D.4 Induction (HL) – IB Physics exam questions with mark scheme

1HardHL only5 marks

A coil of 100 turns has a magnetic flux Φ passing through each turn that varies with time as shown.

(a) Calculate the magnitude of the induced emf between 0 and 0.20 s. [2]

(b) State the induced emf between 0.20 s and 0.40 s and give a reason. [1]

(c) Calculate the magnitude of the induced emf between 0.40 s and 0.50 s and state how its polarity compares with that in (a). [2]

(a) ε = N ΔΦ/Δt (Faraday's law), using gradient 4.0 × 10⁻³ Wb ÷ 0.20 s(1)
(a) ε = 100 × 0.020 = 2.0 V(1)
(b) Zero, because the flux is constant so there is no change in flux linkage(1)
(c) ε = 100 × 4.0 × 10⁻³ ÷ 0.10 = 4.0 V(1)
(c) Polarity is opposite (flux is decreasing rather than increasing, so Lenz's law gives a reversed current/emf)(1)
Answer: (a) 2.0 V; (b) 0 V; (c) 4.0 V, opposite polarity
Examiner tip: Convert mWb to Wb and remember the number of turns – emf depends on flux linkage NΦ.

2HardHL only5 marks

A rectangular coil of 50 turns and area 0.020 m² rotates at a constant frequency of 50 Hz in a uniform magnetic field of flux density 0.15 T, forming a simple AC generator. The axis of rotation is perpendicular to the field.

(a) Calculate the maximum magnetic flux linkage through the coil. [1]

(b) The coil turns through a quarter of a revolution, starting from the position where its plane is perpendicular to the field. Calculate the average emf induced during this quarter turn. [2]

(d) The frequency of rotation is doubled. State the effect on the peak emf and on the period of the output. [1]

(a) NΦ = NBA = 50 × 0.15 × 0.020 = 0.15 Wb (accept Wb-turns)(1)
(b) Time for a quarter turn = T/4 = (1/50)/4 = 5.0 × 10⁻³ s and ΔΦ goes from BA to 0 (Δ(NΦ) = 0.15 Wb)(1)
(b) ε = NΔΦ/Δt = 0.15 ÷ 5.0 × 10⁻³ = 30 V (allow ECF from (a))(1)
(c) The plane of the coil is parallel to the field (accept: normal to the coil perpendicular to B; flux linkage zero but changing at its fastest rate)(1)
(d) Peak emf doubles and the period halves (both needed)(1)
Answer: (a) 0.15 Wb; (b) 30 V; (c) coil plane parallel to B; (d) peak emf ×2, period ÷2
Examiner tip: A quarter turn takes T/4 = 1/(4f); the emf is greatest when the flux linkage is zero, because that is when it is changing fastest.

3CoreHL only4 marks

A flat circular coil of 200 turns has radius 4.0 cm. It is placed in a uniform magnetic field of flux density 0.25 T so that the normal to the plane of the coil makes an angle of 60° with the field.

(a) Calculate the magnetic flux through one turn of the coil. [2]

(b) The coil is rotated in 0.050 s until the normal to its plane is parallel to the field. Calculate the average emf induced in the coil during the rotation. [2]

(a) Area $A=\pi r^2=\pi(0.040)^2=5.03\times10^{-3}$ m$^2$ and uses $\Phi=BA\cos\theta$ with $\theta=60°$(1)
(a) $\Phi=0.25\times5.03\times10^{-3}\times\cos60°=6.3\times10^{-4}$ Wb (award [2] for a bare correct answer; using $\sin60°$ giving $1.1\times10^{-3}$ Wb scores [1] max; radius in cm scores [1] max)(1)
(b) Change in flux per turn $\Delta\Phi=BA-BA\cos60°=6.3\times10^{-4}$ Wb (allow ECF from (a))(1)
(b) $\varepsilon=N\Delta\Phi/\Delta t=200\times6.28\times10^{-4}/0.050=2.5$ V (ignore a negative sign; using final flux only, giving 5.0 V, scores [0] for this mark)(1)
Answer: (a) $6.3\times10^{-4}$ Wb (b) 2.5 V
Examiner tip: $\theta$ in $\Phi=BA\cos\theta$ is measured from the normal to the coil, not from its plane; and the emf depends on the change in flux, not the final flux.

