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IB Physics Higher Level · Topic D.3

D.3 Motion in electromagnetic fields

79 free, high-quality past-paper-style questions on Motion in electromagnetic fields for IB Physics HL, each with a full mark scheme. Covers: Force on a moving charge; Charged particles in electric fields; Direction of the magnetic force; and more. For when you have finished the past papers, or want to practise one topic at a time.D.3 at Standard Level →Read the D.3 notes →My revision plan →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.354 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked

1D-1A-50Force on a moving charge Paper 1AEasy1 mark

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?

B = 0.050 Tα particlev = 1.5 × 10⁷ m s⁻¹40°
Diagram not to scale
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Marking pointMarkNotes
Step 1F = qvB sin θ, where θ is the angle between the velocity and the field.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2The alpha particle carries charge +2e = 3.20 × 10−19 C.—
Step 3F = 3.20 × 10−19 × 1.5 × 107 × 0.050 × sin 40° = 1.5 × 10−13 N.✓ 1Answer 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.
2D-1A-66Charged particles in electric fields Paper 1AEasy1 mark

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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Marking pointMarkNotes
Step 1The force on the electron, eE, is opposite to the field and so opposite to its velocity: it decelerates with a = eE/m.—All 3 steps must be completed — there is no mark for a part-answer.
Step 20 = v² − 2as gives s = v²/(2a) = mv²/(2eE).—
Step 3Alternatively, the work done against the electric force equals the initial kinetic energy: eEs = ½mv².✓ 1Answer 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.
3IBP2H-033-01-01Direction of the magnetic force Paper 1AEasy1 mark

In a laboratory, a uniform magnetic field points vertically upwards. An electron moves horizontally due east through the field.
What is the direction of the magnetic force on the electron?

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Marking pointMarkNotes
For a positive charge, F = qv × B. With v due east and B vertically up, v × B points due south. The electron is negative, so the force is reversed and points north. The force cannot be along B (upwards) or opposite to v (west), because the magnetic force is always perpendicular to both v and B.✓ 1Answer B

Answer: B

Common error: Applying the right-hand rule and forgetting to reverse the result for a negative charge gives 'south', the answer for a positive charge.

4IBP2H-033-01-02Sign of a charge from its path Paper 1AEasy1 mark

In a storage-ring experiment, a muon moves at constant speed in a circle in a uniform magnetic field. The muon's velocity is perpendicular to the field.
Which quantity remains constant?

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Marking pointMarkNotes
The magnetic force is always perpendicular to the velocity, so it does no work on the muon. The speed, and therefore the kinetic energy \(\frac{1}{2}mv^2\), stays constant. Velocity and momentum are vectors along the direction of motion, and that direction changes continuously. The acceleration is a vector directed towards the centre of the circle, and that direction also rotates. None of these three is constant, even though their magnitudes are.✓ 1Answer C

Answer: C

Exam tip: Constant speed is not the same as constant velocity. In uniform circular motion, the velocity, momentum and centripetal acceleration keep their magnitudes but continuously change direction, so none of them is constant.

5IBP2H-033-00-01Force on a moving charge Paper 1AEasy1 mark

A cosmic-ray proton passes through a laboratory magnet at a speed of 2.0 × 106 m s−1. Its velocity makes an angle of 30° with a uniform magnetic field of flux density 0.50 T.
What is the magnitude of the magnetic force on the proton? (e = 1.6 × 10−19 C)

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Marking pointMarkNotes
F = qvB sin θ, where θ is the angle between the velocity and the field. Only the velocity component perpendicular to B contributes. F = 1.6 × 10−19 × 2.0 × 106 × 0.50 × sin 30° = 8.0 × 10−14 N.
1.4 × 10−13 N comes from using cos 30°. 1.6 × 10−13 N ignores the angle. Zero would only be correct if the proton moved parallel to the field.
✓ 1Answer A

Answer: A

Common error: In F = qvB sin θ the angle is measured between v and B. Check whether a question gives the angle to the field or to its normal.

6IBP2H-033-00-08Mass spectrometer Paper 1AEasy1 mark

A charged dust grain in space moves in a region containing only a uniform magnetic field. The grain does not move parallel to the field.
Which quantity remains constant?

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Marking pointMarkNotes
The magnetic force is always perpendicular to the velocity, so it does no work. The kinetic energy, and therefore the speed, is constant.
The direction of motion changes continuously, so the velocity and momentum (both vectors) change. The acceleration has constant magnitude, but its direction rotates, so the acceleration vector is not constant either.
✓ 1Answer D

Answer: D

Exam tip: A magnetic force can change a particle's direction but never its speed. Only an electric field can change the kinetic energy of a charge.

7IBP3H-D3-20r42-01Forces between parallel currents Paper 1AEasy1 mark

Two long, straight, parallel cables on a pumping-station wall carry currents in the same direction. Which statement about the forces between the cables is correct?

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Marking pointMarkNotes
Each cable sits in the magnetic field of the other. For currents in the same direction the force on each cable points towards the other, so they attract. By Newton's third law the two forces are equal in magnitude, whatever the currents. Repulsion is the result for opposite currents, and the forces are never unequal because they form an interaction pair.✓ 1Answer A

Answer: A

Common error: Parallel currents in the same direction attract, which is the opposite of the rule for like electric charges.

8IBP3H-D3-20r42-03Force on a moving charge Paper 1AEasy1 mark

A proton of charge \(q\) moves with speed \(v\) through a uniform magnetic field of flux density \(B\). Its velocity makes an angle of \(30^\circ\) to the field direction. What is the magnitude of the magnetic force on the proton?

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Marking pointMarkNotes
The force is \(qvB\sin\theta\) with \(\theta\) the angle between velocity and field, so \(F=qvB\sin30^\circ=0.5\,qvB\). Using \(qvB\) assumes a right angle; \(0.87\,qvB\) uses the cosine, which is the component of velocity along the field; zero would apply only to motion parallel to the field.✓ 1Answer B

Answer: B

Common error: Only the velocity component perpendicular to the field produces a force, so the sine of the angle between \(v\) and \(B\) is needed.

9IBP3H-D3-20r42-08Sign of a charge from its path Paper 1AEasy1 mark

A charged particle moves in a circular path of period \(T\) in a uniform magnetic field, with its velocity perpendicular to the field. The speed of the particle is doubled and the field is unchanged. What is the new period of the circular motion?

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Marking pointMarkNotes
The period is \(T=\frac{2\pi m}{qB}\), which does not contain the speed. Doubling the speed doubles the radius but also doubles the distance travelled in each revolution, so the time per revolution is unchanged. \(2T\) comes from thinking the period grows with radius, \(T/2\) from thinking faster means shorter period, and \(4T\) from squaring the speed factor.✓ 1Answer A

Answer: A

Common error: A larger speed gives a larger circle in the same time, not a shorter period, because \(r\) is proportional to \(v\).

10D-1A-05Charged particles in magnetic fields Paper 1AMedium1 mark

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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Marking pointMarkNotes
Step 1eV = ½mv², so v ∝ √V: quadrupling V doubles the speed.—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.—
Step 3The radius doubles: 2r.✓ 1Answer 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.
11D-1A-06Velocity selector Paper 1AMedium1 mark

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?

Epositive ionv× : magnetic field B into the page
Diagram not to scale
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Marking pointMarkNotes
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.—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.—
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.✓ 1Answer 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.
12D-1A-08Charged particles in magnetic fields Paper 1AMedium1 mark

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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Marking pointMarkNotes
Step 1Radius r = mv/(qB) and period T = 2πr/v.—All 4 steps must be completed — there is no mark for a part-answer.
Step 2Combining: T = 2πm/(qB) — independent of the speed.—
Step 3Alpha: mass 4mp, charge 2e, so m/q = 4/2 = 2.—
Step 4Tα/Tp = (4/2)/(1/1) = 2.✓ 1Answer 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.
13D-1A-11Force on a current-carrying conductor Paper 1AMedium1 mark

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?

θrod (end view)threadsuniform vertical field B
Seen along the rod; not to scale
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Marking pointMarkNotes
Step 1The rod is perpendicular to the vertical field, so the magnetic force BIL is horizontal (perpendicular to both the rod and the field).—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.—
Step 3Dividing: tan θ = BIL/mg, so I = mg tan θ/(BL).✓ 1Answer 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.
14D-1A-20Charged particles in electric fields Paper 1AMedium1 mark

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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Marking pointMarkNotes
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.—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.—
Step 3Gain in kinetic energy = e × V/2, so the final kinetic energy is Ek + eV/2.✓ 1Answer 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.
15D-1A-24Forces between parallel currents Paper 1AMedium1 mark

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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Marking pointMarkNotes
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.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Acceleration = (force per unit length)/(mass per unit length).—
Step 3Ratio = (mass per unit length of Y)/(mass per unit length of X) = 1/2.✓ 1Answer 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.
16D-1A-28Sign of a charge from its path Paper 1AMedium1 mark

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?

vFOmagnetic field out of the page
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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Marking pointMarkNotes
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.—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.—
Step 3r = mv/qB ∝ v: doubling the speed doubles the radius (the period 2πm/qB stays the same).✓ 1Answer 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.
17D-1A-52Specific charge from a circular path Paper 1AMedium1 mark

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?

B = 0.20 Tion sourceE = 3.0 × 10⁴ V m⁻¹B = 0.060 Tdetectorslit5.2 cm
Path in the magnetic field; not to scale
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Marking pointMarkNotes
Step 1Selector: qE = qvB, so v = E/B = 3.0 × 104/0.060 = 5.0 × 105 m s−1.—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.—
Step 3r = mv/(qB) gives q/m = v/(Br) = 5.0 × 105/(0.20 × 0.026) = 9.6 × 107 C kg−1 (these are protons).✓ 1Answer C

Answer: C  ·  3 stages of work, one mark

Every option, and why

  • 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.
18D-1A-53Force between parallel wires Paper 1AMedium1 mark

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?

