Magnetism and Electromagnetism: Unit Diagnostic (Physics 2 Unit 12)

What this diagnostic is for

Use this short check to find which topics in Unit 12 (Magnetism and Electromagnetism) you should revisit. These are original Marlbridge practice questions, not past exam questions. They are not calibrated against real exam results, and your total does not predict an exam score. What matters is which questions you miss.

How to take it:

Work without notes, in about 30 minutes. Write a reason for each answer.
A scientific calculator is allowed. Data: μ₀/(2π) = 2 × 10⁻⁷ T·m/A; e = 1.60 × 10⁻¹⁹ C; electron mass = 9.11 × 10⁻³¹ kg. Wires are long and straight; current means conventional current; “up” and “down” mean towards the top and bottom of the page.
Open each answer only after you have committed to your own.

Question 1 (multiple choice · 12.1)

A bar magnet lies across the page with its north pole at the right-hand end. Point X is inside the magnet, at its centre. What is the direction of the magnetic field at X?

(A) To the left, towards the south pole
(B) To the right, towards the north pole
(C) There is no field at X, because the fields of the two poles cancel
(D) Up the page, at right angles to the magnet

Answer and explanation

Answer: (B). Magnetic field lines are closed loops. Outside the magnet they run from N round to S, so inside they run from S back to N, to the right.

(A) copies the outside rule to the inside, as if lines started and stopped at the poles.
(C) treats the poles like separate charges.
(D) Inside a bar magnet the field runs along its length.

If you missed this: read “Dipoles, poles and field lines” in the Topic 12.1 study guide.

Question 2 (multiple choice · 12.1)

An unmagnetised iron nail is placed near the south pole of a magnet. End A of the nail is closest to the magnet. Which statement is correct?

(A) End A becomes a south pole and the nail is repelled.
(B) Nothing happens, because the nail is not a magnet.
(C) The nail is attracted because the magnet gives it a net electric charge.
(D) End A becomes a north pole and the nail is attracted.

Answer and explanation

Answer: (D). Iron is ferromagnetic. The field lines its domains up, so the nail becomes an induced magnet with N nearest the S pole. Unlike poles attract.

(A) In iron the induced pole facing the magnet is always the opposite one, so both poles attract iron.
(B) ignores induced magnetism.
(C) mixes up magnetic and electric effects.

If you missed this: read “Permanent and induced magnetism” in the Topic 12.1 study guide.

Question 3 (multiple choice · 12.2)

An electron moves into the page through a uniform magnetic field that points to the right. Which way does the magnetic force on it point?

(A) Up the page
(B) Down the page
(C) To the right
(D) Into the page

Answer and explanation

Answer: (A). For a positive charge: fingers into the page (v), curl towards the right (B), thumb points down. The electron is negative, so its force is reversed: up.

(B) is the force on a positive charge.
(C) points along B, and (D) along v. The magnetic force is at right angles to both.

If you missed this: read “Direction: the right-hand rule” in the Topic 12.2 study guide.

Question 4 (multiple choice · 12.2)

A proton moves in a circle at right angles to a uniform magnetic field. Its speed is doubled; the field is unchanged. What happens to the radius and to the time for one orbit?

(A) The radius doubles; the time doubles.
(B) The radius is multiplied by 4; the time doubles.
(C) The radius doubles; the time is unchanged.
(D) The radius is unchanged; the time halves.

Answer and explanation

Answer: (C). qvB = mv²/r gives r = mv/(qB), so r ∝ v. The time for one orbit is 2πr/v = 2πm/(qB): twice the distance at twice the speed takes the same time.

(A) forgets that the proton also travels twice as fast.
(B) treats r as proportional to v².
(D) assumes the field alone fixes the size of the circle.

If you missed this: read “Circular motion” in the Topic 12.2 study guide.

Question 5 (multiple choice · 12.3)

Two long parallel wires are 6.0 cm apart, at right angles to the page. One carries 3.0 A out of the page; the other carries 6.0 A into the page. What is the size of the field midway between them?

(A) 2.0 × 10⁻⁵ T
(B) 6.0 × 10⁻⁵ T
(C) 3.0 × 10⁻⁵ T
(D) 4.0 × 10⁻⁵ T

Answer and explanation

Answer: (B). Each wire is 3.0 cm from the midpoint. B = μ₀I/(2πr) gives 2.0 × 10⁻⁵ T (3.0 A wire) and 4.0 × 10⁻⁵ T (6.0 A wire). With the currents in opposite directions, the right-hand rule gives both fields the same direction between the wires, so they add.

(A) subtracts. Fields oppose between the wires only when the currents are in the same direction.
(C) uses the 6.0 cm separation as r for both wires.
(D) includes only the larger current.

If you missed this: read “Fields from more than one wire” in the Topic 12.3 study guide.

Question 6 (multiple choice · 12.3)

A 0.15 m length of wire in the plane of the page carries 4.0 A to the right. It is in a uniform 0.050 T field pointing out of the page. What is the magnetic force on it?

(A) 0.030 N, up the page
(B) 0.030 N, out of the page
(C) 0 N, because the field is at right angles to the wire
(D) 0.030 N, down the page

Answer and explanation

Answer: (D). The current is at right angles to B, so F = IℓB = (4.0 A)(0.15 m)(0.050 T) = 0.030 N. Fingers right along the current, curl out of the page towards B: the thumb points down.

(A) reverses the rule, often by using the left hand.
(B) puts the force along B; it is at right angles to both.
(C) has the angle rule backwards. The force is largest at 90°.

