b)(ii)
r ∝ IY, so r = 1.2 × 9.62 = 11.5 mm
✓ 1
Accept 11.5–11.6 mm. Allow ECF from (b)(i).
Part (b)(iii)
5.2 × 10−3 N m−1, vertically downwards (Newton's third law: equal and opposite to the force on Y)
✓ 1
Part (c)(i)
The magnetic force is inversely proportional to r, so when Y is closer to X the upward force increases above the weight
✓ 1
The resultant force is then upwards, towards the equilibrium position — a restoring force
✓ 1
Part (c)(ii)
Acceleration a = force per unit length ÷ λ = −(g/r)x: proportional to displacement and opposite in direction, so SHM with ω² = g/r
✓ 1
T = 2π√(r/g) = 2π√(9.62 × 10−3/9.81) = 0.20 s
✓ 1
Accept 0.196–0.20 s. Allow ECF from the value of r in (b)(i).
Part (c)(iii)
If Y moves slightly sideways, the repulsive force (along the line from X to Y) acquires a horizontal component
✓ 1
This component points away from the vertical through X, pushing Y further sideways, so the sideways equilibrium is unstable and Y slides off / falls
✓ 1
Answers: (a)(i) to the right · (b)(i) 9.6 mm · (b)(ii) 11.5 mm · (b)(iii) 5.2 × 10−3 N m−1 downwards · (c)(ii) 0.20 s (the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the force per unit length between parallel wires as given by F/L = μ0I1I2/(2πr); the direction of the magnetic field of a straight wire; the direction of the force on a current-carrying conductor; C.1 — conditions for simple harmonic motion Command term: Explain
45D-2-28
Charged particles in electric fields·D.3 Motion in electromagnetic fields
Paper 2Easy13 marks
Short answer & extended response8 steps to full marksCalculate
In a continuous ink-jet printer, ink droplets of mass 1.5 × 10−11 kg are each given a negative charge of magnitude 2.0 × 10−13 C. Each droplet travels horizontally at 18 m s−1 and enters, midway between them, the region between two horizontal deflecting plates. The plates are 1.2 cm long and 3.0 mm apart, and the potential difference between them is 900 V. The paper is 1.5 cm beyond the end of the plates. Assume the field between the plates is uniform and zero outside them.
(a)
The field and the force.
(i)
Calculate the electric field strength between the plates.
(1)
(ii)
Show that the acceleration of a droplet between the plates is about 4.0 × 103 m s−2.
(2)
(iii)
Explain why the effect of gravity on the droplet can be neglected.
(1)
(b)
The deflection.
(i)
Calculate the time a droplet spends between the plates.
(1)
(ii)
Determine the vertical deflection of the droplet as it leaves the plates.
(2)
(iii)
Determine the total vertical deflection of the droplet when it reaches the paper.
(3)
(c)
Path and energy.
(i)
State the shape of the path of the droplet between the plates and beyond them.
(1)
(ii)
Determine the increase in kinetic energy of the droplet between entering and leaving the plates.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
E = V/d = 900/3.0 × 10−3 = 3.0 × 105 V m−1
✓ 1
Part (a)(ii)
F = qE = 2.0 × 10−13 × 3.0 × 105 = 6.0 × 10−8 N
✓ 1
Allow ECF from (a)(i).
a = F/m = 6.0 × 10−8/1.5 × 10−11 = 4.0 × 103 m s−2
✓ 1
Part (a)(iii)
g ≪ a: the weight is about 410 times smaller than the electric force
✓ 1
Allow ECF from (a)(ii).
Part (b)(i)
t = 0.012/18 = 6.7 × 10−4 s
✓ 1
Part (b)(ii)
y = ½at² = ½ × 4.0 × 103 × (6.67 × 10−4)²
✓ 1
Allow ECF from (b)(i).
y = 8.9 × 10−4 m (0.89 mm)
✓ 1
Accept 0.88–0.90 mm.
Part (b)(iii)
Vertical velocity leaving the plates vy = at = 2.67 m s−1
✓ 1
Allow ECF from (b)(i).
Beyond the plates there is no force, so the droplet moves in a straight line: extra deflection = vy × (0.015/18) = 2.2 × 10−3 m
✓ 1
Total = 3.1 × 10−3 m (3.1 mm)
✓ 1
Allow ECF from (b)(ii). Award [2 max] for 4.5 mm (the field taken to extend to the paper).
