These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Data and assumptions for every question: e = 1.60 × 10⁻¹⁹ C; fields are uniform; gravity is ignored; “up” and “down” mean towards the top and bottom of the page; × is a field into the page and · is a field out of the page. A scientific calculator is assumed. Round only at the end.
Question 1 (multiple choice · foundation)
A particle with charge +3.2 × 10⁻¹⁹ C moves at 2.0 × 10⁶ m/s through a uniform 0.40 T field. Its velocity points in exactly the opposite direction to the field. What is the size of the magnetic force on it?
- (A) 2.56 × 10⁻¹³ N
- (B) 1.28 × 10⁻¹³ N
- (C) 0 N
- (D) −2.56 × 10⁻¹³ N
Answer and explanation
Answer: (C). The angle between v and B is 180°, and sin 180° = 0, so F = qvB sin θ = 0. A charge moving along a field line, either way, feels no magnetic force.
- (A) is qvB = 2.56 × 10⁻¹³ N, the force at 90°. It ignores the angle.
- (B) is half of that, as if sin θ were 0.5. That would need an angle of 30° or 150°, not 180°.
- (D) treats “opposite direction” as a negative force. The size of a force cannot be negative, and here it is zero anyway.
Question 2 (multiple choice · core)
An electron moves towards the top of the page through a uniform field that points out of the page. Which way does the magnetic force on the electron point?
- (A) To the left
- (B) To the right
- (C) Into the page
- (D) Towards the top of the page
Answer and explanation
Answer: (A). For a positive charge: fingers up the page (v), curl out of the page (B), thumb points to the right. The electron is negative, so its force is reversed: to the left.
- (B) is the force on a positive charge. It forgets to reverse the rule for the electron.
- (C) is along the field. The magnetic force is always at right angles to B.
- (D) is along the velocity. The magnetic force is always at right angles to v, so it cannot speed the electron up.
Question 3 (multiple choice · core)
An electron moves to the right along a straight line across the page. Point P is a short distance directly below the electron. What is the direction of the magnetic field that the electron produces at P?
- (A) Into the page
- (B) Out of the page
- (C) To the right
- (D) There is no field at P, because the electron is negative
Answer and explanation
Answer: (B). For a positive charge moving to the right, the right-hand rule (thumb along v, fingers curl round the line of motion) gives a field into the page below the line. An electron is negative, so its field is reversed: out of the page at P.
- (A) is the field of a positive charge. It forgets the reversal.
- (C) points along the velocity. The field of a moving charge is at right angles to both v and the line from the charge to P.
- (D) is wrong: any moving charge, positive or negative, makes a magnetic field. The field is zero only at points straight ahead of or behind the charge.
Question 4 (multiple choice · core)
A charged particle moves in a circle at right angles to a uniform magnetic field. Its speed is doubled and, at the same time, the field is halved. By what factor does the radius of its circle change?
- (A) ×1
- (B) ×2
- (C) ×4
- (D) ×¼
Answer and explanation
Answer: (C). r = mv/(|q|B), so r ∝ v/B. Doubling v doubles r; halving B doubles r again. The radius changes by 2 × 2 = 4.
- (A) assumes the two changes cancel. They would cancel only if r depended on the product vB.
- (B) includes only one of the two changes.
- (D) inverts the relationship, as if r were proportional to B/v.
Question 5 (calculation · core)
A singly charged positive ion of mass 1.2 × 10⁻²⁶ kg enters a uniform 0.50 T field at 4.0 × 10⁵ m/s, moving at right angles to the field.
(a) Calculate the magnetic force on the ion. (b) Calculate the radius of its circular path. (c) The ion leaves the field after half a circle. Calculate the time it spends in the field. (d) State the work done on the ion by the magnetic force during this time, and its speed as it leaves. Justify.
Worked solution
- (a) F = |q|vB = (1.60 × 10⁻¹⁹ C)(4.0 × 10⁵ m/s)(0.50 T) = 3.2 × 10⁻¹⁴ N.
- (b) r = mv/(|q|B) = (1.2 × 10⁻²⁶ kg)(4.0 × 10⁵ m/s) ÷ [(1.60 × 10⁻¹⁹ C)(0.50 T)] = 0.060 m (6.0 cm).
- (c) Half a circle has length πr. t = πr/v = π(0.060 m) ÷ (4.0 × 10⁵ m/s) = 4.71 × 10⁻⁷ s.
- (d) Zero work, and the ion leaves at 4.0 × 10⁵ m/s. The magnetic force is always at right angles to the velocity, so it has no component along the motion and does no work. With no work done, the kinetic energy and the speed are unchanged.
Suggested mark points (4): 1 each for (a), (b) and (c) with units; 1 for (d) with the “force at right angles to velocity” reason.
Common errors: using the full circumference in (c), which gives 9.42 × 10⁻⁷ s; and saying the speed changes because “there is a force”. A force changes velocity, but here only its direction.
Question 6 (constructed response · core)
Two parallel plates are 1.2 cm apart, with the top plate at +240 V and the bottom plate at 0 V. A uniform magnetic field of 0.080 T fills the region between them. Protons enter from the left, moving to the right.
