The questions in this review are original Marlbridge practice questions, not past exam questions, and each combines two or more topics. The rubrics are a suggested Marlbridge rubric, not official scoring. Data: μ₀/(2π) = 2 × 10⁻⁷ T·m/A; e = 1.60 × 10⁻¹⁹ C. Wires are long and straight; current means conventional current. All data are invented.
Big ideas of the unit
- A magnetic field is a vector field made by moving charge. Its lines are closed loops (no monopoles), and a magnet is many electron dipoles lined up (Topic 12.1).
- One force law for moving charge. F = |q|vB sin θ, at right angles to v and B. It does no work, so it bends paths into circles (Topic 12.2).
- Electric and magnetic forces act independently. Balancing them gives v = E/B and the Hall potential difference ΔV_H = vBw.
- A wire is many moving charges. Adding up their fields gives circles with B = μ₀I/(2πr); adding up their forces gives F = IℓB sin θ (Topic 12.3).
- Fields superpose. Fields from wires, magnets and Earth add as vectors; a sensor reads the total.
- A changing flux induces an emf. Φ = BA cos θ; |ε| = N|ΔΦ|/Δt; a moving rod gives ε = Bℓv (Topic 12.4).
- Lenz’s law ties it together. The induced current’s field opposes the change, and the force on it opposes the motion.
Key relationships and methods
| Idea | Relationship or method | Watch out for |
|---|---|---|
| Field lines (12.1) | Closed loops; outside N → S, inside S → N | Lines that start or stop |
| Materials (12.1) | Ferro: strong, can stay; para: weak attraction; dia: weak repulsion | μ of a material is not constant |
| Force on a charge (12.2) | F = |q|vB sin θ, ⊥ to v and B | Reverse for negative charges; F = 0 along B |
| Circular path (12.2) | r = mv/(|q|B); period 2πm/(|q|B) | Magnetic force does no work |
| Crossed fields, Hall (12.2) | v = E/B; ΔV_H = vBw | Carriers of either sign go to the same edge |
| Field of a wire (12.3) | B = μ₀I/(2πr); circles round the wire | 1/r, not 1/r² |
| Force on a wire (12.3) | F = IℓB sin θ; parallel currents attract | ℓ is only the length in the field |
| Flux, Faraday (12.4) | Φ = BA cos θ; |ε| = N|ΔΦ|/Δt; I = ε/R | θ from the area vector; flux can be negative |
| Lenz, moving rod (12.4) | Induced field opposes the change; ε = Bℓv | Opposes the change, not the field |
Question 1 (multiple choice · mixed)
Topics 12.1, 12.2. Model Earth’s field as a dipole. Protons arrive from space moving vertically downward, all at the same speed. Where is the magnetic force on such a proton largest, just above the ground?
- (A) Near the magnetic poles, because the field is strongest there
- (B) Everywhere the same, because the protons all have the same speed
- (C) Near the magnetic equator, because there the field is roughly horizontal, at right angles to the proton’s velocity
- (D) Nowhere: the force is zero everywhere, because Earth’s field does no work on the protons
Answer and explanation
Answer: (C). F = qvB sin θ. In the dipole model the field near the magnetic equator is nearly horizontal, so θ ≈ 90°. Near the poles it is nearly vertical, almost parallel to v, so sin θ ≈ 0.
- (A) The field is stronger near the poles, but sin θ is small there.
- (B) ignores the angle.
- (D) confuses “no work” with “no force”. The force changes the direction, not the speed.
Question 2 (multiple choice · mixed)
Topics 12.1, 12.3, 12.4. A bar magnet is dropped, north pole down, through a horizontal copper ring. Which statement is correct?
- (A) The ring pushes up on the magnet as it approaches and pulls it down as it leaves.
- (B) The ring pushes up on the magnet both as it approaches and as it leaves; the current in the ring reverses between the two stages.
- (C) There is no force, because copper is not ferromagnetic.