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Marking pointMarkNotes
Step 1F/L = μ0I1I2/(2πr) = 4π × 10−7 × (2.0 × 103)²/(2π × 0.25) = 3.2 N m−1.—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.—
Step 3The currents are in opposite directions, so the force is repulsive: the supports must hold the busbars apart.✓ 1Answer 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.
19D-1A-54Direction of the magnetic force Paper 1AMedium1 mark

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?

Ilong straight wireelectronvrplane of the page
The wire and the electron's velocity both lie in the plane of the page.
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Marking pointMarkNotes
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.—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).—
Step 3The electron is negative, so the force on it is reversed: away from the wire.✓ 1Answer 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.
20D-1A-67Charged particles in electric fields Paper 1AMedium1 mark

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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Marking pointMarkNotes
Step 1E = V/d, so doubling d halves the field strength and therefore halves the acceleration a = eV/(md).—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.—
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.)✓ 1Answer 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.
21D-1A-68Charged particles in magnetic fields Paper 1AMedium1 mark

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?

××××××××××××××××××××××××××××protonθw
Not to scale. The magnetic field (×) is directed into the page.
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Marking pointMarkNotes
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.—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°.✓ 1Answer 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.
22IBP2S-032-01-07Forces between parallel currents Paper 1AMedium1 mark

Two long parallel wires carry currents I and 3I in the same direction. Which statement describes the magnetic forces between the wires?

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Marking pointMarkNotes
The force per unit length is F/L = μ0I1I2/(2πr). This expression is the same for both wires, as Newton's third law requires. Currents in the same direction attract. The options with unequal forces ignore Newton's third law. The 'repulsive' options confuse same-direction currents with opposite-direction currents.✓ 1Answer B

Answer: B

Common error: The wire with the larger current does not feel the larger force. The two forces are a Newton's third law pair.

23IBP2S-032-01-08Forces between parallel currents Paper 1AMedium1 mark

Two parallel power cables in an electrical substation carry currents and exert a force per unit length F on each other. Both currents are doubled. The separation of the cables is also doubled. What is the new force per unit length?

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Marking pointMarkNotes
The force per unit length is F/L ∝ I1I2/r. Doubling both currents multiplies the force by 4. Doubling the separation divides it by 2. The net factor is 4/2 = 2, so the new force is 2F. The option 4F ignores the separation. The option F treats the force as inverse-square in r, so the factors appear to cancel.✓ 1Answer C

Answer: C

Common error: The force between long parallel wires varies as 1/r, not 1/r².

24IBP2H-033-01-03Sign of a charge from its path Paper 1AMedium1 mark

A particle detector contains a uniform magnetic field directed out of the plane of the page. A charged particle leaves a track that spirals clockwise and inwards, so its radius gets steadily smaller.
What is the sign of the particle's charge, and what is happening to its kinetic energy?

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Marking pointMarkNotes
Consider a positive charge moving down the page on the right-hand side of a circle, with B out of the page. F = qv × B then points to the left, towards the centre, so a positive charge circulates clockwise. The particle is therefore positive.
The radius is r = mv/(qB). With q and B fixed, a shrinking radius means v is falling. The particle is losing kinetic energy, for example through collisions with the detector material. The magnetic force itself does no work.
✓ 1Answer A

Answer: A

Common error: An inward spiral is not evidence that the field is pulling the particle in more strongly. The field is uniform, so a smaller radius means less momentum.

25IBP2H-033-01-04Charged particles in magnetic fields Paper 1AMedium1 mark

In a beamline for a radiotherapy research facility, protons and alpha particles (He2+) start from rest. Both are accelerated through the same potential difference. They then enter the same uniform magnetic field at right angles to it.
What is \(\frac{r_\alpha}{r_p}\), the ratio of the radius of the alpha particle's path to that of the proton's path? Take the mass of an alpha particle as 4 times the proton mass.

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Marking pointMarkNotes
Accelerating through p.d. V gives qV = ½mv², so the momentum is mv = √(2mqV).
The radius is r = mv/(qB) = √(2mV/q)/B, so r ∝ √(m/q).
For the proton, m/q = mp/e. For the alpha particle, m/q = 4mp/2e = 2mp/e.
The ratio is therefore √2.
The distractors come from three mistakes:
  • '2' assumes both particles have the same speed.
  • '1' assumes both have the same kinetic energy, ignoring that the alpha particle gains 2eV.
  • '1/√2' inverts the ratio.
✓ 1Answer D

Answer: D

Exam tip: When particles are accelerated through the same p.d., write r in terms of V first. Do not assume the particles have equal speeds or equal kinetic energies.

26IBP2H-033-01-05Force on a moving charge Paper 1AMedium1 mark

A straight wire in a test rig makes an angle of 30° with a uniform magnetic field. The magnetic force on the wire is F.
The wire is rotated until it is perpendicular to the field, and at the same time the current is halved.
What is the new magnetic force on the wire?

Show mark scheme
Marking pointMarkNotes
Use F = BIL sin θ.
Initially, F = BIL sin 30° = 0.5BIL.
Afterwards, F' = B(I/2)L sin 90° = 0.5BIL = F.
The doubling from sin 30° to sin 90° exactly cancels the halving of the current. The distractors come from three mistakes:
  • '0' uses cos θ instead of sin θ.
  • 'F/2' ignores the change of angle.
  • '2F' ignores the halved current.
✓ 1Answer C

Answer: C

Common error: θ is the angle between the wire and the field, and the force uses sin θ. A wire parallel to the field feels no force.

27IBP2H-033-01-06Forces between parallel currents Paper 1AMedium1 mark

Two long, straight, parallel cables in a data centre are a distance d apart. One carries current I and the other carries current 3I, in the same direction. The magnetic force per unit length on the cable carrying I is F.
What is the force per unit length on the cable carrying 3I, and what is its nature?

Show mark scheme
Marking pointMarkNotes
The force per unit length is μ0I1I2/(2πd). This expression is symmetric in the two currents, so both cables experience the same magnitude F. This is also required by Newton's third law.
Currents in the same direction attract.
The distractors come from two mistakes:
  • '3F' assumes the larger current feels the larger force.
  • 'Repulsive' reverses the rule for parallel currents.
✓ 1Answer A

Answer: A

Exam tip: Forces between the two wires form a Newton's third law pair, so they are always equal in magnitude, whatever the currents.

28IBP2H-033-00-02Velocity selector Paper 1AMedium1 mark

A velocity selector uses perpendicular uniform electric and magnetic fields. Singly charged ions moving at speed v pass through it undeflected.
A doubly charged ion enters the same selector at speed 2v. What happens to this ion?

Show mark scheme
Marking pointMarkNotes
An ion is undeflected only when qE = qvB, which gives v = E/B. Both the charge and the mass cancel, so the selected speed is fixed by the fields alone. At speed 2v the magnetic force q(2v)B is twice the electric force qE, whatever the charge. The resultant force is therefore in the direction of the magnetic force.
Options A and D correctly state that charge and mass cancel, but wrongly conclude that the ion is undeflected. Only the speed determines balance. Option B reverses which force is larger.
✓ 1Answer C

Answer: C

Exam tip: A selector picks out a speed, not a charge or a mass. Compare qvB with qE: faster particles bend towards the magnetic force, slower ones towards the electric force.

29IBP2H-033-00-03Forces between parallel currents Paper 1AMedium1 mark

Two long, parallel power cables carry currents of 2I and I in opposite directions. Their separation is r, and the force per unit length on each cable has magnitude F.
Both currents are doubled and the separation is increased to 3r. What is the new force per unit length?

Show mark scheme
Marking pointMarkNotes
F/L = μ0I1I2/(2πr). Doubling both currents multiplies the force by 4. Tripling r divides it by 3, not by 9. The new force is therefore 4F/3. Antiparallel currents repel.
4F/9 treats the force as inverse-square. 2F/3 doubles only one current. 'Attractive' reverses the rule that like (parallel) currents attract.
✓ 1Answer D

Answer: D

Common error: The field of a long straight wire falls as 1/r, so the force between two wires also varies as 1/r, not 1/r².

30IBP2H-033-00-05Charged particles in electric fields Paper 1AMedium1 mark

An electron enters the uniform field between two parallel deflecting plates. Its initial velocity is parallel to the plates. The potential difference between the plates is then doubled, and the electron's entry speed is unchanged.
What happens to the time the electron spends between the plates and to its sideways deflection when it leaves the plates?

Show mark scheme
Marking pointMarkNotes
The force is perpendicular to the plates, so the velocity component along the plates is unchanged. The time in the field is therefore t = L/v, which does not change. Doubling V doubles E and hence doubles the acceleration a = eE/m. The deflection ½at2 is therefore doubled.
'Quadrupled' wrongly squares the change in field. The other options wrongly assume the field affects the motion parallel to the plates.
✓ 1Answer A

Answer: A

Exam tip: Treat deflection in a uniform field like projectile motion: constant velocity along the plates, uniform acceleration across them.

31IBP2H-033-00-06Charged particles in magnetic fields Paper 1AMedium1 mark

In a cyclotron used to produce medical isotopes, a proton moves in a circle perpendicular to a uniform magnetic field. The proton's speed is doubled while it stays in the same field.
What happens to the radius and to the period of its circular motion?

Show mark scheme
Marking pointMarkNotes
From qvB = mv2/r, the radius is r = mv/(qB), so doubling v doubles r. The period is T = 2πr/v = 2πm/(qB), which does not depend on speed. The period is therefore unchanged. This is why a fixed-frequency alternating voltage can keep accelerating the protons.
'Period doubled' accounts for the larger radius but not the higher speed. 'Radius quadrupled' links the radius to kinetic energy rather than to momentum.
✓ 1Answer C

Answer: C

Exam tip: T = 2πm/(qB) depends only on the specific charge and the field. This result can be derived quickly when needed.

32IBP3H-D3-20r42-02Forces between parallel currents Paper 1AMedium1 mark

Two long parallel wires carry currents \(I_1\) and \(I_2\) at separation \(d\), with force per unit length \(F\) on each wire. Both currents are doubled and the separation is also doubled. What is the new force per unit length?