If you missed this: read “The force on a current-carrying wire” in the Topic 12.3 study guide.

Question 7 (multiple choice · 12.4)

A long straight wire in the plane of the page carries a current up the page. A small square loop lies flat on the page to the right of the wire. The wire’s current is increasing. Which statement is correct?

(A) A clockwise current is induced in the loop while the wire’s current increases.
(B) No current is induced, because the wire does not touch the loop.
(C) An anticlockwise current is induced in the loop while the wire’s current increases.
(D) An anticlockwise current is induced, and it keeps flowing after the wire’s current becomes steady.

Answer and explanation

Answer: (C). (1) To the right of an upward current, the wire’s field points into the page. (2) The flux into the page increases. (3) The induced field opposes the increase, so it points out of the page. (4) Thumb out of the page: the current is anticlockwise.

(A) is right only for a decreasing flux.
(B) Induction needs only a changing flux, not contact.
(D) misses that a steady flux induces nothing.

If you missed this: read “Lenz’s law: which way does the current go?” in the Topic 12.4 study guide.

Question 8 (multiple choice · 12.4)

A flat coil of 40 turns, each of area 0.025 m², has its area vector along a uniform 0.60 T field. It is flipped over through 180° in 0.15 s. What is the size of the average induced emf?

(A) 8.0 V
(B) 4.0 V
(C) 0.20 V
(D) 0 V

Answer and explanation

Answer: (A). The flux through one turn starts at +BA = +0.015 Wb. After the flip the area vector points against B, so Φ = −0.015 Wb and ΔΦ = −0.030 Wb. |ε| = N|ΔΦ|/Δt = 40 × 0.030 ÷ 0.15 = 8.0 V.

(B) forgets that the flux changes sign, so it uses 0.015 Wb.
(C) leaves out the 40 turns.
(D) ignores the sign of the flux.

If you missed this: read “Magnetic flux” and “Faraday’s law” in the Topic 12.4 study guide.

Question 9 (short answer · 12.2)

An electron moves at 4.0 × 10⁶ m/s at right angles to a uniform magnetic field, in a circle of radius 2.5 mm. It goes round clockwise as seen from the front of the page.

(a) Does the field point into or out of the page? Justify. (b) Calculate the strength of the field. (c) State the work done on the electron by the magnetic force in one orbit, and explain.

Worked answer

(a) Try B into the page with the electron moving right. A positive charge would be pushed up, so the electron is pushed down, towards a centre below it. Moving right at the top of a circle is clockwise, as observed, so B points into the page.
(b) evB = mv²/r, so B = mv/(er) = (9.11 × 10⁻³¹)(4.0 × 10⁶) ÷ [(1.60 × 10⁻¹⁹)(2.5 × 10⁻³)] = 9.1 × 10⁻³ T.
(c) Zero. The force is always at right angles to the velocity, so it does no work and the speed stays 4.0 × 10⁶ m/s.

If you missed this: read “Circular motion” and Worked example 1 in the Topic 12.2 study guide.

Question 10 (short answer · 12.3)

A compass lies flat on a bench and points north, where the horizontal part of Earth’s field is 2.0 × 10⁻⁵ T. A long wire runs north–south, 4.0 cm directly above the compass, carrying 5.0 A north.

(a) Calculate the size of the wire’s field at the compass and state its direction. (b) Calculate the angle of the needle from north.

Worked answer

(a) B = μ₀I/(2πr) = (2 × 10⁻⁷)(5.0) ÷ 0.040 = 2.5 × 10⁻⁵ T. Thumb north along the current: below the wire the fingers curl towards the west.
(b) The needle lines up with the total horizontal field: 2.0 × 10⁻⁵ T north plus 2.5 × 10⁻⁵ T west. tan θ = 2.5/2.0, so θ = 51° west of north.

If you missed this: read “The field around a long straight wire” in the Topic 12.3 study guide.

Question 11 (short answer · 12.4)

A student pushes the north pole of a magnet quickly into a coil whose ends are joined. She then cuts the wire, so the ends are no longer joined, and repeats the push at the same speed.

(a) Compare the force needed in the two cases. Explain using Lenz’s law. (b) Is there an emf in the coil in the second case? Explain. (c) In the first case, where does the extra work she does go?

Worked answer

(a) With the ends joined, the induced current’s field opposes the increasing flux: the end facing the magnet acts as a north pole and repels it, so she must push harder. With the ends cut, no current flows and there is no opposing force.
(b) Yes. The flux still changes, so an emf is induced; there is just no complete path for a current.
(c) It becomes thermal energy in the coil’s resistance. Lenz’s law is energy conservation.

If you missed this: read “Lenz’s law” in the Topic 12.4 study guide.

Your next step

Count a short answer as wrong if any part is missing, and any answer as wrong if it was a guess.

Topic	Question(s)	If you missed it, read
12.1 Magnetic fields	1, 2	Topic 12.1 study guide
12.2 Magnetism and moving charges	3, 4, 9	Topic 12.2 study guide
12.3 Magnetism and current-carrying wires	5, 6, 10	Topic 12.3 study guide
12.4 Electromagnetic induction and Faraday’s law	7, 8, 11	Topic 12.4 study guide

How to use your result

Start with the topic where you missed the most. Read its study guide, then do that topic’s practice questions.
Look at why. If most misses were directions (Questions 3, 6, 7, 9(a), 10(a)), practise the right-hand rules across all four topics before re-reading whole guides.
Missed nothing? Go straight to the Unit 12 mixed review.