IB Physics HL D.4 Induction Questions

SLSL HLHL

Practise IB Physics HL D.4 by applying magnetic flux, Faraday’s law, Lenz’s law and induced-emf evidence to moving conductors and coils.

84 questions in this syllabus scope, with question parts counted together.

Exam points & syllabus

Exam points

Define magnetic flux as the normal field component multiplied by surface area, Φ = BA cosθ, with θ measured from the surface normal.
Distinguish single-turn flux BA from N-turn linkage NBA when the field is normal to the coil plane.
Keep signed flux and flux magnitude distinct: signed extrema are −BA and +BA, while the magnitude ranges from zero to BA.
For a fixed ring approached by a magnet, explain increasing flux by the increasing field at the ring as the magnet gets closer.
State that induced emf equals the rate of change of magnetic flux linkage, ε = −NΔΦ/Δt; a flux change alone does not specify emf.
For a fixed coil in a changing uniform field, use ε = NA cosθ ΔB/Δt, using the normal angle and the supplied field-time data.
Read emf magnitude from the steepest absolute flux–time gradient, convert graph units, and multiply by N when the graph shows per-turn flux.
A measured emf gives the flux-linkage rate directly; divide by N only when the rate of flux change through one turn is requested.
Construct signed emf from the negative flux gradient: emf vanishes at flux extrema and is greatest where flux changes fastest.
If flux endpoints and resistance are supplied, calculate average emf from ΔΦ/Δt, then current from I = ε/R.
Use signed flux for field reversal: changing B to −B changes flux by 2BA, giving average emf magnitude 2BA/Δt.
For a rectangular N-turn coil turned through 180° from normal alignment, |ΔΦ| = 2BA and average |ε| = 2NBA/Δt.
At the same flux rate, induced emf scales with turn count; at fixed flux change it scales inversely with the time taken.
For two requested emf factors, relate each to flux change, time interval or linkage rate; listing factors or saying “more emf” is insufficient.
If a field-current relation is supplied, find ΔB and the fixed coil’s ΔΦ before NΔΦ/Δt; retain any uniform-field approximation.
For a straight rod or rectangular loop entering B, flux rate equals B times swept-area rate; compare the transverse edge lengths.
A rectangular loop has constant emf during entry and exit, opposite signs on the two stages, and zero emf while wholly inside a uniform field.
Explain motional emf through magnetic forces on conduction electrons, their separation along a conductor and the resulting potential difference.
Derive ε = BvL by balancing qvB with qE, then using E = V/L; the charge cancels in the steady state.
Alternatively derive ε = BvL from swept area LvΔt, flux change BLvΔt and Faraday’s rate of change of flux.
Use electron motion to identify polarity: the electron-rich end is negative, while the positive ions remain in the metal lattice.
Apply BvL with the perpendicular field component and active length; convert μT, mT, mV, cm and km before calculating.
When inferring B from motional emf, show B = ε/(vL) and a correctly aligned substitution if both method steps are required.
Check whether the question asks for one wire or an entire coil: one active wire uses BvL without multiplying by the coil’s turn count.
For an N-turn entering rectangular coil, multiply the one-turn emf by N and use the transverse side length, not the perimeter.
For a voltage-versus-active-length graph at fixed B and v, the gradient equals Bv; divide it by B to obtain the speed.
For fixed B and L, emf follows velocity, so read the displacement-time gradient rather than copying the displacement graph.
Find speed before applying BvL: for a vertical rod moving around a horizontal circle, use v = 2πr/T and the perpendicular-motion maximum.
Sketch the orbiting rod’s emf as a sinusoid with the motion period, showing at least half a cycle and any phase when the scheme permits it.
An open coil can have emf without current. For the falling square coil, find its free-fall speed and use one active edge in BvL.
At terminal speed balance mg = BIL, combine I = BvL/R and solve v = mgR/(B²L²), using SI loop data.
For a rod moving steadily through a closed resistive circuit, mechanical input power equals electrical power, P = (BvL)²/R.
For an entering and leaving loop, calculate dissipated energy only over the intervals with induced current, using E = Fvt or E = VIt.
In the supplied induction apparatus, use the given magnetic force to obtain rod current from I = F/(BL), showing the required substitution.
Explain secondary emf through changing primary current, changing field and changing secondary flux linkage; current needs a closed secondary circuit.
A steady primary current gives no secondary emf; switching produces an emf only while linked flux changes.
The core transfers flux to the secondary; flux leakage reduces linked flux and induced emf, rather than explaining core heating.
Explain qualitative self-induction by a changing coil current changing its own flux and inducing emf; no RC transient equation is required.
For a spring magnet, induced current is zero at a turning point; higher frequency gives larger peaks and a shorter period.
Reducing the magnet’s oscillation amplitude gives smaller induced-current peaks with the same period in the supplied spring model.
A falling magnet gives opposite emf pulses on entry and exit; greater exit speed makes the second pulse taller and narrower.
The signed emf-time area gives flux-linkage change: equal return changes give equal opposite areas even when the pulse heights differ.
A lower magnet drop height makes the interaction longer and the emf peaks smaller, while the pulse areas stay the same.
State Lenz’s law as opposition to the change producing induced emf or current, and link it to conservation of energy.
Identify the external flux change first, then the opposing induced field and conventional current; a decreasing flux is reinforced, not opposed.
Use relative motion: the induced force repels an approaching magnet and attracts a receding one; comoving magnet and coil need no induced current.
Check local field direction for each loop: inside and outside a source loop can require opposite induced currents.
For field entry and exit, reverse the induced current to oppose the changing flux; magnetic braking opposes the motion in both stages.
Closing an open falling coil allows induced current and an upward magnetic force, increasing its entry time compared with the open coil.
Explain damping through a complete force chain or energy chain: induced current produces opposing force or dissipates mechanical energy as heat.
For a moving loudspeaker coil, changing flux induces back emf opposing the driving voltage and reducing net current.
In two-coil switching, increasing primary current repels the other coil, decreasing current attracts it, and the induced force is transient.
A stationary ring in a uniform increasing field can carry induced current while its net magnetic force is zero, because segment forces cancel.
For a cell-driven rod, back emf rises with speed until it cancels the cell emf, giving zero current and net force at terminal speed.
For a rod on closed rails, connect increasing flux to induced current and magnetic force opposing the motion or the external driving force.
For a rectangular coil, emf and flux are a quarter-cycle out of phase; field initially parallel to the coil surface gives an emf extremum.
Inspect the rotation axis: rotation about an axis parallel to B can leave flux constant and produce zero emf.
Match the actual field direction and initial orientation to the graph: a quarter-turn can take emf from zero to maximum, or from maximum to zero.
Use T = 2π/ω for the emf period and start the sinusoid at the correct initial extremum or zero crossing.
At fixed resistance, P = ε²/R gives two power peaks per revolution, so rotation period is twice the power-graph period.
Find average power from energy divided by time using the supplied power-time area; do not divide peak power by √2.
For the same generator, peak emf scales with rotation frequency and period scales inversely; doubling frequency doubles the peak and halves T.
Apply fractional frequency changes consistently: at 3ω/4 the emf amplitude is 3/4 as large and the period becomes 4T/3.
Calculate rotation frequency as revolutions divided by elapsed time before comparing emf graphs.
At fixed resistance peak power scales as peak emf squared, so a lower rotation frequency gives lower power peaks and longer repetition time.