Show mark scheme
Marking pointMarkNotes
Force per unit length is proportional to \(I_1I_2/d\). Doubling both currents gives a factor 4 in the numerator and doubling \(d\) divides by 2, so the result is \(2F\). Quadrupling would result from forgetting the change in \(d\); leaving the force at \(F\) would come from treating the force as inversely proportional to \(d^2\); \(F/2\) would come from inverting the effect of the currents.✓ 1Answer C

Answer: C

Exam tip: Write the proportionality first, then substitute each factor change separately.

33IBP3H-D3-20r42-04Sign of a charge from its path Paper 1AMedium1 mark

A velocity selector has an electric field pointing towards the top of the page and a magnetic field pointing out of the page. A positive ion moves to the right through the selector at a speed greater than \(E/B\). Towards which direction is the ion deflected?

Show mark scheme
Marking pointMarkNotes
For a positive ion moving right in a field pointing out of the page, the magnetic force is towards the bottom of the page, opposite to the electric force, which acts towards the top. At speed greater than \(E/B\) the magnetic force \(qvB\) exceeds the electric force \(qE\), so the net force, and the deflection, is towards the bottom. Towards the top would result if the electric force were wrongly taken to be larger. The magnetic force is perpendicular to both \(v\) and \(B\), so it can never act into or out of the page here, which rules out the other two.✓ 1Answer D

Answer: D

Exam tip: Find the direction of each force separately, then compare their sizes using \(qE\) and \(qvB\).

34IBP3H-D3-20r42-05Charged particles in magnetic fields Paper 1AMedium1 mark

Electrons are accelerated from rest through a potential difference \(V\) in the electron gun of a retro arcade monitor. They then move in a circle of radius \(r\) in a uniform magnetic field \(B\) perpendicular to their velocity. Which expression gives the specific charge \(e/m\) of the electron?

Show mark scheme
Marking pointMarkNotes
From \(eV=\frac12mv^2\) we get \(v^2=2eV/m\); from \(evB=mv^2/r\) we get \(v=eBr/m\). Eliminating \(v\) gives \(\frac{e}{m}=\frac{2V}{B^2r^2}\). \(\frac{V}{B^2r^2}\) comes from writing \(eV=mv^2\) and so losing the factor of 2. The last two expressions do not square \(B\) and \(r\) as the algebra requires, so they cannot be correct.✓ 1Answer B

Answer: B

Exam tip: Eliminate the speed between the energy equation and the circular-motion equation; the result must contain \(B\) and \(r\) squared.

35IBP3H-D3-20r42-07Force on a moving charge Paper 1AMedium1 mark

A straight section of a pump cable of length 0.50 m carries a current of 3.0 A. It lies in a uniform magnetic field of flux density 0.20 T, at \(30^\circ\) to the field direction. What is the magnitude of the magnetic force on the cable?

Show mark scheme
Marking pointMarkNotes
\(F=BIL\sin\theta=0.20\times3.0\times0.50\times\sin30^\circ=0.15\) N. The value 0.30 N omits the sine factor, 0.26 N uses the cosine, and 0.60 N divides by \(\sin30^\circ\) instead of multiplying.✓ 1Answer D

Answer: D

Common error: The angle in \(BIL\sin\theta\) is between the current direction and the field, and the sine always multiplies.

36D-1A-07Force between parallel wires Paper 1AHard1 mark

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?

XIYIZ2Id2d⊗ into the page ⊙ out of the page
Three long parallel wires perpendicular to the page (⊗ current into the page, ⊙ current out of the page).
Show mark scheme
Marking pointMarkNotes
Step 1X and Y carry currents in the same direction, so X attracts Y (towards X): F/L = μ0I·I/(2πd) = μ0I²/(2πd).—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).—
Step 3Both forces act towards X, so they add: μ0I²/(2πd) + μ0I²/(2πd) = μ0I²/(πd).✓ 1Answer 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.
37D-1A-27Direction of the magnetic force Paper 1AHard1 mark

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?

BIPQRS
A current-carrying rectangular loop in a uniform magnetic field in the plane of the page.
Show mark scheme
Marking pointMarkNotes
Step 1PQ and SR are parallel (or antiparallel) to the field, so F = BIL sin 0° = 0 on them.—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.—
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).✓ 1Answer 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.
38D-1A-51Charged particles in electric fields Paper 1AHard1 mark

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
Show mark scheme
Marking pointMarkNotes
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.—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.—
Step 3Substituting v² = 2qV/m: y = EL²/(4V) — independent of q and m, so the deflections are equal.✓ 1Answer 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.
39D-1A-55Charged particles in magnetic fields Paper 1AHard1 mark

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
Marking pointMarkNotes
Step 1r = mv/(qB) and the momentum is mv = √(2mEk), so r = √(2mEk)/(qB) ∝ √m/q at the same Ek and B.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Deuteron: √2/1 = √2 times the proton value.—
Step 3Alpha particle: √4/2 = 1, the same as the proton. So rp = rα < rd.✓ 1Answer 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/(√m q)): 1 : 1/√2 : 1/4.
40IBP2H-033-01-07Specific charge from a circular path Paper 1AHard1 mark

In a mass spectrometer used to check the purity of a pharmaceutical sample, positive ions first pass undeflected through a velocity selector. The selector has an electric field of strength E and a magnetic field of strength B1, perpendicular to each other.
The ions then enter a region containing only a uniform magnetic field of strength B2. They complete a semicircle and strike a detector a distance D from the entry slit, where D is the diameter of the semicircle.
What is the specific charge q/m of the ions?

Show mark scheme
Marking pointMarkNotes
In the selector, qE = qvB1, so v = E/B1.
In the second field, the radius is r = D/2 = mv/(qB2).
Rearranging gives q/m = v/(B2r) = 2v/(B2D) = 2E/(B1B2D).
The distractors come from two mistakes:
  • Option A treats the diameter as the radius.
  • Options C and D use the wrong selector speed, v = EB1 or v = B1/E.
✓ 1Answer B

Answer: B

Common error: The detector position measures the diameter of the semicircle, not the radius. Halve it before using r = mv/(qB).

41IBP2H-033-01-08Direction of the magnetic force Paper 1AHard1 mark

A long straight wire carries a steady conventional current. An electron moves parallel to the wire, a short distance from it, in the same direction as the conventional current.
What is the direction of the magnetic force on the electron?

Show mark scheme
Marking pointMarkNotes
The field of the wire forms circles around it, so at the electron's position B is perpendicular both to the wire and to the radial direction.
A positive charge moving in the direction of the current behaves like a parallel current, so it is attracted towards the wire.
The electron has negative charge, so the force on it is reversed: it is pushed away from the wire. Equivalently, the electron's motion corresponds to a conventional current flowing opposite to the wire's current, and antiparallel currents repel.
There is a force because v is perpendicular to B, even though v is parallel to the wire. The force must be perpendicular to v, so it cannot act along the wire.
✓ 1Answer D

Answer: D

Exam tip: Treat a moving charge as a small current element. Remember that an electron's conventional current points opposite to its velocity.

42IBP2H-033-00-04Charged particles in magnetic fields Paper 1AHard1 mark

In a radiotherapy research facility, a proton and an alpha particle have the same kinetic energy. Each moves perpendicular to the same uniform magnetic field. The alpha particle has 4 times the mass and 2 times the charge of the proton.
What is the ratio (radius of alpha particle's path)/(radius of proton's path)?

Show mark scheme
Marking pointMarkNotes
The radius is r = mv/(qB) = p/(qB), and the momentum is p = √(2mEk). So r = √(2mEk)/(qB). At equal Ek, the ratio is √(mα/mp) × (qp/qα) = √4 × ½ = 1.
A ratio of 2 assumes equal speeds, giving m/q. √2 uses the square root of the mass ratio but ignores the charge. 1/2 considers only the charge.
✓ 1Answer B

Answer: B

Exam tip: When kinetic energies (or accelerating voltages) are equal, rewrite mv as √(2mEk) before comparing radii.

43IBP2H-033-00-07Charged particles in magnetic fields Paper 1AHard1 mark

In a fine-beam tube, electrons are accelerated from rest through a potential difference V. They then move in a circle of radius r in a uniform magnetic field of flux density B.
The accelerating potential difference is increased to 4V. By what factor must B be multiplied to keep the radius equal to r?

Show mark scheme
Marking pointMarkNotes
From eV = ½mv2, the speed is v = √(2eV/m). From evB = mv2/r, the radius is r = mv/(eB) = √(2mV/e)/B. Keeping r constant requires B ∝ √V. Quadrupling V therefore requires B to be multiplied by 2.
A factor of 4 assumes B ∝ V. √2 takes the square root twice. 1/2 has the dependence inverted: a faster electron needs a stronger field, not a weaker one.
✓ 1Answer B

Answer: B

Common error: Speed depends on √V, not V. Track square roots carefully when combining energy and circular-motion equations.

44IBP3H-D3-20r42-06Charged particles in magnetic fields Paper 1AHard1 mark

In a mass spectrometer used to analyse a drug metabolite, ion P has mass \(m\) and charge \(+q\); ion Q has mass \(4m\) and charge \(+2q\). Both are accelerated from rest through the same potential difference and then enter the same uniform magnetic field at right angles to it. What is the ratio \(\frac{\text{radius of path of Q}}{\text{radius of path of P}}\)?

Show mark scheme
Marking pointMarkNotes
Since \(qV=\frac12mv^2\) and \(r=\frac{mv}{qB}\), the radius is \(r=\frac1B\sqrt{\frac{2mV}{q}}\), so \(r\propto\sqrt{m/q}\). For Q, \(m/q\) is twice that of P, so the ratio is \(\sqrt2\). The value 2 assumes equal speeds, giving \(r\propto m/q\). The value 4 uses the mass ratio alone. The value \(1/\sqrt2\) inverts the ratio.✓ 1Answer C

Answer: C

Common error: The ions do not have equal speeds after acceleration through the same potential difference, so \(r\propto m/q\) cannot be used directly.