Part (c)(i)
A parabola between the plates (constant force perpendicular to the initial velocity), then a straight line
✓ 1
Part (c)(ii)
Work done by the field = Fy = 6.0 × 10−8 × 8.9 × 10−4
✓ 1
Allow ECF from (b)(ii). ALT: ½mvy² = ½ × 1.5 × 10⁻¹¹ × 2.67².
ΔEk = 5.3 × 10−11 J
✓ 1
Accept 5.2–5.4 × 10⁻¹¹ J.
Answers: (a)(i) 3.0 × 105 V m−1 · (b)(i) 6.7 × 10−4 s · (b)(ii) 0.89 mm · (b)(iii) 3.1 mm · (c)(ii) 5.3 × 10−11 J (the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the motion of a charged particle in a uniform electric field; D.2 — the uniform electric field E = V/d between parallel plates and E = F/q; A.1 — the behaviour of projectiles, motion resolved into horizontal and vertical components Command term: Calculate
46D-2-33
Force on a current-carrying conductor·D.3 Motion in electromagnetic fields
Paper 2Easy10 marks
Short answer & extended response7 steps to full marksExplain
An electrodynamic shaker is used to vibrate small components for testing. A circular coil of 80 turns and radius 1.8 cm is fixed to a light table. The coil sits in the narrow circular gap of a permanent magnet, where the magnetic field is radial, has a flux density of 0.65 T and is everywhere perpendicular to the wire of the coil, as shown in the top view.
The table and coil have a total mass of 0.12 kg. They are supported by flexible mounts that behave as a spring of spring constant 3.0 × 103 N m−1.
Top view, not to scale. The current in the coil is anticlockwise as seen from above.
(a)
The force on the coil.
(i)
State and explain the direction of the resultant magnetic force on the coil.
(2)
(ii)
Show that the magnetic force on the coil is about 8.8 N when the current is 1.5 A.
(2)
(iii)
Calculate the initial acceleration of the table when a current of 1.5 A is switched on.
(1)
(b)
Vibrating the table.
(i)
The current of 1.5 A is kept on until the table is at rest again. Determine the displacement of the table from its original position.
(1)
(ii)
Calculate the natural frequency of oscillation of the table on its mounts.
(2)
(iii)
A sinusoidal current of constant amplitude 0.20 A is now passed through the coil, and its frequency is slowly increased from 5 Hz to 60 Hz. Explain how the amplitude of vibration of the table varies.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
On every part of the coil the force is perpendicular to both the current (along the wire) and the radial field, so it is along the axis of the coil; these forces all act in the same direction and add
✓ 1
Vertically downwards (into the page in the top view), by Fleming's left-hand rule
✓ 1
Accept "away from the observer looking down".
Part (a)(ii)
Length of wire in the field L = 80 × 2π × 0.018 = 9.05 m
✓ 1
F = BIL = 0.65 × 1.5 × 9.05 = 8.82 N
✓ 1
Full substitution or an answer to at least 3 s.f. is required.
Part (a)(iii)
a = F/m = 8.82/0.12 = 74 m s−2
✓ 1
Allow ECF from (a)(ii). Accept 73–74 m s⁻².
Part (b)(i)
x = F/k = 8.82/3.0 × 103 = 2.9 mm
✓ 1
Allow ECF from (a)(ii). Accept 2.9–3.0 mm.
Part (b)(ii)
T = 2π√(m/k) = 2π√(0.12/3.0 × 103) = 0.0397 s
✓ 1
f = 1/T = 25 Hz
✓ 1
Accept 25 Hz.
Part (b)(iii)
The coil exerts a periodic driving force of constant amplitude at the frequency of the current
✓ 1
The amplitude increases to a maximum when the driving frequency equals the natural frequency of about 25 Hz (resonance), and then decreases
✓ 1
Allow ECF from (b)(ii). The frequency of the maximum must be stated.
Answers: (a)(ii) 8.82 N · (a)(iii) 74 m s−2 · (b)(i) 2.9 mm · (b)(ii) 25 Hz (the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the magnitude and direction of the force on a current-carrying conductor in a magnetic field as given by F = BIL sin θ; A.2 — Newton's second law and Hooke's law; C.1 — the time period of a mass–spring system; C.4 — resonance Command term: Explain
47D-2-47
Ion thruster·D.3 Motion in electromagnetic fields
Paper 2Easy14 marks
Short answer & extended response9 steps to full marksDetermine
A spacecraft is driven by an ion thruster. Singly charged xenon ions, each of mass 131 u, are accelerated from rest through a potential difference of 1.10 kV by the uniform electric field between two parallel grids. The ions leave the thruster as a beam; the beam current is 1.60 A. Electrons are injected into the beam as it leaves.