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4StandardHL only3 marks

An aircraft with a wingspan of 36 m flies horizontally at 240 m s⁻¹ in a region where the vertical component of the Earth's magnetic field is $4.5\times10^{-5}$ T.

(a) Calculate the emf induced between the wingtips. [2]

(b) An engineer connects a voltmeter between the two wingtips using wires that run along the wings inside the aircraft. Suggest why the voltmeter reads zero. [1]

(a) Uses $\varepsilon=BvL$ with the vertical component of the field(1)
(a) $\varepsilon=4.5\times10^{-5}\times240\times36=0.39$ V (accept 0.389 V; award [2] for a bare correct answer)(1)
(b) The connecting wires move through the same field at the same speed, so the same emf is induced in them, which cancels/opposes the wing's emf around the circuit (accept: no change of flux through the circuit formed by the wing and wires, so no net emf)(1)
Answer: (a) 0.39 V (b) the leads cut the field too, so their emf cancels the wing's
Examiner tip: Only the field component perpendicular to both the wing and the velocity counts; here that is the vertical component.

5StandardHL only6 marks

A metal rod rests across two parallel horizontal rails 0.20 m apart. The rails are joined at one end by a resistor of resistance 0.80 Ω. A uniform magnetic field of flux density 0.50 T acts into the page. The rod is pulled to the right at a constant speed of 3.0 m s⁻¹. The resistance of the rod and rails is negligible and friction can be ignored.

(a) Calculate the emf induced across the rod. [1]

(b) Calculate the current in the resistor. [1]

(c) State the direction of the current in the rod, up the page or down the page. Explain your answer using Lenz's law. [2]

(d) Calculate the external force needed to keep the rod moving at constant speed. [1]

(e) Show that the rate at which this force does work equals the power dissipated in the resistor. [1]

(a) $\varepsilon=BvL=0.50\times3.0\times0.20=0.30$ V(1)
(b) $I=\varepsilon/R=0.30/0.80=0.375$ A (accept 0.38 A; allow ECF from (a))(1)
(c) Current is up the page in the rod (anticlockwise around the circuit as viewed)(1)
(c) Area and hence flux (into page) through the circuit increases, so the induced current produces a field out of the page inside the circuit to oppose the change (accept: force on the rod must oppose its motion, i.e. act to the left, which needs current up the page by the motor rule) (do not award this mark if the direction is wrong)(1)
(d) $F=BIL=0.50\times0.375\times0.20=0.0375$ N (accept 0.038 N; allow ECF from (b))(1)
(e) $Fv=0.0375\times3.0=0.1125$ W and $I^2R=0.375^2\times0.80=0.1125$ W, so they are equal (both values needed; accept $\varepsilon I=0.30\times0.375$)(1)
Answer: (a) 0.30 V (b) 0.375 A (c) up the page (d) 0.038 N (e) both 0.11 W
Examiner tip: For Lenz's law say what the change in flux is and that the induced current's field opposes that change; just writing 'it opposes the motion' without a direction earns little.

6HardHL only6 marks

A strong cylindrical magnet is released from rest at the top of a long vertical copper tube. It takes much longer to reach the bottom than an identical magnet released inside a plastic tube of the same dimensions. The magnet does not touch the tube walls.