45D-1B-10Specific charge of the electron Paper 1BEasy7 marks

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 / V100200300400500
r / cm3.44.85.86.77.5
r² / cm²11.623.044.956.3
0100200300400500V / V0102030405060r² / cm²
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
Marking pointMarkNotes
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✓ 1Allow ECF from (a)(i).
Gradient = 0.113 cm² V−1 = 1.13 × 10−5 m² V−1✓ 1Accept 1.09–1.16 × 10⁻⁵ m² V⁻¹. The conversion from cm² to m² is required for this mark.
Part (b)(i)
Gradient = 2m/(eB²), so e/m = 2/(gradient × B²)✓ 1
e/m = 2/(1.13 × 10−5 × (1.00 × 10−3)²) = 1.8 × 1011 C kg−1✓ 1Allow 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 %✓ 1Adding 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✓ 1Allow 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)

46D-1B-18Force between parallel wires Paper 1BEasy7 marks

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 / AI² / A²Δm / gF / mN
101000.0800.78
152250.1811.78
204000.3273.21
256250.5074.97
309000.734
upper rod (fixed)lower rodIIrtop-pan balancerods connected in series (leads not shown)
Not to scale. The overlap length of the rods is 0.400 m and r is the distance between their centres.
01002003004005006007008009001000I² / A²012345678F / mN
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 pointMarkNotes
Part (a)(i)
0.734 × 10−3 × 9.81 = 7.20 mN✓ 1Accept 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✓ 1Allow ECF from (a)(i).
Gradient = 8.0 × 10−6 N A−2✓ 1Accept 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✓ 1Allow 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✓ 1Allow 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)

47D-1B-04Force on a current-carrying conductor Paper 1BMedium7 marks

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 / A1.02.03.04.05.06.0
m / g0.430.841.251.662.062.47
01234567I / A0.00.51.01.52.02.53.0m / g
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
Marking pointMarkNotes
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✓ 1Both 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✓ 1The line need not pass through the origin.
Gradient = 0.408 g A−1 (4.08 × 10−4 kg A−1)✓ 1Accept 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✓ 1Award 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✓ 1Allow ECF from (a)(ii). Accept 0.078–0.082 T. The conversion from g to kg is required.
Part (b)(ii)
Percentage uncertainty = 2 % + (0.10/5.00 × 100 =) 2 % = 4 %, so ΔB = 0.04 × 0.080 ≈ 0.003 T✓ 1Allow 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✓ 1Allow 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)

48D-1B-12Charged particles in electric fields Paper 1BMedium7 marks

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 / Vy / mm(1/Va) / 10−3 V−1
40041.82.50
50034.32.00
60028.4
80022.21.25
100017.91.00
120015.40.83
0.00.51.01.52.02.53.0(1/Va) / 10⁻³ V⁻¹01020304050y / mm
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 pointMarkNotes
Part (a)(i)
1/600 = 1.67 × 10−3 V−1✓ 1Accept 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✓ 1Allow ECF from (a)(i).
Gradient = 15.9 mm per 10−3 V−1 = 15.9 V m✓ 1Accept 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✓ 1Allow 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✓ 1The comparison must use the candidate's own value and uncertainty (ECF).
Part (b)(ii)
Line extrapolated to 1/Va = 0: intercept ≈ +2 mm✓ 1Accept 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✓ 1Both 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)

49D-1B-23Velocity selector Paper 1BHard8 marks

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 / kVVp / VVp² / 10³ V²
0.50816.6
1.0011112.3
1.5013919.3
2.0016125.9
2.5017731.3
3.0019738.8
0.00.51.01.52.02.53.03.5Va / kV051015202530354045Vp² / 10³ V²
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 pointMarkNotes
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✓ 1The 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✓ 1Accept 12.4–13.2 V. The unit V (or 10³ V² kV⁻¹) is required for this mark.
Part (b)(i)
e/m = gradient/(2d²B²) = 12.8/(2 × (5.0 × 10−3)² × (1.20 × 10−3)²) = 1.78 × 1011 C kg−1✓ 1Allow 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 %✓ 1Adding 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✓ 1Allow 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 %)✓ 1Allow 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)

50D-2-28Charged particles in electric fields Paper 2Easy13 marks

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.

paperdroplet18 m s⁻¹1.2 cm1.5 cm3.0 mm
Diagram not to scale
(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 pointMarkNotes
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✓ 1Allow 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✓ 1Allow 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✓ 1Allow ECF from (b)(i).
y = 8.9 × 10−4 m (0.89 mm)✓ 1Accept 0.88–0.90 mm.
Part (b)(iii)
Vertical velocity leaving the plates vy = at = 2.67 m s−1✓ 1Allow 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)✓ 1Allow 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✓ 1Allow ECF from (b)(ii). ALT: ½mvy² = ½ × 1.5 × 10⁻¹¹ × 2.67².
ΔEk = 5.3 × 10−11 J✓ 1Accept 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)

51D-2-33Force on a current-carrying conductor Paper 2Easy10 marks

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.

Nouter pole (S)coil, 80 turns,radius 1.8 cmB (radial)I
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 pointMarkNotes
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✓ 1Accept "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✓ 1Full 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✓ 1Allow ECF from (a)(ii). Accept 73–74 m s⁻².
Part (b)(i)
x = F/k = 8.82/3.0 × 103 = 2.9 mm✓ 1Allow 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✓ 1Accept 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✓ 1Allow 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)

52IBP2H-033-00-09Force on a moving charge Paper 2Easy5 marks

An engineering student builds a model rail launcher. A metal bar of mass 0.050 kg rests across two horizontal rails that are 0.12 m apart. A current of 25 A flows along the rails and through the bar. A uniform vertical magnetic field of flux density 0.40 T acts perpendicular to the bar. Ignore friction and any changes in the current.

supplybar, 0.050 kg25 Auniform vertical magnetic field B = 0.40 T0.12 m0.80 m
Top view; not to scale
(a)

Calculate the magnitude of the magnetic force on the bar.

(2)
(b)

The bar starts from rest. Determine its speed after it has moved 0.80 m along the rails.

(2)
(c)

State one change to the arrangement that would double the force on the bar.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Uses F = BIL, with the field perpendicular to the bar .✓ 1
F = 0.40 × 25 × 0.12 = 1.2 N .✓ 1
Part (b)
a = F/m = 1.2/0.050 = 24 m s−2 .✓ 1
v = √(2as) = √(2 × 24 × 0.80) = 6.2 m s−1 .✓ 1
ECF from (a).—
Part (c)
Any one of: double the current; double the magnetic flux density; double the separation of the rails, so that the length of bar carrying current doubles .✓ 1

Exam tip: In F = BIL, L is the length of conductor that carries current inside the field. Here that is the rail separation, not the full length of the bar.

53IBP3H-D3-20r42-11Mass spectrometer Paper 2Easy7 marks

A singly charged ion of a drug molecule has mass \(5.4\times10^{-25}\) kg. It moves at \(2.5\times10^{4}\) m s−1 at right angles to a uniform magnetic field of flux density 0.40 T. Use \(e=1.60\times10^{-19}\) C.

(a)

Calculate the magnitude of the magnetic force on the ion.

(2)
(b)

Calculate the radius of the circular path of the ion.

(3)
(c)

Explain why the speed of the ion does not change while it is in the field.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(F=qvB=1.60\times10^{-19}\times2.5\times10^4\times0.40\)✓ 1
A1 \(1.60\times10^{-15}\) N✓ 1
Part (b)
M1 \(qvB=\frac{mv^2}{r}\)✓ 1
M1 \(r=\frac{mv}{qB}=\frac{5.4\times10^{-25}\times2.5\times10^4}{1.60\times10^{-19}\times0.40}\)✓ 1
A1 0.211 m✓ 1
Part (c)
the magnetic force is always perpendicular to the velocity✓ 1
so it does no work on the ion, and the kinetic energy and speed stay constant✓ 1

Common error: The magnetic force changes the direction of motion but not the speed, so do not describe it as accelerating the ion along its path.

54D-2-05Specific charge and de Broglie wavelength Paper 2Medium13 marks

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.

electron gun3.00 kV9.5 cmuniform B = 1.95 mT(⟂ to the page)
Path of the electron beam; not to scale
(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 pointMarkNotes
Part (a)(i)
v = √(2eV/m) = √(2 × 1.60 × 10−19 × 3000/9.11 × 10−31) = 3.25 × 107 m s−1✓ 1Full 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✓ 1
Part (b)(i)
evB = mv²/r gives v = eBr/m✓ 1
eV = ½m(eBr/m)², so e/m = 2V/(B²r²)✓ 1
Part (b)(ii)
e/m = 2 × 3000/((1.95 × 10−3)² × 0.095²) = 1.75 × 1011 C kg−1✓ 1Allow ECF from (b)(i).
Data booklet: 1.60 × 10−19/9.11 × 10−31 = 1.76 × 1011 C kg−1, so the two agree to within 1 %✓ 1The 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✓ 1The 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✓ 1Allow ECF from (a)(i).
λ = h/p = 6.63 × 10−34/2.96 × 10−23 = 2.2 × 10−11 m✓ 1
Part (c)(ii)
d sin θ = λ: sin θ = 2.2 × 10−11/2.1 × 10−10 = 0.11, so θ ≈ 0.11 rad (about 6°)✓ 1Allow 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✓ 1Allow 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)

55D-2-08The cyclotron Paper 2Medium12 marks

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.

DDalternating p.d. across the gap; protons start at the centreuniform B = 1.50 T, perpendicular to the plane
Top view; not to scale
(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 pointMarkNotes
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✓ 1
Part (a)(ii)
evB = mv²/r gives r = mv/(eB)✓ 1
T = 2πr/v = 2πm/(eB): v cancels✓ 1
Part (a)(iii)
f = eB/(2πm) = 1.60 × 10−19 × 1.50/(2π × 1.67 × 10−27) = 2.29 × 107 Hz✓ 1Allow ECF from (a)(ii).
Part (b)(i)
Two crossings per revolution: 100 keV per revolution, so 10.0 MeV/100 keV = 100 revolutions✓ 1Award [1 max] for 200 revolutions (one crossing per revolution) carried through correctly.
t = 100/2.29 × 107 = 4.4 × 10−6 s✓ 1Allow 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✓ 1ALT: 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✓ 1Allow 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✓ 1Do 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)

56D-2-09Charged particles in electric and magnetic fields Paper 2Medium14 marks

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.