- (D) The ring pushes up on the magnet both times, and the current flows the same way both times.
Answer and explanation
Answer: (B). The magnet’s field points down through the ring throughout. On approach the downward flux increases, so the induced field points up (anticlockwise from above): the ring’s top face acts as N and repels the falling N pole. As the magnet leaves, the flux decreases, so the induced field points down (clockwise): the ring’s lower face acts as N and pulls the magnet’s S pole up. Both forces oppose the motion.
- (A) is right on approach but not as the magnet leaves. A force helping the fall would create energy from nothing.
- (C) The force comes from the induced current.
- (D) misses that increasing and decreasing flux need opposite induced fields.
Question 3 (constructed response · mixed)
Topics 12.2, 12.3. Two long parallel wires, P and Q, are at right angles to the page, 10 cm apart, with Q to the right of P. P carries 12 A out of the page; Q carries 4.0 A into the page. M is the midpoint.
(a) Calculate the size and direction of the total magnetic field at M. (b) Calculate the force per metre on each wire and state whether they attract or repel. (c) Find the point on the line through P and Q where the total field is zero. (d) A proton passes through M moving out of the page at 3.0 × 10⁵ m/s. Calculate the magnetic force on it and state its direction.
Worked solution and suggested Marlbridge rubric
(a) At M (5.0 cm from each): B_P = (2 × 10⁻⁷)(12) ÷ 0.050 = 4.8 × 10⁻⁵ T and B_Q = (2 × 10⁻⁷)(4.0) ÷ 0.050 = 1.6 × 10⁻⁵ T. P’s current is out of the page (anticlockwise circles) and M is to its right, where the circle runs up. Q’s current is into the page (clockwise) and M is to its left, where the circle also runs up. Total: 6.4 × 10⁻⁵ T, up the page.
(b) F/ℓ = μ₀I₁I₂/(2πd) = (2 × 10⁻⁷)(12)(4.0) ÷ 0.10 = 9.6 × 10⁻⁵ N/m on each wire (a Newton’s third law pair). Opposite currents repel.
(c) Between the wires the fields add. Left of P, P’s field is always larger. Right of Q the fields oppose and the weaker current is closer. At x beyond Q: 12/(0.10 + x) = 4.0/x, so x = 0.050 m. The field is zero 5.0 cm to the right of Q (check: 1.6 × 10⁻⁵ T from each).
(d) F = evB = (1.60 × 10⁻¹⁹)(3.0 × 10⁵)(6.4 × 10⁻⁵) = 3.1 × 10⁻¹⁸ N. Fingers out of the page (v), curl up the page (B): the thumb points left, towards P. Like a parallel current, it is attracted to P.
| Point | What earns it |
|---|---|
| 1 | B_P = 4.8 × 10⁻⁵ T and B_Q = 1.6 × 10⁻⁵ T |
| 1 | Both fields up at M, with reasons; total 6.4 × 10⁻⁵ T |
| 1 | 9.6 × 10⁻⁵ N/m on each wire, repulsion |
| 1 | Zero point outside the pair, on Q’s side, with a reason |
| 1 | 5.0 cm to the right of Q |
| 1 | Proton force 3.1 × 10⁻¹⁸ N |
| 1 | Direction towards P, with the right-hand rule stated in words |
Total: 7 points. Carry forward an error in (a) into (d) once.
Question 4 (constructed response · mixed)
Topics 12.1, 12.2, 12.3. A long straight wire lies flat on a horizontal bench (the page) and carries 15 A up the page. A Hall sensor strip lies flat on the bench 3.0 cm to the right of the wire. The sensor’s own current flows to the right, carried by positive charge carriers drifting at 12 m/s. The strip is 4.0 mm wide (measured up the page).