Explain, with reference to Faraday's law, Lenz's law and energy, why the magnet falls more slowly in the copper tube and why it reaches a constant speed. [6]

As the magnet falls, the magnetic flux through each ring/section of the tube (below and above the magnet) changes(1)
By Faraday's law an emf is induced in the copper; copper is a conductor so (eddy) currents flow around the tube(1)
Plastic is an insulator so no current flows (an emf may be induced but there is no current), so the magnet falls freely with acceleration g(1)
By Lenz's law the currents produce magnetic fields that oppose the change causing them: repel the approaching magnet below it / attract the receding magnet above it(1)
So there is an upward (retarding) magnetic force on the magnet(1)
The rate of change of flux, and so the emf, current and retarding force, increase with the magnet's speed(1)
Terminal speed is reached when the magnetic force equals the weight, so resultant force and acceleration are zero(1)
Energy: gravitational potential energy is transferred to internal (thermal) energy of the tube via the currents, not to kinetic energy; at terminal speed the loss of GPE per second equals the power dissipated in the tube(1)
Lenz's law is a consequence of energy conservation: if the induced force aided the fall, energy would be created(1)

Level	Marks	What it looks like
Level 3	5–6	Clear, logically ordered explanation linking the changing flux to induced currents in the copper (and none in plastic), the direction of the opposing force from Lenz's law, the force increasing with speed until it balances the weight, and the energy transfer to internal energy. Correct terms used throughout.
Level 2	3–4	Explains that currents are induced in the copper and that they oppose the motion, with some link to terminal speed or energy, but the chain of reasoning has gaps (e.g. no reason why the force grows, or no contrast with plastic).
Level 1	1–2	Isolated relevant points, e.g. 'copper is a conductor' or 'currents are induced', with little or no explanation of why the magnet slows.
Level 0	0	No relevant content.

Answer: Changing flux induces eddy currents in the copper (none in plastic); by Lenz's law their fields oppose the magnet's motion with a force that grows with speed until it equals the weight; GPE becomes internal energy of the tube.
Examiner tip: Top answers link the size of the retarding force to speed; without that link, 'terminal speed' is just asserted and the level drops.

7HardHL only5 marks

A coil rotates at constant angular speed in a uniform magnetic field. The graph shows how the magnetic flux linkage $N\Phi$ through the coil varies with time $t$.

(a) Determine the frequency of rotation of the coil. [1]

(b) Determine the peak emf induced in the coil. [2]

(a) Period read as 20 ms, so $f=1/0.020=50$ Hz(1)
(b) $\omega=2\pi f=314$ rad s$^{-1}$ (allow ECF from (a)) and peak emf $=N\Phi_{max}\,\omega$ with $N\Phi_{max}=0.080$ Wb (or BAN ω seen)(1)
(b) $\varepsilon_0=0.080\times314=25$ V (accept 25.1 V; accept 24–26 V from the maximum gradient of a tangent drawn at 5 ms or 15 ms)(1)
(c) Any one of $t=0$, 10 ms or 20 ms (accept any time where $N\Phi$ is at a maximum or minimum)(1)
(c) emf is proportional to the rate of change of flux linkage / gradient of the graph, and the gradient is zero at a maximum or minimum of $N\Phi$ (do not award for 'flux linkage is maximum' with no reference to rate of change or gradient)(1)
Answer: (a) 50 Hz (b) 25 V (c) e.g. 10 ms: the gradient of the graph is zero there
Examiner tip: Emf follows the gradient of the flux-linkage graph, not its value: it is zero where $N\Phi$ peaks and greatest where $N\Phi$ crosses zero.

8HardHL only6 marks

A student investigates the emf induced when a bar magnet is dropped from rest, north pole first, through a coil connected to a data logger. The magnet falls a height $h$ before it reaches the coil. The data logger shows a positive peak as the magnet enters the coil and a negative peak as it leaves.

(a) Explain why the emf changes sign as the magnet passes through the coil. [2]

(b) Suggest why the negative peak has a larger magnitude than the positive peak. [1]

$h$ / m	0.10	0.20	0.30	0.40	0.50
$\varepsilon$ / V	0.16	0.22	0.28	0.31	0.36

The student predicts that $\varepsilon$ is proportional to $\sqrt{h}$. Using the data, evaluate this prediction. [3]