+240 V0 Velectrons1.2 cmplates 4.0 cm long
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 pointMarkNotes
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✓ 1Allow 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)✓ 1Award 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✓ 1Allow 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✓ 1As 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✓ 1The 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✓ 1Allow 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)

57IBP2H-033-01-10The cyclotron Paper 2Medium7 marks

A hospital cyclotron produces protons for making medical isotopes used in PET scanning. Inside the cyclotron, protons move in semicircular paths in a uniform magnetic field of strength 1.4 T. Each time a proton crosses the gap between the two halves, it is accelerated by an alternating potential difference. Relativistic effects may be ignored.
Mass of proton = 1.67×10−27 kg; e = 1.60×10−19 C.

(a)

Show that the time for a proton to complete one revolution does not depend on its speed.

(2)
(b)

Calculate the frequency of the alternating potential difference needed to keep the protons in step.

(2)
(c)

Protons leave the cyclotron when the radius of their path is 0.50 m. Determine their kinetic energy in MeV.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
qvB = mv²/r, so r = mv/(qB)✓ 1
T = 2πr/v = 2πm/(qB). The speed v cancels, so T is independent of speed✓ 1
Part (b)
f = qB/(2πm) = (1.60×10−19 × 1.4)/(2π × 1.67×10−27)✓ 1
f = 2.1×107 Hz✓ 1
Part (c)
v = qBr/m = (1.60×10−19 × 1.4 × 0.50)/(1.67×10−27) = 6.71×107 m s−1✓ 1
Ek = ½mv² = 3.76×10−12 J✓ 1
Ek = 3.76×10−12/1.60×10−13 = 23.5 MeV. Accept 23–24 MeV, which allows for rounding of intermediate values✓ 1

Exam tip: Ek = (qBr)²/(2m) gives the final kinetic energy in a single calculation. The final energy depends on the extraction radius, not on the size of the accelerating p.d. Carry unrounded values through to the final line.

58IBP2H-033-01-11Mass spectrometer Paper 2Medium7 marks

In a semiconductor ion implanter, singly charged positive ions are accelerated from rest through a potential difference of 20.0 kV. They then enter an analysing magnet. Only ions that follow a path of radius 0.400 m reach the exit slit. The magnetic field strength B is adjusted to select each ion. The table shows the field needed for three ions.

IonB / T
10B+0.161
11B+0.169
unknown X+0.357

e = 1.60×10−19 C; 1 u = 1.66×10−27 kg; mass of 11B+ = 1.83×10−26 kg.

analysing magnetion source20.0 kVaccelerated throughexit slitslit0.400 m
Path in the magnetic field; not to scale
(a)

Show that the speed of 11B+ ions entering the magnet is about 6×105 m s−1.

(2)
(b)

Show that the mass of an ion reaching the exit slit is given by m = qB²r²/(2V).

(2)
(c)

Determine the mass of ion X+ in atomic mass units. Hence suggest its identity, given that the source gas contains boron-11 and fluorine-19.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
qV = ½mv²✓ 1
v = √(2 × 1.60×10−19 × 2.00×104/1.83×10−26) = 5.9×105 m s−1✓ 1
Part (b)
qvB = mv²/r gives r = mv/(qB)✓ 1
Substituting v = √(2qV/m) gives r² = 2mV/(qB²), so m = qB²r²/(2V)✓ 1
Part (c)
m = 1.60×10−19 × 0.357² × 0.400²/(2 × 2.00×104) = 8.16×10−26 kg✓ 1
m = 8.16×10−26/1.66×10−27 = 49 u✓ 1
X+ is BF2+, because 11 + 2 × 19 = 49✓ 1
ECF from (b)—

Exam tip: For a fixed radius and p.d., B ∝ √m. Checking that the two boron rows obey this ratio is a quick sanity test of the data.

59IBP2H-033-01-12Force between parallel wires Paper 2Medium6 marks

In an electric-vehicle battery pack, two long, straight, parallel copper busbars are 12 mm apart. Each is 0.80 m long. In normal operation, each busbar carries 300 A, and the currents are in opposite directions.
μ0 = 4π×10−7 T m A−1.

(a)

Outline why the busbars repel each other.

(2)
(b)

Calculate the force on one busbar in normal operation.

(2)
(c)

During a short-circuit fault, the current in each busbar briefly rises to 6.0 kA. Determine the force on one busbar and comment on the design implication.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Each busbar lies in the magnetic field produced by the other✓ 1
For opposite currents, the right-hand rule shows that the force F = BIL on each busbar points away from the other busbar, so they repel✓ 1
Part (b)
F = μ0I1I2L/(2πd) = (4π×10−7 × 300 × 300 × 0.80)/(2π × 0.012)✓ 1
F = 1.2 N✓ 1
Part (c)
Either use F ∝ I2, a factor of (6000/300)2 = 400, so F = 400 × 1.2 = 480 N, or recalculate directly from F = μ0I2L/(2πd) to get 480 N (allow ECF from (b))✓ 1
This is comparable to the weight of about 50 kg, so the busbars must be rigidly clamped or supported to withstand the fault forces✓ 1

Common error: Both currents increase, so the force scales with I², not I. Multiplying by 20 instead of 400 is a frequent slip.

60IBP2H-033-01-13Charged particles in magnetic fields Paper 2Medium8 marks

A solar-wind proton enters a region of the Earth's outer magnetosphere where the magnetic field strength is 30 nT. The proton moves at 4.0×105 m s−1, perpendicular to the field.
Mass of proton = 1.67×10−27 kg; e = 1.60×10−19 C.

(a)

Show that the radius of the proton's circular path is about 1.4×105 m.

(2)
(b)

Calculate the time for the proton to complete one orbit.

(2)
(c)

Explain why the kinetic energy of the proton does not change while it moves in this field.

(2)
(d)

The proton drifts into a region nearer the Earth where the field is stronger. Describe and explain how the radius and the period of its motion change.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
r = mv/(qB)✓ 1
r = (1.67×10−27 × 4.0×105)/(1.60×10−19 × 30×10−9) = 1.39×105 m✓ 1
Part (b)
T = 2πr/v = 2π × 1.39×105/4.0×105✓ 1
T = 2.2 s✓ 1
ECF from (a)—
Part (c)
The magnetic force is always perpendicular to the velocity✓ 1
So no work is done on the proton and its kinetic energy (speed) stays constant✓ 1
Part (d)
r = mv/(qB) with v constant, so the radius decreases as B increases✓ 1
T = 2πm/(qB), so the period also decreases✓ 1

Exam tip: Convert nT to T before substituting: 30 nT = 30×10⁻⁹ T.

61IBP2H-033-00-10Force between parallel wires Paper 2Medium8 marks

In a data centre, power is distributed by two long, straight, parallel copper busbars. The busbars are 4.0 cm apart. In normal operation the supply bar and the return bar each carry 1200 A, in opposite directions. Treat each busbar as a long thin wire.
μ0 = 4π × 10−7 T m A−1

(a)

State and explain whether the busbars attract or repel each other.

(2)
(b)

Calculate the magnetic flux density produced by one busbar at the position of the other.

(2)
(c)

Calculate the force on a 1.5 m length of one busbar.

(2)
(d)

During a short-circuit fault, the current in both bars briefly rises to 25 kA. Determine the factor by which the force between the bars increases.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
They repel .✓ 1
The field of one bar at the second bar is perpendicular to that bar's current. Because the current in the second bar is in the opposite direction, the force on it (from F = BIL and the left-hand rule) points away from the first bar. Accept the equivalent statement that antiparallel currents repel .✓ 1
Part (b)
B = μ0I/(2πr) = (4π × 10−7 × 1200)/(2π × 0.040) .✓ 1
B = 6.0 × 10−3 T .✓ 1
Part (c)
F = BIL = 6.0 × 10−3 × 1200 × 1.5 .✓ 1
F = 11 N (accept 10.8 N) .✓ 1
Allow ECF from (b).—
Part (d)
The force is proportional to I1I2 = I2, because both bars carry the same current .✓ 1
Factor = (25 000/1200)2 ≈ 430 (accept 434 or 4.3 × 102) .✓ 1

Common error: When the same current flows in both conductors, the force depends on I². Scaling by the current ratio only once underestimates fault forces badly.

62IBP2H-033-00-12Specific charge of the electron Paper 2Medium8 marks

A student measures the specific charge of the electron using a fine-beam tube. Electrons are accelerated from rest through a potential difference V. They then move in a circle of radius r in a uniform magnetic field of flux density 0.80 mT, perpendicular to their velocity.

V / V : 150, 200, 250, 300
r / cm : 5.2, 6.0, 6.6, 7.3

(a)

Show that the specific charge of the electron is given by e/m = 2V/(B2r2).

(2)
(b)

Explain why the speed of an electron is constant while it moves in the magnetic field.

(1)
(c)

Calculate e/m using the data for V = 300 V.

(2)
(d)

A graph of r2 against V is a straight line through the origin. Use the first and last rows of the data to determine the gradient of this line, and hence determine e/m.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
Uses eV = ½mv2 for the acceleration and evB = mv2/r for the circular motion .✓ 1
Substitutes v = eBr/m into the energy equation: eV = ½m(eBr/m)2, which rearranges to e/m = 2V/(B2r2) .✓ 1
Part (b)
The magnetic force is always perpendicular to the velocity, so it does no work and the kinetic energy (and hence the speed) is unchanged .✓ 1
Part (c)
e/m = 2 × 300/((0.80 × 10−3)2 × (0.073)2) .✓ 1
e/m = 1.8 × 1011 C kg−1 (accept 1.76 × 1011) .✓ 1
Part (d)
r2 values: 2.70 × 10−3 m2 and 5.33 × 10−3 m2 .✓ 1
Gradient = (5.33 − 2.70) × 10−3/150 = 1.75 × 10−5 m2 V−1 .✓ 1
The gradient equals 2/(B2(e/m)), so e/m = 2/((0.80 × 10−3)2 × 1.75 × 10−5) = 1.8 × 1011 C kg−1 .✓ 1
Allow error carried forward from incorrect r2 values or gradient.—

Exam tip: To find a quantity from a gradient, first rearrange the equation into the form y = mx. Here r² = [2/(B²(e/m))]V, so the bracket is the gradient.