(a)
(i)
Show that the speed of the ions leaving the grids is about 4.0 × 104 m s−1.
(2)
(b)
(i)
Calculate the number of ions leaving the thruster per second.
(1)
(ii)
Determine the thrust on the spacecraft.
(2)
(iii)
Show that the thrust per unit of beam power is 2/v.
(2)
(iv)
The accelerating potential difference is doubled and the beam power is unchanged. Deduce the new thrust.
(1)
(c)
The mass of the spacecraft is 600 kg.
(i)
Determine the time, in days, for the thruster to change the speed of the spacecraft by 1.00 km s−1. Assume the mass of the spacecraft stays constant.
(2)
(ii)
Determine the mass of xenon used in this time and comment on the assumption in (c)(i).
(2)
(d)
(i)
Suggest why electrons are injected into the beam as it leaves the thruster.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
eV = ½mv², with m = 131 × 1.66 × 10−27 = 2.17 × 10−25 kg
✓ 1
v = √(2 × 1.60 × 10−19 × 1100/2.17 × 10−25) = 4.02 × 104 m s−1
✓ 1
Full substitution or an answer to at least 3 s.f. is required.
Part (b)(i)
N = I/e = 1.60/1.60 × 10−19 = 1.0 × 1019 s−1
✓ 1
Part (b)(ii)
Thrust = rate of change of momentum of the ions = Nmv
✓ 1
Reference to the momentum given to the ions per second is required.
Allow ECF from (a)(i) and (b)(i). Accept 0.087–0.088 N.
Part (b)(iii)
Beam power P = IV = NeV and thrust = Nmv with N = I/e
✓ 1
ALT: power = N × ½mv².
Thrust/P = mv/(eV) = mv/(½mv²) = 2/v
✓ 1
Use of eV = ½mv² must be seen.
Part (b)(iv)
v ∝ √V, so v increases by √2 and the thrust (= 2P/v) falls to 0.087/√2 = 0.062 N
✓ 1
Allow ECF from (b)(ii) and (b)(iii). Accept 0.062 N.
Part (c)(i)
t = m∆v/F = 600 × 1000/0.0875 = 6.9 × 106 s
✓ 1
Allow ECF from (b)(ii).
= 79 days
✓ 1
Accept 78–80 days.
Part (c)(ii)
Mass = Nmt = 1.0 × 1019 × 2.17 × 10−25 × 6.9 × 106 = 15 kg
✓ 1
Allow ECF from (b)(i) and (c)(i). Accept 14–15 kg.
This is only about 2.5 % of 600 kg, so treating the mass of the spacecraft as constant is reasonable (the true time is very slightly shorter)
✓ 1
The comment must be consistent with their mass.
Part (d)(i)
Positive charge leaves the spacecraft in the beam, so without the electrons the spacecraft would become more and more negatively charged
✓ 1
The (electric) attraction between the negative spacecraft and the positive ions would pull the ions back, reducing the thrust; the electrons neutralise the beam so this does not happen
✓ 1
"To keep the spacecraft neutral" with no consequence scores [1 max].
Answers: (a)(i) 4.02 × 104 m s−1 · (b)(i) 1.0 × 1019 s−1 · (b)(ii) 0.087 N · (b)(iv) 0.062 N · (c)(i) 79 days · (c)(ii) 15 kg (the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the motion of a charged particle in a uniform electric field; D.2 (HL) — the work done in moving a charge q in an electric field as given by W = q∆Ve; B.5 — electric current as the rate of flow of charge; A.2 — force as the rate of change of momentum; A.3 — power Command term: Determine
48D-2-48
Current balance in a solenoid·D.3 Motion in electromagnetic fields
Paper 2Medium13 marks
Short answer & extended response8 steps to full marksDetermine
A current balance is used to measure the magnetic field inside a long solenoid. A light rectangular wire frame is pivoted on a horizontal axis outside the solenoid. The two arms of the frame are horizontal and parallel to the axis of the solenoid; the end section XY, of length 5.0 cm, is inside the solenoid, perpendicular to its axis, at a horizontal distance of 15.0 cm from the pivot axis. A rider of adjustable mass can be placed on the arms 10.0 cm from the pivot axis, on the other side. With no currents the frame is balanced horizontally.
The magnetic flux density inside a long solenoid is B = μ0nIs, where n is the number of turns per unit length and Is is the current in the solenoid. The diagram shows a top view.