(a) Calculate the wire’s field at the sensor and state its direction. (b) State which edge (top or bottom) is at the higher potential, with a reason. Calculate the Hall potential difference. (c) The sensor is moved to 6.0 cm from the wire and the current is raised to 45 A. Predict the new Hall potential difference. (d) The sensor responds to the total field at right angles to the bench. Explain why Earth’s field affects the readings, and describe how the student can remove its effect.
Worked solution and suggested Marlbridge rubric
(a) B = (2 × 10⁻⁷)(15) ÷ 0.030 = 1.0 × 10⁻⁴ T. Thumb up the page: on the right of the wire the field points into the page.
(b) Positive carriers move right; with B into the page the force on them points up the page (fingers right, curl into the page, thumb up), so they collect on the top edge, which is at the higher potential. ΔV_H = vBw = (12)(1.0 × 10⁻⁴)(4.0 × 10⁻³) = 4.8 × 10⁻⁶ V (4.8 μV).
(c) B ∝ I/r: the current is ×3 and the distance ×2, so B is ×1.5. ΔV_H ∝ B, so ΔV_H = 7.2 μV.
(d) At most places Earth’s field has a vertical part of tens of microtesla, the same order as the wire’s 100 μT. Read the sensor with the wire’s current off and subtract that value from each reading.
| Point | What earns it |
|---|---|
| 1 | B = 1.0 × 10⁻⁴ T |
| 1 | Into the page, from the right-hand rule for a wire |
| 1 | Top edge higher, because the force on positive carriers points up |
| 1 | ΔV_H = 4.8 μV |
| 1 | 7.2 μV from B ∝ I/r and ΔV_H ∝ B |
| 1 | Earth’s vertical field is comparable in size, plus a valid method to remove it |
Total: 6 points.
Question 5 (constructed response · mixed)
Topics 12.3, 12.4. A wire in the plane of the page carries a steady 30 A up the page. A rectangular loop lies flat on the page to the right of the wire. Its two vertical sides are 0.10 m long. At one instant the near side is 2.0 cm from the wire and the far side is 6.0 cm from it. The loop is moving directly away from the wire at 0.50 m/s. Its resistance is 0.020 Ω.
(a) State the direction of the wire’s field through the loop, and the direction of the induced current. Justify with Lenz’s law. (b) Treating each vertical side as a rod moving across a field, calculate the emf in each, then the net emf and the current at this instant. (c) Calculate the net magnetic force on the loop and state its direction. (d) The loop instead moves parallel to the wire. Is a current induced? Explain.
Worked solution and suggested Marlbridge rubric
(a) To the right of an upward current, the field points into the page. B ∝ 1/r, so as the loop moves away the flux into the page decreases. The induced field opposes the decrease, so it points into the page inside the loop: the current is clockwise.
(b) Field at the near side: (2 × 10⁻⁷)(30) ÷ 0.020 = 3.0 × 10⁻⁴ T. At the far side: (2 × 10⁻⁷)(30) ÷ 0.060 = 1.0 × 10⁻⁴ T. emfs: Bℓv = (3.0 × 10⁻⁴)(0.10)(0.50) = 1.5 × 10⁻⁵ V and (1.0 × 10⁻⁴)(0.10)(0.50) = 5.0 × 10⁻⁶ V. Both push positive charge up their side, so around the loop they oppose. Net emf = 1.0 × 10⁻⁵ V, the near side winning (clockwise, as in (a)). I = ε/R = 1.0 × 10⁻⁵ ÷ 0.020 = 5.0 × 10⁻⁴ A.
(c) Near side: current up, field into the page, so F = IℓB = (5.0 × 10⁻⁴)(0.10)(3.0 × 10⁻⁴) = 1.5 × 10⁻⁸ N to the left. Far side: current down, 5.0 × 10⁻⁹ N to the right. The top and bottom forces cancel. Net: 1.0 × 10⁻⁸ N towards the wire, opposing the motion.