63IBP2H-033-00-14Mass spectrometer Paper 2Medium9 marks

A solar-wind proton travels at 4.0 × 105 m s−1 into a region of Earth's magnetosphere. There the magnetic field is uniform, with flux density 30 nT. At first the proton's velocity is perpendicular to the field.
e = 1.60 × 10−19 C; mp = 1.67 × 10−27 kg

(a)

Calculate the magnitude of the magnetic force on the proton.

(1)
(b)

Explain why the proton moves in a circle.

(2)
(c)

Calculate the radius of the proton's path.

(2)
(d)

Determine the time the proton takes to complete one orbit.

(2)
(e)

Another proton has a velocity component parallel to the field as well as one perpendicular to it. Describe its motion.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
F = qvB = 1.60 × 10−19 × 4.0 × 105 × 30 × 10−9 = 1.9 × 10−21 N .✓ 1
Part (b)
The magnetic force is always perpendicular to the velocity, so it does no work and the speed stays constant .✓ 1
With v and B constant, the force has constant magnitude and always acts perpendicular to the velocity in one plane. It therefore acts as a centripetal force, giving uniform circular motion .✓ 1
Part (c)
qvB = mv2/r, so r = mv/(qB) = (1.67 × 10−27 × 4.0 × 105)/(1.60 × 10−19 × 30 × 10−9) .✓ 1
r = 1.4 × 105 m (about 140 km) .✓ 1
Part (d)
Use T = 2πr/v = 2πm/(qB) .✓ 1
T = 2π × 1.67 × 10−27/(1.60 × 10−19 × 30 × 10−9) = 2.2 s .✓ 1
Allow ECF from (c) if 2πr/v is used.—
Part (e)
The parallel component experiences no magnetic force, so the proton moves along the field direction at constant speed .✓ 1
Combined with circular motion in the perpendicular plane, the path is a helix with its axis along the field direction .✓ 1

Exam tip: Split the velocity into components parallel and perpendicular to B. Only the perpendicular component experiences a force, so it alone sets the radius.

64IBP2H-033-00-16The cyclotron Paper 2Medium8 marks

In the injector of a proton-therapy accelerator, protons start from rest and are accelerated across a 4.0 cm gap. The gap lies between two parallel electrodes with a potential difference of 50 kV between them. Assume the field in the gap is uniform and non-relativistic mechanics applies.
e = 1.60 × 10−19 C; mp = 1.67 × 10−27 kg

4.0 cmprotonV50 kV
Diagram not to scale
(a)

Calculate the electric field strength in the gap.

(1)
(b)

Calculate the force on a proton in the gap.

(1)
(c)

State the kinetic energy gained by a proton crossing the gap, in eV and in J.

(2)
(d)

Calculate the speed of a proton as it leaves the gap.

(2)
(e)

Determine the time a proton takes to cross the gap.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
E = V/d = 50 × 103/0.040 = 1.3 × 106 V m−1 (accept 1.25 × 106) .✓ 1
Part (b)
F = eE = 1.60 × 10−19 × 1.25 × 106 = 2.0 × 10−13 N .✓ 1
ECF from (a).—
Part (c)
Ek = 50 keV .✓ 1
Ek = eV = 1.60 × 10−19 × 50 × 103 = 8.0 × 10−15 J .✓ 1
Part (d)
Using ½mv2 = 8.0 × 10−15 J gives v = √(2 × 8.0 × 10−15/1.67 × 10−27) .✓ 1
v = 3.1 × 106 m s−1 .✓ 1
ECF from (c).—
Part (e)
Award the first mark for either valid method :✓ 1
• the acceleration is uniform from rest, so the average speed is v/2 and t = 2d/v;—
• or use a = F/m = 2.0 × 10−13/1.67 × 10−27 = 1.2 × 1014 m s−2 and t = √(2d/a).—
t = 2.6 × 10−8 s .✓ 1
ECF from (b) or (d).—

Common error: The time to cross the gap is not d/v using the final speed. The proton starts from rest, so use the average speed v/2 or s = ½at².

65IBP3H-D3-20r42-09Forces between parallel currents Paper 2Medium7 marks

Two long straight cables carry currents in the same direction. They are parallel and 0.080 m apart. Cable P carries a current of 60 A. The force per unit length on cable Q is measured as \(2.4\times10^{-3}\) N m−1. Use \(\mu_0=4\pi\times10^{-7}\) T m A−1.

(a)

Explain why the cables attract each other.

(2)
(b)

Calculate the current in cable Q.

(3)
(c)

State the magnitude and the direction of the force on cable P per unit length, compared with that on Q.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
cable P produces a magnetic field at cable Q (field lines circle P)✓ 1
the current in Q is at right angles to this field, giving a force on Q (right-hand rules) directed towards P; the same argument applies to P✓ 1
Part (b)
M1 \(\frac{F}{L}=\frac{\mu_0I_PI_Q}{2\pi d}\)✓ 1
M1 \(I_Q=\frac{2\pi d(F/L)}{\mu_0I_P}=\frac{2\pi(0.080)(2.4\times10^{-3})}{4\pi\times10^{-7}\times60}\)✓ 1
A1 16 A✓ 1
Part (c)
magnitude is the same, \(2.4\times10^{-3}\) N m−1✓ 1
direction is opposite (towards Q), by Newton's third law✓ 1

Exam tip: The equal-and-opposite forces on the two cables hold for any pair of currents, even when the currents are different.

66IBP3H-D3-20r42-10Specific charge and de Broglie wavelength Paper 2Medium7 marks

In a demonstration of the specific charge of the electron, electrons are accelerated from rest through a potential difference of 1.20 kV and enter a region of uniform magnetic field of flux density 1.80 mT perpendicular to their velocity. Their path is a circle of radius 6.49 cm.

(a)

Show that the specific charge of the electron is given by \(\frac{e}{m}=\frac{2V}{B^2r^2}\).

(3)
(b)

Use the data to calculate the specific charge of the electron.

(2)
(c)

The accelerating potential difference is increased to 4.80 kV with the field unchanged. Calculate the new radius of the path, in cm.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(eV=\frac12mv^2\)✓ 1
M1 \(evB=\frac{mv^2}{r}\), so \(v=\frac{eBr}{m}\)✓ 1
A1 substitutes \(v\) and rearranges to \(\frac{e}{m}=\frac{2V}{B^2r^2}\)✓ 1
Part (b)
M1 \(\frac{2\times1200}{(1.80\times10^{-3})^2(0.0649)^2}\)✓ 1
A1 \(1.76\times10^{11}\) C kg−1✓ 1
Part (c)
M1 \(r\propto\sqrt V\), so the radius doubles for a fourfold increase in \(V\)✓ 1
A1 12.98 cm, which is 13.0 cm to 3 s.f.✓ 1

Exam tip: If a quantity changes by a simple factor, use a proportionality rather than recalculating from scratch.

67IBP3H-D3-20r42-15Mass spectrometer Paper 2Medium7 marks

Two singly charged ions of a radioactively labelled drug fragment are produced in a mass spectrometer. The lighter ion has mass \(1.99\times10^{-26}\) kg and the heavier ion has mass \(2.32\times10^{-26}\) kg. Each is accelerated from rest through a potential difference of 3.00 kV and then enters a uniform magnetic field of flux density 0.200 T at right angles to the field, and travels in a semicircle. Use \(e=1.60\times10^{-19}\) C.
Use unrounded values in later parts.

B = 0.200 Tlighter ionheavier ionion source3.00 kVaccelerated throughdetectorslit
Path in the magnetic field; not to scale
(a)

Calculate the speed of the lighter ion as it enters the magnetic field.

(2)
(b)

Calculate the radius of the path of the lighter ion.

(2)
(c)

Calculate the distance between the points where the two ions hit the detector, which lies at the end of each semicircle.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(eV=\frac12mv^2\)✓ 1
A1 \(v=\sqrt{\frac{2\times1.60\times10^{-19}\times3000}{1.99\times10^{-26}}}=2.20\times10^5\) m s−1✓ 1
Part (b)
M1 \(r=\frac{mv}{eB}\)✓ 1
A1 0.137 m✓ 1
Part (c)
M1 radius of the heavier ion found in the same way, 0.147 m✓ 1
M1 separation \(=2(r_2-r_1)\) since the detector lies at the diameter of each semicircle✓ 1
A1 0.0218 m✓ 1

Exam tip: The ions land a full diameter from the entry point, so the separation is twice the difference in radius; keep unrounded radii because the difference is small.

68IBP3H-D3-20r42-16Velocity selector Paper 2Medium8 marks

An electromagnetic flow meter measures the speed of water in a pipe of internal diameter 0.12 m. The pipe is non-magnetic and non-conducting. A uniform magnetic field of flux density 0.080 T is applied across the pipe, perpendicular to the flow. Electrodes on opposite sides of the pipe, at the ends of a diameter that is perpendicular to both the flow and the field, measure a potential difference caused by dissolved ions.

electrodeelectrode0.12 mwater flows⟂ to the pageVuniform B = 0.080 T across the pipe,perpendicular to the flow and to the electrodes
Cross-section of the pipe; not to scale
(a)

Explain why a potential difference develops between the electrodes.

(3)
(b)

Calculate the potential difference for water flowing at 1.5 m s−1.

(2)
(c)

A potential difference of 9.6 mV is measured. Calculate the volume of water passing through the pipe per second.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
moving positive and negative ions experience magnetic forces in opposite directions across the pipe✓ 1
charge builds up at the two electrodes, producing an electric field across the pipe✓ 1
the potential difference rises until the electric force \(qE\) equals the magnetic force \(qvB\) on an ion, after which it is steady✓ 1
Part (b)
M1 \(qE=qvB\) with \(E=\frac Vd\), so \(V=Bvd\)✓ 1
A1 \(0.080\times1.5\times0.12=0.0144\) V (14.4 mV)✓ 1
Part (c)
M1 \(v=\frac{V}{Bd}=\frac{9.6\times10^{-3}}{0.080\times0.12}=1.0\) m s−1✓ 1
M1 volume per second \(=\pi\left(\frac{0.12}{2}\right)^2v\)✓ 1
A1 \(1.13\times10^{-2}\) m3 s−1 (0.0113 m3 s−1)✓ 1

Common error: The distance in \(V=Bvd\) is the separation of the electrodes, which here is the pipe diameter, not the radius.