Top view of the current balance: the field inside the solenoid points to the right; XY is the end section of the frame (not to scale).Mass of rider needed for balance against the square of the current (drawn to scale).
(a)
The solenoid has 2400 turns per metre and carries a current of 3.0 A.
(i)
Show that the magnetic flux density inside the solenoid is about 9 mT.
(1)
(ii)
Explain why the magnetic forces on the arms of the frame produce no turning effect about the pivot.
(2)
(b)
There is a current of 4.0 A in the frame.
(i)
Using the diagram, state the direction of the current in XY needed for the magnetic force on XY to be vertically downwards.
(1)
(ii)
Determine the mass of the rider needed to keep the frame horizontal.
(3)
(c)
The balance is moved to a different solenoid. The solenoid and the frame are now connected in series, so that they carry the same current I. The graph shows the mass of the rider needed for balance against I².
(i)
Show that the mass m of the rider needed is given by m = (μ0nLx/(gy))I², where L = 5.0 cm is the length of XY, x = 15.0 cm and y = 10.0 cm.
(2)
(ii)
Use the graph to determine the number of turns per unit length of this solenoid.
(2)
(d)
(i)
The connections to the supply are reversed, so that the current in both the solenoid and the frame is reversed. Explain whether the rider needs to be moved.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
B = 4π × 10−7 × 2400 × 3.0 = 9.05 × 10−3 T
✓ 1
An answer to at least 2 s.f. (9.0 mT) or full substitution is required.
Part (a)(ii)
Inside the solenoid the arms are parallel to the field, so F = BIL sin 0 = 0
✓ 1
Outside the solenoid the field is very weak (negligible), so the parts of the frame there experience negligible force
✓ 1
Accept "only XY is perpendicular to the field".
Part (b)(i)
From X to Y (towards the top of the page in the top view)
✓ 1
From F = IL × B or the left-hand rule.
Part (b)(ii)
F = BIL = 9.05 × 10−3 × 4.0 × 0.050 = 1.8 × 10−3 N
✓ 1
Allow ECF from (a)(i).
Moments about the pivot: 1.8 × 10−3 × 0.150 = mg × 0.100
✓ 1
Torques must be balanced; using equal forces: [0] for this mark.
m = 2.8 × 10−4 kg (0.28 g)
✓ 1
Accept 0.27–0.28 g.
Part (c)(i)
F = BIL with B = μ0nI, so F = μ0nI²L
✓ 1
Moments: μ0nI²L × x = mg × y, so m = (μ0nLx/(gy))I²
✓ 1
Allow ECF from the moment equation in (b)(ii).
Part (c)(ii)
Gradient of the line through the origin = 29.3 mg A−2 = 2.93 × 10−5 kg A−2
Allow ECF from (c)(i) and their gradient. Accept 2900–3150 m−1.
Part (d)(i)
Both the field in the solenoid and the current in XY reverse
✓ 1
The force depends on the product of I and B (F = IL × B), so it is still downwards with the same magnitude: the rider does not need to be moved
✓ 1
Allow ECF from (c)(i): m ∝ I² does not depend on the sign of I. Award this mark only with a reason.
Answers: (a)(i) 9.05 mT · (b)(i) from X to Y · (b)(ii) 0.28 g · (c)(ii) 3.0 × 103 m−1 · (d)(i) no (the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the magnitude and direction of the force on a current-carrying conductor in a magnetic field as given by F = BIL sin θ; D.2 — magnetic field patterns of a solenoid; A.4 (HL) — torque τ = Fr sin θ and rotational equilibrium; Tools 3 — gradient of a linear graph Command term: Determine
49D-2-49
Deflecting a proton beam back across a boundary·D.3 Motion in electromagnetic fields
Paper 2Medium13 marks
Short answer & extended response8 steps to full marksDetermine
Protons of kinetic energy 20.0 keV pass through a fine grid G into a region where there is a uniform field. Each proton crosses G at O, moving at an angle α = 40° to the grid, as shown. Two designs are compared for turning the protons back so that they cross G again.
Design E: a uniform electric field of strength 4.0 × 105 V m−1, perpendicular to G and directed towards G. Design M: a uniform magnetic field, parallel to the plane of G and perpendicular to the plane of the diagram. Gravitational effects are negligible.
A proton crossing the grid G at O at an angle α = 40° (not to scale).
(a)
Design E is used.
(i)
State the shape of the path of a proton in the field region, giving a reason.