Covered syllabus points

D.4.1 (HL)—Magnetic flux D.4.2 (HL)—Faraday’s law D.4.3 (HL)—Motional emf D.4.4 (HL)—Lenz’s law D.4.5 (HL)—Rotating-coil emf D.4.6 (HL)—Rotation frequency and emf

Exam points

Define magnetic flux as the magnetic field component normal to a surface multiplied by its area, Φ = BA cosθ.
Measure θ from the surface normal: field parallel to the coil surface gives zero flux, while field normal to the surface gives maximum magnitude BA.
Distinguish flux through one turn from flux linkage: for N identical turns normal to the field, the linkage is NBA.
Keep signed flux and flux magnitude distinct: signed extrema are −BA and +BA, while the magnitude ranges from zero to BA.
For a fixed ring approached by a magnet, explain increasing flux by the increasing magnetic field at the ring as the magnet gets closer.

[Maximum number: 1]

A coil is rotating in a region of magnetic field with angular speed $12.56 \mathrm{rad} \mathrm{s}^{-1}$ . At t=0, the field is parallel to the surface of the coil.

State the magnetic flux linkage through the coil at t=0.

[Maximum number: 5]

A rod, R , lies perpendicular to a uniform magnetic field B of strength 0.50 T directed into the plane of the page. R is connected to a circuit and the electric current $I_{\mathrm{R}}$ is 2.0 A .

(a)

A small coil of wire of radius 2.0 cm and 20 turns is now located with its centre 10.0 cm from R. The current in R is kept constant at 2.0 A .

[ 5 ]

(i)

Explain why there is no current induced in the coil.

[ 2 ]

the magnetic field produced by the wire $/ \mathrm{I}_{\mathrm{R}}$ is constant.

therefore the «magnetic» flux/field through the coil is constant.

(ii)

The current in R increases at a constant rate from 2.0 A to 10.0 A in 0.5 seconds.

Calculate the emf induced in the coil.

[ 3 ]

$\begin{aligned} \Delta \mathrm{B}=\frac{\mu_{0}}{2 \pi}\left(\frac{10-2}{0.1}\right) \text { OR } 2 \times 10^{-5}-4 \times 10^{-6}=1.6 \times 10^{-5}<\mathrm{T} \gg \Delta \phi=A \Delta \mathrm{~B}=(\pi)\left(0.02^{2}\right)\left(\left(2 \times 10^{-5}\right)-\left(4 x 10^{-6}\right)\right)=2.5 \times 10^{-8}-5.03 \times 10^{-9} =2.01 \times 10^{-8}<\mathrm{Wb}> \varepsilon=\frac{n \Delta \phi}{\Delta t}=\frac{(20)\left(2.01 \times 10^{-8}\right)}{0.5}=8.0 \times 10^{-7}<\mathrm{V} \gg \end{aligned}$