69D-2-03Mass spectrometer Paper 2Hard10 marks

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.

B = 0.400 T²⁴Mg²⁶Mgion source2.00 kVaccelerated throughdetectorslit
Path in the magnetic field; not to scale
(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.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
v = √(2qV/m) = √(2 × 1.60 × 10−19 × 2000/(24 × 1.66 × 10−27)) = 1.27 × 105 m s−1✓ 1Full substitution or an answer to at least 3 s.f. is required.
Part (a)(ii)
The magnetic force is always perpendicular to the velocity, so it does no work and the speed (kinetic energy) is constant✓ 1
A force of constant magnitude qvB perpendicular to the velocity provides a centripetal force, so the path is a circle✓ 1
Part (b)(i)
r = mv/(qB) = 24 × 1.66 × 10−27 × 1.27 × 105/(1.60 × 10−19 × 0.400) = 7.9 × 10−2 m✓ 1Allow ECF from (a)(i).
Part (b)(ii)
r = mv/(qB) and both isotopes have the same kinetic energy qV = ½mv², so r ∝ √m: r26 = 0.0789 × √(26/24) = 0.0821 m✓ 1Allow ECF from (b)(i). Accept a separate calculation of v and r for 26Mg.
Each ion strikes the detector a full diameter, 2r, from the point of entry✓ 1
Separation = 2(0.0821 − 0.0789) = 6.4 × 10−3 m (6.4 mm)✓ 1Accept 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)✓ 1Accept 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✓ 1From 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)

70D-2-27Force between parallel wires Paper 2Hard13 marks

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.

X (50 A, into the page)Y (5.0 A, out of the page)glass guidesr
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 pointMarkNotes
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✓ 1Must see the substitution or an unrounded 9.62 mm.
Part (b)(ii)
r ∝ IY, so r = 1.2 × 9.62 = 11.5 mm✓ 1Accept 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✓ 1Accept 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)

71D-2-38A magnetic spectrometer for alpha particles Paper 2Hard19 marks

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.

sourceSdetector plate Dpath of alpha particles
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 pointMarkNotes
Part (a)(i)
Nucleon number 237 and proton number 93✓ 1Both required.
Part (a)(ii)
v = √(2Ek/m) = √(2 × 5.49 × 1.60 × 10−13/6.64 × 10−27) = 1.63 × 107 m s−1✓ 1The 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✓ 1
EkNp = 5.49 MeV × 6.64 × 10−27/(237 × 1.66 × 10−27) = 93 keV✓ 1The answer must be in keV. Accept 92–93 keV.
Part (b)(i)
Into the page (plane of the diagram)✓ 1From 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✓ 1Award [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✓ 1Allow 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✓ 1Allow reference to the reasoning in (a)(iii).
Same momentum and same charge (2e), so the radius is the same: 0.42 m✓ 1Allow 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)✓ 1Accept 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✓ 1
Part (c)(ii)
Photon energy ≈ 5.49 − 5.44 = 0.05 MeV = 8.0 × 10−15 J✓ 1Accept 50–51 keV (including the recoil correction).
λ = hc/E = 6.63 × 10−34 × 3.00 × 108/8.0 × 10−15 = 2.5 × 10−11 m = 25 pm✓ 1Allow 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✓ 1Allow 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)

72IBP2H-033-01-14Force between parallel wires Paper 2Hard7 marks

A small-scale model of an electromagnetic aircraft-launch system uses two horizontal parallel rails 0.25 m apart and 3.0 m long. A conducting armature of mass 0.40 kg slides along the rails and completes the circuit. A uniform magnetic field of 2.0 T acts perpendicular to the plane of the rails. The current through the armature is 5.0 kA. Ignore friction and air resistance.

supplyarmature, 0.40 kgI = 5.0 kAuniform field B = 2.0 T perpendicular to the plane of the rails3.0 m0.25 m
Top view; not to scale
(a)

Calculate the magnetic force on the armature.

(2)
(b)

Determine the speed of the armature as it leaves the rails, assuming it starts from rest.

(2)
(c)

In a redesign, the field at the armature is produced by the current in the rails themselves. The field is 2.0 T when the current is 5.0 kA and is proportional to the current. The current is increased to 10 kA. Determine the new exit speed.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
F = BIL = 2.0 × 5.0×103 × 0.25✓ 1
F = 2.5×103 N✓ 1
Part (b)
a = F/m = 2.5×103/0.40 = 6.25×103 m s−2 OR work done Fs = 2.5×103 × 3.0 = 7.5×103 J = ½mv2✓ 1
v = √(2 × 6.25×103 × 3.0) = 1.9×102 m s−1 (allow 194 m s−1)✓ 1
Allow error carried forward from the force in the first part.—
Part (c)
Both B and I double, so the force increases by a factor of 4 OR B = 4.0 T and F = 4.0 × 1.0×104 × 0.25 = 1.0×104 N✓ 1
v ∝ √F (from v = √(2as) or Fs = ½mv2), so v doubles OR v = √(2 × 1.0×104 × 3.0 / 0.40)✓ 1
v = 3.9×102 m s−1 (accept 3.8–3.9×102 m s−1)✓ 1
Allow error carried forward from the speed in the previous part.—

Common error: Treating the force as proportional to I alone is wrong when the field comes from the same current. Here B ∝ I, so F ∝ I². Doubling the current quadruples the force, and the exit speed only doubles because v ∝ √F.

73IBP2H-033-01-15Specific charge of the electron Paper 2Hard8 marks

A battery-recycling plant uses a simple magnetic analyser to check recovered material. Singly charged positive ions start from rest and are accelerated through a potential difference V. They then enter a uniform magnetic field of 0.120 T at right angles to it, where they move in circular paths of radius r. The table shows measurements of r for different values of V.

V / Vr / mm
20044.9
40063.5
60077.8
80089.8

e = 1.60×10−19 C; 1 u = 1.66×10−27 kg.

(a)

Show that r² = 2mV/(qB²).

(2)
(b)

Explain whether the data support the relationship r² = 2mV/(qB²).

(2)
(c)

Determine the specific charge q/m of the ions.

(2)
(d)

Identify the ion.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
qV = ½mv², so v = √(2qV/m)✓ 1
qvB = mv²/r gives r = mv/(qB), so r² = m²(2qV/m)/(q²B²) = 2mV/(qB²)✓ 1
Part (b)
r²/V = 10.1 mm² V−1 (1.01×10−5 m² V−1) for every row, e.g. 89.8²/800 = 10.08✓ 1
r²/V is constant, so r² ∝ V, as predicted, with constant = 2m/(qB²)✓ 1
Part (c)
q/m = 2V/(B²r²) = 2/(0.120² × 1.008×10−5)✓ 1
q/m = 1.38×107 C kg−1✓ 1
Part (d)
m = 1.60×10−19/1.38×107 = 1.16×10−26 kg✓ 1
m = 7.0 u, so the ion is lithium-7 (Li+)✓ 1
Allow follow-through from the calculated value of q/m.—

Exam tip: Use all the data, through a constant ratio or a gradient, rather than a single row. This shows the relationship holds and reduces random error.

74IBP2H-033-01-16Mass spectrometer Paper 2Hard8 marks

An ion analyser on a magnetosphere research satellite contains a velocity selector. The selector has two parallel plates 20 mm apart with a potential difference of 120 V between them. A uniform magnetic field of 0.015 T acts perpendicular to both the electric field and the ion beam.
Ions that pass through undeflected enter a second region containing only a uniform magnetic field of 0.25 T, perpendicular to their velocity. A group of singly charged positive atomic ions (single atoms that have each lost one electron) follows a semicircular path of radius 0.266 m.
e = 1.60×10−19 C; 1 u = 1.66×10−27 kg.

B = 0.25 Tion sourceplates 20 mm apart, 120 VB = 0.015 Tdetectorslitr = 0.266 m
Path in the magnetic field; not to scale
(a)

Calculate the electric field strength between the plates.

(1)
(b)

Show that ions pass undeflected when v = E/B. Hence calculate this speed.

(2)
(c)

Explain why the selector passes ions of this speed whatever the sign or magnitude of their charge.

(2)
(d)

Determine the mass of the ions in atomic mass units and identify the element.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
E = V/d = 120/0.020 = 6.0×103 V m−1✓ 1
Part (b)
For no deflection the electric and magnetic forces are equal and opposite: qE = qvB, so v = E/B✓ 1
v = 6.0×103/0.015 = 4.0×105 m s−1✓ 1
Allow ECF from (a).—
Part (c)
Both forces are proportional to q, so reversing the sign of the charge reverses both forces, which therefore still oppose each other✓ 1
q cancels in qE = qvB, so the balance condition depends only on v and not on the charge (or mass)✓ 1
Part (d)
r = mv/(qB), so m = qrB/v✓ 1
m = (1.60×10−19 × 0.266 × 0.25)/(4.0×105) = 2.66×10−26 kg✓ 1
m = 2.66×10−26/1.66×10−27 ≈ 16 u, so the ions are oxygen, O+✓ 1
Allow ECF from (b).—

Common error: Use the second field strength, 0.25 T, for the radius calculation, not the selector field of 0.015 T. The two magnetic fields do different jobs.

75IBP2H-033-00-11Mass spectrometer Paper 2Hard10 marks

In a radiocarbon-dating laboratory, singly charged carbon ions (charge +1.60 × 10−19 C) first pass through a velocity selector. In the selector, two parallel plates 5.0 mm apart have a potential difference of 200 V between them. A perpendicular uniform magnetic field of 0.080 T also acts. The selected ions then enter a second uniform magnetic field of 0.50 T, which is perpendicular to their velocity. They travel through a semicircle in this field and strike a detector.
Mass of a 12C ion = 12u; mass of a 14C ion = 14u; u = 1.66 × 10−27 kg.