(1)
(ii)
Show that the greatest distance of a proton from G is about 21 mm.
(2)
(iii)
Show that the distance from O to the point where the proton crosses G again is s = 2Ek sin 2α/(eE).
(2)
(iv)
Calculate this distance.
(1)
(b)
Design M is used instead. The protons must cross G again at the same point as in design E.
(i)
State the direction of the magnetic field needed, relative to the plane of the diagram.
(1)
(ii)
Determine the magnetic flux density needed.
(3)
(iii)
Compare, for the two designs, the time a proton spends in the field region.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
A parabola: the force eE is constant and is not parallel to the initial velocity
✓ 1
The reason is required.
Part (a)(ii)
Component of kinetic energy associated with the velocity perpendicular to G = Ek sin² α; at the greatest distance this has all been transferred to electric potential energy
✓ 1
ALT: v⊥ = v sin 40° = 1.26 × 106 m s−1 with v = √(2Ek/m) = 1.96 × 106 m s−1, and d = v⊥²/(2eE/m).
d = Ek sin² α/(eE) = 20.0 × 103 × sin² 40°/(4.0 × 105) = 2.07 × 10−2 m
✓ 1
Full substitution or an answer to at least 3 s.f. (20.7 mm) is required.
Part (a)(iii)
Time in the field: t = 2v sin α/a with a = eE/m
✓ 1
s = v cos α × t = 2v² sin α cos α/a = v² sin 2α × m/(eE) = 2Ek sin 2α/(eE)
✓ 1
Use of Ek = ½mv² must be seen.
Part (a)(iv)
s = 2 × 20.0 × 103 × sin 80°/(4.0 × 105) = 0.0985 m
✓ 1
Allow ECF from (a)(iii). Accept 98–99 mm.
Part (b)(i)
Out of the plane of the diagram
✓ 1
For a positive charge moving up and to the right, the force must be downwards and to the right (towards G).
Part (b)(ii)
The proton turns through 2α = 80° on a circular arc; the chord from O to the exit point is 2r sin α
✓ 1
The hidden geometric step. A diagram with the chord and the radius is acceptable.
r = 0.0985/(2 sin 40°) = 0.0766 m
✓ 1
Allow ECF from (a)(iv).
B = mv/(er) = 1.67 × 10−27 × 1.96 × 106/(1.60 × 10−19 × 0.0766) = 0.267 T
✓ 1
Accept 0.26–0.27 T. Taking the chord as a diameter (0.41 T) or as the radius (0.21 T): [2 max].
Part (b)(iii)
Design E: t = 2v sin α/a = 2 × 1.26 × 106/3.83 × 1013 = 6.6 × 10−8 s
✓ 1
Allow ECF from the speed in (a)(ii).
Design M: t = (2α)/ω with ω = eB/m: t = 1.40 rad/2.56 × 107 rad s−1 = 5.5 × 10−8 s
The proton spends longer in the electric field (by about 20 %)
✓ 1
The comparison must be consistent with their values.
Answers: (a)(i) parabola · (a)(ii) 20.7 mm · (a)(iv) 98.5 mm · (b)(i) out of the plane · (b)(ii) 0.267 T · (b)(iii) E: 6.6 × 10−8 s; M: 5.5 × 10−8 s (the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the motion of a charged particle in a uniform electric field; the motion of a charged particle in a uniform magnetic field; the magnitude and direction of the force on a charge moving in a magnetic field as given by F = qvB sin θ; Guidance: the kinetic energy of a charged particle stays constant in a magnetic field; A.1 — the behaviour of projectiles; A.2 — circular motion Command term: Determine
50D-2-50
A wire repelled by a parallel current·D.3 Motion in electromagnetic fields
Paper 2Hard13 marks
Short answer & extended response9 steps to full marksDetermine
A long straight horizontal wire P is fixed. A straight wire Q, of length 0.50 m and mass 5.1 g, hangs from two light vertical conducting threads, each 0.30 m long, so that it is parallel to P and level with it. With no current, the separation of the centres of P and Q is 4.0 mm. A current of 10 A is then passed through P and through Q, in opposite directions. Q swings away from P to a new equilibrium position in which the threads make a small angle θ with the vertical.
The graph shows how the magnetic force on Q varies with the separation r of the wires for a current of 10 A in each.
Magnetic force on Q against the separation r for currents of 10 A (drawn to scale).
(a)
(i)
Outline why Q moves away from P.
(1)
(ii)
Show that, when Q has moved a small horizontal distance x from its original position, the horizontal force needed to hold it there is approximately mgx/l, where m is the mass of Q and l is the length of the threads.