(a) As the magnet enters, the flux linkage through the coil increases; as it leaves, the flux linkage decreases(1)
(a) The emf is proportional to $-\Delta(N\Phi)/\Delta t$, so the rate of change of flux has opposite signs and so does the emf (accept Lenz's law argument: induced current must repel the entering magnet and attract the leaving one, so it reverses)(1)
(b) The magnet is accelerating, so it is moving faster when it leaves; the flux changes more rapidly, giving a larger emf(1)
(c) Prediction justified: speed at coil $v=\sqrt{2gh}$ and $\varepsilon\propto$ rate of change of flux $\propto v$, so $\varepsilon\propto\sqrt{h}$ (accept: suitable test proposed, e.g. plot $\varepsilon$ against $\sqrt{h}$, or calculate $\varepsilon/\sqrt{h}$)(1)
(c) Correct values for at least three points, e.g. $\varepsilon/\sqrt{h}$ = 0.51, 0.49, 0.51, 0.49, 0.51 V m$^{-1/2}$ (or gradient of $\varepsilon$–$\sqrt{h}$ graph ≈ 0.51 V m$^{-1/2}$ with intercept ≈ 0)(1)
(c) Conclusion: ratio constant to within about 2–4% (or straight line through the origin), so the data supports the prediction (a conclusion with no supporting values scores [0] for this mark)(1)
Answer: (a) flux linkage increases then decreases, so its rate of change reverses (b) magnet faster on exit (c) $\varepsilon/\sqrt{h}\approx0.50$ V m$^{-1/2}$, constant, so supported
Examiner tip: When testing proportionality, use all the data: a constant ratio or a straight line through the origin is the evidence, not just 'both increase'.

9StretchHL only6 marks

A square coil of 20 turns and side 0.10 m has a total resistance of 2.0 Ω. It moves at a constant speed of 0.50 m s⁻¹ towards a region of uniform magnetic field of width 0.30 m, as shown. The field has flux density 0.40 T and is directed into the page. The plane of the coil is perpendicular to the field. Time $t=0$ is when the leading edge of the coil reaches the field.

(a) Calculate the emf induced in the coil while it is entering the field. [1]

(b) Describe how the induced emf varies from $t=0$ until the coil has completely left the field. Give the relevant times. [3]

(a) $\varepsilon=NBLv=20\times0.40\times0.10\times0.50=0.40$ V (using $N=1$, giving 0.020 V, scores [0])(1)
(b) 0 to 0.20 s: emf constant at 0.40 V while the coil enters (allow ECF from (a))(1)
(b) 0.20 s to 0.60 s: emf zero, because the coil is entirely in the field and the flux linkage is constant(1)
(b) 0.60 s to 0.80 s: emf of the same magnitude, 0.40 V, but opposite sign/polarity as the coil leaves(1)
(c) $I=\varepsilon/R=0.40/2.0=0.20$ A and power $=\varepsilon I=0.080$ W (or $I^2R$)(1)
(c) Current flows for a total of 0.40 s, so energy $=0.080\times0.40=0.032$ J (accept via work done: $F=NBIL=0.16$ N over a distance of 0.20 m; allow ECF from (a); using 0.80 s giving 0.064 J scores [1] max)(1)
Answer: (a) 0.40 V (b) 0.40 V for 0–0.20 s; 0 for 0.20–0.60 s; −0.40 V for 0.60–0.80 s (c) 0.032 J
Examiner tip: The emf is zero while the whole coil is inside the field: uniform field means constant flux linkage, even though the coil is moving.

10CoreHL only3 marks

A light aluminium ring hangs from a thread. The north pole of a bar magnet is pushed quickly towards the ring along its axis.

(a) State Lenz's law. [1]

(b) State and explain the direction in which the ring moves. [2]

(a) The direction of the induced emf/current is such that its effects oppose the change that produced it (accept: opposes the change in magnetic flux)(1)
(b) The ring moves away from the magnet (is repelled)(1)
(b) The induced current makes the face of the ring nearest the magnet a north pole, which repels the approaching north pole / opposes the increasing flux (do not award for 'the ring is magnetised' without reference to an induced current)(1)
Answer: (a) induced emf opposes the change causing it (b) away from the magnet: the induced current creates a north pole facing the magnet
Examiner tip: Lenz's law is about opposing the change in flux, not the field itself; 'opposes the magnet' without saying what change earns no credit.

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