B = 0.50 T¹²C¹⁴Cion sourceplates 5.0 mm apart, 200 VB = 0.080 Tdetectorslit
Path in the magnetic field; not to scale
(a)

Explain why only ions of one particular speed leave the velocity selector undeflected.

(2)
(b)

Calculate the speed of the ions that leave the selector.

(2)
(c)

Show that the radius of the path of a 12C ion in the 0.50 T field is about 0.12 m.

(2)
(d)

Determine the distance between the points where 12C ions and 14C ions strike the detector.

(3)
(e)

Outline why the velocity selector is needed before the ions enter the 0.50 T field.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
The electric force qE and the magnetic force qvB act in opposite directions .✓ 1
They balance only when qE = qvB, that is, when v = E/B. Faster or slower ions experience a resultant force and are deflected .✓ 1
Part (b)
E = V/d = 200/(5.0 × 10−3) = 4.0 × 104 V m−1 .✓ 1
v = E/B = 4.0 × 104/0.080 = 5.0 × 105 m s−1 .✓ 1
Part (c)
r = mv/(qB) .✓ 1
r = (12 × 1.66 × 10−27 × 5.0 × 105)/(1.60 × 10−19 × 0.50) = 0.1245 m ≈ 0.12 m .✓ 1
Part (d)
Radius for 14C: r = (14 × 1.66 × 10−27 × 5.0 × 105)/(1.60 × 10−19 × 0.50) = 0.1453 m .✓ 1
After a semicircle, each ion lands a distance 2r from the entry point .✓ 1
Separation = 2(0.1453 − 0.1245) = 0.0415 m ≈ 0.042 m (4.2 cm) .✓ 1
Equivalent: separation = 2Δm v/(qB). Allow ECF from (b). Accept 0.040–0.042 m if rounded radii are used consistently.—
Part (e)
All ions then enter the field with the same speed, so, since they have the same charge, the radius depends only on mass. Each isotope then lands at a single, well-defined position .✓ 1

Exam tip: A semicircular path displaces a particle by a diameter, 2r, from its entry point. The separation between two isotopes is therefore twice the difference in their radii. Keep the unrounded radii until the final step, because the difference between them is small.

76IBP2H-033-00-13Charged particles in electric fields Paper 2Hard8 marks

In a continuous inkjet printer, charged ink droplets are steered by an electric field. Each droplet has a mass of 1.2 × 10−10 kg and a charge of magnitude 2.0 × 10−13 C. A droplet enters the region between two parallel deflecting plates at 20 m s−1, moving parallel to the plates. The plates are 1.5 cm long and the uniform field between them is 1.6 × 106 V m−1. The paper is 1.0 cm beyond the end of the plates. Assume the field is zero outside the plates.

paperdroplet20 m s⁻¹1.5 cm1.0 cm
Diagram not to scale
(a)

Show that the weight of the droplet can be neglected.

(2)
(b)

Calculate the sideways deflection of the droplet as it leaves the plates.

(3)
(c)

Determine the total sideways deflection of the droplet when it reaches the paper.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
Weight = mg = 1.2 × 10−10 × 9.8 = 1.2 × 10−9 N .✓ 1
Electric force = qE = 2.0 × 10−13 × 1.6 × 106 = 3.2 × 10−7 N, which is about 270 times greater than the weight .✓ 1
Part (b)
a = qE/m = 3.2 × 10−7/1.2 × 10−10 = 2.7 × 103 m s−2 .✓ 1
Time between plates: t = 0.015/20 = 7.5 × 10−4 s .✓ 1
y = ½at2 = 7.5 × 10−4 m (0.75 mm) .✓ 1
Part (c)
Sideways velocity at exit: vy = at = 2.7 × 103 × 7.5 × 10−4 = 2.0 m s−1 .✓ 1
Time from plates to paper = 0.010/20 = 5.0 × 10−4 s, giving an extra deflection of 2.0 × 5.0 × 10−4 = 1.0 × 10−3 m .✓ 1
Total deflection = 0.75 + 1.0 = 1.8 mm (1.75 × 10−3 m) .✓ 1
ECF from (b).—

Common error: Beyond the plates the droplet moves in a straight line. Do not keep applying the acceleration after the droplet has left the field.

77IBP2H-033-00-15Force between parallel wires Paper 2Hard8 marks

In a test rig for superconducting fusion-magnet cables, three long, straight, parallel conductors P, Q and R lie in the same plane. P and Q are 0.30 m apart and carry currents of 40 A and 10 A respectively, in the same direction. Conductor R carries a current of 5.0 A in the same direction as P and Q. R can be placed anywhere on the line between P and Q.
μ0 = 4π × 10−7 T m A−1

P40 AQ10 A0.30 mR (5.0 A) lies on the line between P and Qall three currents in the same direction
Cross-section; conductors perpendicular to the page; not to scale
(a)

State the direction of the force on Q due to P.

(1)
(b)

Calculate the force per unit length between P and Q, with R absent.

(2)
(c)

Determine the distance from P at which R experiences no resultant magnetic force.

(3)
(d)

R is displaced slightly towards P from this position. Explain whether R returns to its equilibrium position.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Towards P, because parallel currents attract .✓ 1
Part (b)
F/L = μ0IPIQ/(2πr) = (4π × 10−7 × 40 × 10)/(2π × 0.30) .✓ 1
F/L = 2.7 × 10−4 N m−1 .✓ 1
Part (c)
The attraction towards P must equal the attraction towards Q: μ0IPIR/(2πx) = μ0IQIR/(2π(0.30 − x)) .✓ 1
40(0.30 − x) = 10x .✓ 1
x = 0.24 m from P. This result is independent of IR .✓ 1
Part (d)
Moving closer to P increases the attraction to P and decreases the attraction to Q .✓ 1
The resultant force on R is therefore away from equilibrium, towards P, so R does not return: the equilibrium is unstable .✓ 1

Exam tip: For a balance point between two wires, set up the force equation with IR on both sides. It cancels, so the current in the third wire is not needed.

78IBP3H-D3-20r42-13Charged particles in electric and magnetic fields Paper 2Hard10 marks

An ion beam passes through a velocity selector made of two parallel plates 4.0 mm apart with a potential difference of 120 V between them, and a uniform magnetic field of flux density 0.15 T perpendicular to the beam and to the electric field. In the selector the electric force on a selected ion acts towards the top of the page and the magnetic force towards the bottom.
Use \(e=1.60\times10^{-19}\) C.

uniform magnetic field B = 0.15 Tionelectric forcemagnetic force4.0 mmp.d. between the plates = 120 V
Diagram not to scale
(a)

Calculate the electric field strength between the plates.

(1)
(b)

Calculate the speed of ions that pass through undeflected.

(2)
(c)

Explain why the selected speed does not depend on the mass or the charge of the ion.

(2)
(d)

An ion in the beam has a speed greater than the selected speed. State, with a reason, the direction in which it is deflected.

(2)
(e)

Ions that pass through the selector enter a second region of uniform magnetic field of flux density 0.30 T only. A singly charged ion of mass \(4.8\times10^{-26}\) kg is selected. Calculate the radius of its circular path.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
\(E=\frac Vd=\frac{120}{4.0\times10^{-3}}=3.0\times10^4\) V m−1✓ 1
Part (b)
M1 \(qE=qvB\), so \(v=\frac EB\)✓ 1
A1 \(2.0\times10^5\) m s−1✓ 1
Part (c)
the electric and magnetic forces are both proportional to the charge \(q\), so \(q\) cancels✓ 1
there is no net force on an undeflected ion, so the acceleration is zero and mass plays no part in the force balance✓ 1
Part (d)
towards the bottom of the page✓ 1
the magnetic force \(qvB\) increases with speed while \(qE\) is unchanged, so the magnetic force is the larger✓ 1
Part (e)
M1 \(evB=\frac{mv^2}{r}\), so \(r=\frac{mv}{eB}\)✓ 1
M1 \(r=\frac{4.8\times10^{-26}\times2.0\times10^5}{1.60\times10^{-19}\times0.30}\)✓ 1
A1 0.200 m (ECF from (b))✓ 1

Exam tip: Use \(E=V/d\) for the plate field; the field is found from the potential difference and the plate separation, not from the plate area.

79IBP3H-D3-20r42-14Force between parallel wires Paper 2Hard9 marks

In a demonstration, a long straight horizontal wire of mass per unit length \(2.5\times10^{-3}\) kg m−1 floats above a fixed identical wire, parallel to it. The wires form parts of one circuit, so both carry the same current \(I\). The separation of the wires is 10 mm at equilibrium. Use \(g=9.8\) m s−2 and \(\mu_0=4\pi\times10^{-7}\) T m A−1.

(a)

State, with a reason, the relative direction of the currents in the two wires.

(2)
(b)

Determine the current \(I\).

(3)
(c)

The upper wire is pushed slightly downwards and released. Explain why it returns towards its equilibrium position.

(2)
(d)

The current is increased to 42 A. Calculate the new equilibrium separation of the wires.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
the currents are in opposite directions✓ 1
the force between the wires must be repulsive to support the weight of the upper wire, and opposite currents repel✓ 1
Part (b)
M1 magnetic force per unit length \(=\frac{\mu_0I^2}{2\pi d}\)✓ 1
M1 equated to weight per unit length: \(\frac{\mu_0I^2}{2\pi d}=2.5\times10^{-3}\times9.8=0.0245\) N m−1✓ 1
A1 \(I=35\) A✓ 1
Part (c)
the separation decreases, so the repulsive force per unit length (proportional to \(1/d\)) becomes greater than the weight✓ 1
there is a net upward force, so the wire is accelerated back upwards✓ 1
Part (d)
M1 at equilibrium \(d\propto I^2\), so \(d=0.010\times\left(\frac{42}{35}\right)^2\)✓ 1
A1 \(1.44\times10^{-2}\) m (0.0144 m) (ECF from (b))✓ 1

Common error: The wires carry the same current, so the force per unit length depends on \(I^2\), not on \(I\).