(d) No. The loop stays the same distance from the wire, so the flux is constant.
| Point | What earns it |
|---|---|
| 1 | Field into the page, from the right-hand rule |
| 1 | Flux decreases because B falls with r, so clockwise current |
| 1 | Fields 3.0 × 10⁻⁴ T and 1.0 × 10⁻⁴ T at the two sides |
| 1 | Net emf 1.0 × 10⁻⁵ V, recognising that the two side emfs oppose |
| 1 | I = 5.0 × 10⁻⁴ A |
| 1 | Net force 1.0 × 10⁻⁸ N towards the wire |
| 1 | (d) No emf, because the flux is unchanged |
Total: 7 points. Accept ε = ℓv(B_near − B_far) from Faraday’s law directly.
Question 6 (constructed response · mixed)
Topics 12.1, 12.3, 12.4. A long coil lies along the page with its axis left to right. Seen from its right-hand end, its current circulates anticlockwise. When the coil is empty, the field inside is 2.0 mT. A small search coil of 100 turns, each of area 1.5 × 10⁻⁴ m², sits inside at the centre, with the same axis.
(a) State the direction of the field inside the long coil, with a reason. (b) The long coil’s current is switched off steadily in 0.030 s. Calculate the average emf in the search coil, and give the direction of the induced current in it as seen from the right-hand end. (c) An iron rod is now placed inside so it fills the long coil, with the search coil wound around it. With the current on, the field in the rod is 0.50 T. Calculate μ/μ₀ for the iron. (d) When the current is switched off in 0.030 s, the field in the rod falls only to 0.020 T. Calculate the average emf, and explain why the field does not fall to zero. (e) A student predicts that doubling the long coil’s current would give 1.0 T in the rod and double the emf. Evaluate the prediction.
Worked solution and suggested Marlbridge rubric
(a) Right fingers curled anticlockwise as seen from the right-hand end put the thumb towards that end: the field points to the right.
(b) Φ per turn = BA = (2.0 × 10⁻³)(1.5 × 10⁻⁴) = 3.0 × 10⁻⁷ Wb, falling to zero. |ε| = N|ΔΦ|/Δt = 100 × 3.0 × 10⁻⁷ ÷ 0.030 = 1.0 × 10⁻³ V. The flux to the right decreases, so the induced field also points right: the current is anticlockwise seen from the right-hand end.
(c) μ/μ₀ = 0.50 T ÷ 2.0 × 10⁻³ T = 250.
(d) |ε| = 100 × (0.50 − 0.020)(1.5 × 10⁻⁴) ÷ 0.030 = 0.24 V. Iron is ferromagnetic: many domains stay lined up after the external field is removed.
(e) The prediction assumes μ is constant, but permeability changes with field strength. With many domains already aligned at 0.50 T, the rod’s field and the emf will most likely rise by less than double.
| Point | What earns it |
|---|---|
| 1 | Field to the right, from the right-hand rule for a loop |
| 1 | emf 1.0 mV |
| 1 | Induced current anticlockwise, with Lenz reasoning |
| 1 | μ/μ₀ = 250 |
| 1 | emf 0.24 V from a 0.48 T change |
| 1 | Field remains because ferromagnetic domains stay partly aligned |
| 1 | Rejects the prediction because μ depends on field strength |
Total: 7 points.
How did you do?
Questions 1–2 are worth 1 point each; Questions 3–6 are worth 27 in total. There is no pass mark: use the checklist matching the points you lost.
- field lines, materials, permeability, Earth’s field (Q1, Q2, 4(d), 6(c)–(e)): Topic 12.1 checklist
- force on a moving charge, Hall effect (Q1, 3(d), 4(b)): Topic 12.2 checklist
- wire fields, adding fields, forces on wires (Q3, 4(a), 5(b)–(c), 6(a)): Topic 12.3 checklist
- flux, Faraday’s and Lenz’s laws (Q2, Q5, 6(b), 6(d)): Topic 12.4 checklist
For a quicker topic-by-topic check, take the Unit 12 diagnostic.