(2)
(b)
(i)
By drawing a suitable line on the graph, determine the equilibrium separation of the wires.
(2)
(ii)
Determine the magnetic flux density due to P at the position of Q in this equilibrium position.
(1)
(iii)
Explain, with reference to the graph, why Q returns to this equilibrium position if it is pushed slightly further away from P and released.
(2)
(c)
The current in each wire is doubled to 20 A.
(i)
Determine the new equilibrium separation of the wires.
(3)
(ii)
Show that the small-angle approximation used in (a)(ii) is still valid.
(1)
(d)
(i)
State the magnitude and direction of the magnetic force on P when the currents are 20 A and Q is in equilibrium.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
The currents are in opposite directions, so the wires repel (currents in the same direction attract)
✓ 1
Accept an argument using the field of P at Q and the left-hand rule.
Part (a)(ii)
Resolving for Q: T cos θ = mg and T sin θ = Fh, so Fh = mg tan θ
✓ 1
T is the total tension in the threads.
For a small angle tan θ ≈ sin θ = x/l, so Fh ≈ mgx/l
✓ 1
Part (b)(i)
Straight line from (4.0 mm, 0) with gradient mg/l = 0.167 N m−1, e.g. through (16 mm, 2.00 mN)
✓ 1
Allow ECF from (a)(ii). The line must start at r = 4.0 mm, not at the origin.
Intersection with the curve: r = 10.0 mm
✓ 1
Accept 9.5–10.5 mm.
Part (b)(ii)
From the graph F = 1.0 mN, so B = F/(IL) = 1.0 × 10−3/(10 × 0.50) = 2.0 × 10−4 T
✓ 1
Allow ECF from (b)(i). Accept 1.9–2.1 × 10−4 T.
Part (b)(iii)
For r greater than the equilibrium value the magnetic force (curve) is smaller than the horizontal restoring force from the threads and weight (line)
✓ 1
Allow ECF from the line drawn in (b)(i).
So the resultant horizontal force is towards P, back towards equilibrium (and for smaller r it is away from P): the equilibrium is stable
✓ 1
Part (c)(i)
At any separation the magnetic force is now 4 times larger: 4C/r = (mg/l)(r − 4.0 mm), where C/r is the force read from the graph
Allow ECF from (b)(i). ALT: substitution of C = 1.0 × 10−5 N m, mg/l = 0.167 N m−1.
r = 17.6 mm
✓ 1
Accept 17–18 mm. Multiplying the displacement by 4 at the original separation (28 mm): [1 max].
Part (c)(ii)
sin θ = (17.6 − 4.0)/300 = 0.045, so θ ≈ 2.6°, for which tan θ and sin θ differ by about 0.1 %
✓ 1
Allow ECF from (c)(i).
Part (d)(i)
4C/r = 4 × 1.0 × 10−5/1.76 × 10−2 = 2.3 mN, horizontally away from Q (Newton's third law: equal and opposite to the force on Q)
✓ 1
Allow ECF from (c)(i). Both magnitude and direction are needed.
Answers: (b)(i) 10.0 mm · (b)(ii) 2.0 × 10−4 T · (c)(i) 17.6 mm · (d)(i) 2.3 mN, away from Q (the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the force per unit length between parallel wires as given by F/L = μ0I1I2/(2πr); Guidance: the force is attractive when the currents are in the same direction; the magnitude and direction of the force on a current-carrying conductor in a magnetic field as given by F = BIL sin θ; A.2 — equilibrium of forces, resolving forces; D.2 — the magnetic field of a straight wire Command term: Determine
51D-2-51
Helical paths and magnetic focusing·D.3 Motion in electromagnetic fields
Paper 2Hard14 marks
Short answer & extended response8 steps to full marksDetermine
Electrons are accelerated from rest through a potential difference of 600 V and leave an electron gun inside a long solenoid, where there is a uniform magnetic field of flux density B directed along the axis of the solenoid. The beam is slightly divergent: the electrons leave the gun at small angles of up to 5.0° to the axis.
(a)
(i)
Show that the speed of the electrons leaving the gun is about 1.5 × 107 m s−1.
(1)
(ii)
An electron leaves the gun at an angle to the axis. Explain why its path is a helix.
(2)
(b)
B = 2.0 mT and the electron leaves the gun at 5.0° to the axis.
(i)
Calculate the radius of the helix.
(2)
(ii)
Determine the distance travelled along the axis while the electron completes one revolution of the helix.