(a) Draw a free-body diagram for a proton between the plates, showing the electric and magnetic forces, so that the proton can pass straight through. State the direction of B. (b) Derive an expression for the speed v at which protons pass straight through, in terms of the potential difference ΔV, the plate spacing d and B. Calculate v. (c) A proton enters at 5.0 × 10⁵ m/s. Calculate the net force on it as it enters, and state its direction. (d) A student claims that electrons entering at the speed found in (b) will be deflected, “because their charge is negative”. Evaluate the claim.
Worked solution and suggested Marlbridge rubric
(a) The electric field points from the top plate down, so the electric force on the proton, qE, points down. The magnetic force must point up, with equal size. Fingers to the right (v), thumb up (F): fingers curl into the page, so B points into the page. The diagram shows a dot with an arrow down labelled F_E and an equal arrow up labelled F_B.
(b) For a straight path the forces balance: qE = qvB, with E = ΔV/d. So v = E/B = ΔV/(dB). E = 240 V ÷ 0.012 m = 2.0 × 10⁴ V/m, so v = (2.0 × 10⁴ V/m) ÷ (0.080 T) = 2.5 × 10⁵ m/s.
(c) Magnetic force: qvB = (1.60 × 10⁻¹⁹)(5.0 × 10⁵)(0.080) = 6.4 × 10⁻¹⁵ N, up. Electric force: qE = (1.60 × 10⁻¹⁹)(2.0 × 10⁴) = 3.2 × 10⁻¹⁵ N, down. Net force = 3.2 × 10⁻¹⁵ N, up (towards the top plate).
(d) The claim is wrong. For an electron both forces reverse: the electric force points up and the magnetic force points down. Their sizes are still |q|E and |q|vB, which are equal at v = E/B because the charge cancels. So electrons at 2.5 × 10⁵ m/s also pass straight through.
| Point | What earns it |
|---|---|
| 1 | (a) Electric force down and magnetic force up, drawn equal and opposite |
| 1 | (a) B into the page, with a correct link to v and the required force direction |
| 1 | (b) Starts from qE = qvB with E = ΔV/d, reaching v = ΔV/(dB) |
| 1 | (b) v = 2.5 × 10⁵ m/s |
| 1 | (c) 3.2 × 10⁻¹⁵ N upwards, from the difference between the two forces |
| 1 | (d) Rejects the claim because both forces reverse and the charge cancels in v = E/B |
Do not award the (d) point for “the charge cancels” with no mention that both forces reverse together.
Question 7 (constructed response · stretch)
A thin metal strip, 4.0 mm wide, lies in the plane of the page and carries a current to the right. A uniform 0.20 T field points into the page. The charge carriers are electrons.
(a) State which edge of the strip (top or bottom) is at the higher potential. Justify by describing the motion of the electrons and the force on them. (b) Derive an expression for the Hall potential difference ΔV_H across the strip in terms of the drift speed v, the field B and the width w. (c) In a test strip made of a different material, the drift speed is 2.5 m/s. Calculate ΔV_H. (d) The field is tripled with everything else unchanged. Predict the new ΔV_H, and state whether the higher-potential edge changes.
Worked solution and suggested Marlbridge rubric
(a) The current is to the right, so the electrons drift to the left. For a positive charge moving left with B into the page, the right-hand rule gives a force down the page; an electron feels the opposite, up. Electrons collect on the top edge, which becomes negative. The bottom edge is left positive, so the bottom edge is at the higher potential.
(b) Charge builds up until the electric force from the separated charges balances the magnetic force on each carrier: qE_H = qvB. With E_H = ΔV_H/w, this gives ΔV_H = vBw.
(c) ΔV_H = (2.5 m/s)(0.20 T)(4.0 × 10⁻³ m) = 2.0 × 10⁻³ V = 2.0 mV.
(d) ΔV_H ∝ B, so tripling B triples ΔV_H: 6.0 mV. The higher-potential edge does not change, because the directions of v and B are unchanged.
| Point | What earns it |
|---|---|
| 1 | (a) Electrons drift opposite to the current (to the left) |
| 1 | (a) Force on electrons up the page, with the reversal for negative charge stated |
| 1 | (a) Concludes the bottom edge is at the higher potential, linked to electrons collecting on the top |
| 1 | (b) Balance of electric and magnetic forces with E_H = ΔV_H/w, giving ΔV_H = vBw |
| 1 | (c) 2.0 mV with the width converted to metres |
| 1 | (d) 6.0 mV from ΔV_H ∝ B, and the same edge stays higher |
Accept in (a) the shortcut “carriers of either sign are pushed to the top edge; electrons make it negative”, if the force direction is shown.
How did you do?
- Q1 or Q5 wrong: re-read “The force on a moving charge” and Worked example 1 in the study guide.
- Q2 or Q3 wrong: practise the two right-hand rules, then reverse them for negative charges.
- Q4 wrong: revisit “Circular motion” and the relationship r = mv/(|q|B).
- Q6 or Q7 incomplete: work through Worked examples 2 and 3 again.
Then tick off the topic checklist.