(3)
(c)
(i)
Explain why all the electrons leaving the gun at small angles return to the axis at very nearly the same point.
(2)
(d)
The arrangement is used to measure the specific charge of the electron. A screen on the axis is a distance D = 0.40 m from the gun, and B is increased from zero until the beam is first focused to a point on the screen.
(i)
Show that e/me = 8π²V/(B²D²), where V is the accelerating potential difference.
(2)
(ii)
Calculate the value of B at the first focus, using the data booklet values of e and me.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
v = √(2eV/me) = √(2 × 1.60 × 10−19 × 600/9.11 × 10−31) = 1.45 × 107 m s−1
✓ 1
Full substitution or an answer to at least 3 s.f. is required.
Part (a)(ii)
The component of velocity parallel to the field gives no force (sin 0 = 0), so the electron moves along the axis at a constant speed
✓ 1
The component of velocity perpendicular to the field gives a force of constant magnitude perpendicular to the velocity, so the motion seen along the axis is a circle; the two motions together give a helix
✓ 1
Part (b)(i)
Perpendicular component of velocity = 1.45 × 107 × sin 5.0° = 1.27 × 106 m s−1
✓ 1
Allow ECF from (a)(i). Using the full speed: [1 max] (4.1 × 10−2 m).
T = 2πme/(eB) = 2π × 9.11 × 10−31/(1.60 × 10−19 × 2.0 × 10−3) = 1.79 × 10−8 s
✓ 1
Component of velocity along the axis = 1.45 × 107 cos 5.0° = 1.45 × 107 m s−1
✓ 1
Allow ECF from (a)(i).
Distance = 1.45 × 107 × 1.79 × 10−8 = 0.259 m
✓ 1
Accept 0.26 m (using the full speed gives 0.260 m, which also scores this mark).
Part (c)(i)
The time for one revolution, 2πm/(eB), does not depend on the perpendicular speed or the angle, so every electron is back on the axis after the same time T
✓ 1
In this time each electron moves v cos θ × T along the axis, and cos θ ≈ 1 for small angles (cos 5° = 0.996), so the distances are almost the same
✓ 1
Allow ECF from (b)(ii).
Part (d)(i)
First focus: D = vT = v × 2πme/(eB), with v² = 2eV/me
B = √(8π²V/((e/me)D²)) with e/me = 1.76 × 1011 C kg−1
✓ 1
Allow ECF from (d)(i).
B = √(8π² × 600/(1.76 × 1011 × 0.40²)) = 1.3 × 10−3 T
✓ 1
Accept 1.3 mT.
Answers: (a)(i) 1.45 × 107 m s−1 · (b)(i) 3.6 mm · (b)(ii) 0.26 m · (d)(ii) 1.30 mT (the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the motion of a charged particle in a uniform magnetic field; the magnitude and direction of the force on a charge moving in a magnetic field as given by F = qvB sin θ; Guidance: the kinetic energy of a charged particle stays constant in a magnetic field; Guidance: the determination of the charge-to-mass ratio for a charged particle by investigating its path in a uniform magnetic field; D.2 (HL) — W = q∆Ve; A.1 — motion with constant velocity; A.2 — circular motion Command term: Determine
52D-2-58
A magnetic spectrometer for alpha particles·D.3 Motion in electromagnetic fields
Paper 2Hard19 marks
Short answer & extended response12 steps to full marksDetermine
Americium-241 (24195Am) decays by alpha emission to an isotope of neptunium (Np). Most of the alpha particles are emitted with a kinetic energy of 5.49 MeV. In a magnetic spectrometer, alpha particles from a thin source enter a vacuum chamber through a slit S. A uniform magnetic field of flux density 0.80 T, perpendicular to the plane of the diagram, makes them travel through a semicircle to a detector plate D, as shown.
Mass of an alpha particle = 6.64 × 10−27 kg; mass of the neptunium nucleus = 237 u.
Not to scale. The magnetic field is not shown.
(a)
The decay.
(i)
Write down the nucleon number and the proton number of the neptunium nucleus.
(1)
(ii)
Show that the speed of a 5.49 MeV alpha particle is about 1.6 × 107 m s−1.
(1)
(iii)
The americium nucleus is at rest before it decays. Determine, in keV, the kinetic energy of the recoiling neptunium nucleus.
(2)
(b)
The spectrometer.
(i)
State the direction of the magnetic field in the chamber.
(1)
(ii)
Calculate the radius of the path of the 5.49 MeV alpha particles.
(2)
(iii)
In a separate experiment, recoiling neptunium ions, each carrying a charge of +2e, enter the same magnetic field moving perpendicular to it. Deduce, without calculating their speed, the radius of their path.
(3)
(iv)
Compare the time taken by a neptunium ion to travel through a semicircle with the time taken by an alpha particle.
(2)
(c)
Nuclear energy levels. A second group of alpha particles from the same source has a kinetic energy of 5.44 MeV.
(i)
Explain how the two groups of alpha particles provide evidence for discrete energy levels in the neptunium nucleus.
(2)
(ii)
Determine, in pm, the wavelength of the gamma photon emitted when the neptunium nucleus returns from this excited state to its ground state.
(2)
(d)
The 5.49 MeV alpha particles are directed head-on at the nuclei in a thin gold foil (Z = 79).
(i)
Estimate the distance of closest approach of an alpha particle to a gold nucleus.
(2)
(ii)
State one assumption made in your estimate.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Nucleon number 237 and proton number 93
✓ 1
Both required.
Part (a)(ii)
v = √(2Ek/m) = √(2 × 5.49 × 1.60 × 10−13/6.64 × 10−27) = 1.63 × 107 m s−1
✓ 1
The conversion from MeV to J must be seen; answer to at least 3 s.f.
Part (a)(iii)
Momentum is conserved, so the two momenta are equal in magnitude; Ek = p²/(2m), so EkNp = Ekα × mα/mNp
From F = qv × B for a positive particle moving up the page and deflected to the left.
Part (b)(ii)
r = mv/(qB) with q = 2e = 3.20 × 10−19 C
✓ 1
Award [1 max] for 0.84 m (charge e used).
r = 6.64 × 10−27 × 1.63 × 107/(3.20 × 10−19 × 0.80) = 0.42 m
✓ 1
Allow ECF from (a)(ii).
Part (b)(iii)
r = mv/(qB) = p/(qB): the radius depends only on the momentum and the charge
✓ 1
Momentum is conserved in the decay of the americium nucleus at rest, so the neptunium nucleus has momentum equal in magnitude to that of the alpha particle
✓ 1
Allow reference to the reasoning in (a)(iii).
Same momentum and same charge (2e), so the radius is the same: 0.42 m
✓ 1
Allow ECF from (b)(ii). A calculation via the speed of the ion scores [2 max].
Part (b)(iv)
t = πr/v; the radii are equal and the momenta are equal, so t ∝ 1/v ∝ m (or t = πm/(qB) with the same q)
✓ 1
tNp/tα = (237 × 1.66 × 10−27)/(6.64 × 10−27) = 59: the neptunium ion takes about 59 times as long (tα = 8.2 × 10−8 s, tNp = 4.8 × 10−6 s)
✓ 1
Accept 59–60. Allow ECF from (b)(ii).
Part (c)(i)
The energy released in the decay is the same each time, so an alpha particle with less kinetic energy leaves the neptunium nucleus in an excited state
✓ 1
Only particular alpha energies are observed, so the nucleus can have only particular (discrete) energies; it then emits a gamma photon of definite energy
Allow ECF from the energy difference used. The answer must be given in pm.
Part (d)(i)
At closest approach all the kinetic energy has become electric potential energy: Ek = k(2e)(79e)/d
✓ 1
d = 8.99 × 109 × 2 × 79 × (1.60 × 10−19)²/8.78 × 10−13 = 4.1 × 10−14 m
✓ 1
Allow ECF from the kinetic energy in joules used in (a)(ii).
Part (d)(ii)
The gold nucleus does not recoil (remains at rest) / only the electric force acts / the alpha particle loses no energy in the foil before the collision
✓ 1
Answers: (a)(i) 237; 93 · (a)(ii) 1.63 × 107 m s−1 · (a)(iii) 93 keV · (b)(i) into the page · (b)(ii) 0.42 m · (b)(iii) 0.42 m (the same as for the alpha particles) · (b)(iv) about 59 times as long · (c)(ii) 25 pm · (d)(i) 4.1 × 10−14 m (the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the motion of a charged particle in a uniform magnetic field; the magnitude and direction of F = qvB sin θ; D.2 (HL) — Ep = kq1q2/r; E.3 — alpha decay equations; the spectrum of alpha and gamma radiations as evidence for discrete nuclear energy levels; E.1 (HL) — the distance of closest approach in head-on scattering; A.2 — conservation of momentum in explosions; E.1 — E = hfCommand term: Determine
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