Exam points

Explain motional emf by magnetic forces on conduction electrons, their separation along the rod, and the resulting work or potential difference.
Derive ε = BvL by balancing qvB with qE and using E = V/L; q cancels in the steady state.
Alternatively derive ε = BvL from swept area LvΔt, flux change BLvΔt and Faraday’s rate of change of flux.
Use the magnetic force on electrons to identify polarity: the electron-rich end is negative; protons do not move through the metal.
Apply ε = BvL using the perpendicular field component and active conducting length; convert μT, mT, mV, cm and km consistently.
When finding B from ε = BvL, show the rearranged equation or a correctly aligned substitution as required by the markscheme.
Check whether the emf is requested for one wire or a whole coil: a single active edge uses BvL without the coil’s turn count.
For an N-turn rectangular loop entering a field, multiply the one-turn emf by N and use the transverse edge length, not the loop perimeter.
For a V-versus-conductor-length graph at constant B and v, the gradient is Bv, so divide the gradient by B to find v.
For constant B and L, emf follows velocity: use the gradient of displacement–time, giving zero emf when displacement is constant.
Find a rod’s speed before applying BvL: for uniform circular motion v = 2πr/T, and maximum emf occurs where motion is perpendicular to B.
Sketch the circular-motion rod emf as a sinusoid with the stated motion period; use at least half a cycle and accept any phase when specified.
An open coil can have induced emf without current. For the falling square coil, find v from the fall distance, then use one active edge in BvL.
At terminal speed balance mg = BIL, combine I = BvL/R, and solve v = mgR/(B²L²), converting the loop data to SI units.
For a rod moving steadily through a closed resistive circuit, mechanical input power equals electrical power: P = (BvL)²/R.

[Maximum number: 3]

A conducting rod of length L is moved with speed v at right angles to a uniform magnetic field of flux density B. The field is directed into the plane of the page.

Show, using Faraday's law or otherwise, that the potential difference, V, established between the ends of the rod is V=v B L.

ALTERNATE 1

In time $\Delta t$ , rod moves a distance $v \Delta t$

Flux increases by $B(L v \Delta t)^{}$

«By Faraday» induced emf is the rate of change of flux: $\frac{B L v \Delta t}{\Delta t}$

«=BLv»

ALTERNATE 2

<<From $\varepsilon=-\frac{N \Delta \Phi}{\Delta t} \gg$

Recognition that $\Phi=B A$

Recognition that N = 1

Recognition that $\frac{\Delta A}{\Delta t}=L v \vee$

<<Leading to V=vBL OR $\varepsilon=\mathrm{BLv}>$

ALTERNATE 3

«In steady state» electrons stop drifting and so net force is zero q v B=q E

Substitution of $E=\frac{V}{L}$ to get result

ALT 1: Correct MP2 scores MP1

ALT2:

V and $\varepsilon$ are interchangeable. Ignore negative sign.

Exam points

State that the induced emf or current acts to oppose the change producing it; identify conservation of energy as the basis of Lenz’s law.
Identify whether the external flux increases or decreases, choose the induced field opposing that change, then use the right-hand rule for current.
Do not assume the induced field always opposes the external field: it reinforces a decreasing flux and opposes an increasing flux.
For a conducting ring near a moving magnet, predict repulsion on approach and attraction on withdrawal, opposing the relative motion.
Use relative motion to predict the shown induced-current direction; magnet and coil moving together at the same velocity give no flux change.
For moving coaxial coils, an approaching coil has induced current opposite to the source coil; a receding coil has current in the same sense.
Check the local field on each side of a source loop: a loop inside it and one outside it can have opposite induced-current directions.
For a loop entering a field, oppose the increasing flux; on exit, reverse the induced current to oppose the decreasing flux.
When a falling magnet passes a coil’s midpoint, the emf is zero; as the magnet recedes, the flux change and induced emf reverse sign.
A closed falling loop has induced current and an upward magnetic force on both entry and exit, reducing its downward acceleration from g.
Distinguish induced emf from current: closing an open falling coil allows current and magnetic braking, increasing its entry time.
For a rod moving on closed rails, connect increasing flux to induced current and a magnetic force opposing the force driving its motion.
Explain magnetic damping with a complete chain: changing flux induces current, whose field produces a force opposing motion.
Alternatively explain damping by induced current dissipating energy as heat, with that energy supplied by the moving body’s mechanical energy.
In a two-coil switching experiment, increasing primary current initially repels the other coil; decreasing primary current attracts it.
A switched-coil force is transient: it initially rises then falls as the linked flux stops changing, rather than staying constant with primary DC.
A stationary ring in a uniform increasing field can carry induced current while having zero net magnetic force, because segment forces cancel.
For a cell-driven rod, back emf BvL increases with speed until it equals the cell emf; current and net force then vanish at terminal speed.

[Maximum number: 2]

A rod, R , lies perpendicular to a uniform magnetic field B of strength 0.50 T directed into the plane of the page. R is connected to a circuit and the electric current $I_{\mathrm{R}}$ is 2.0 A .

A small coil of wire of radius 2.0 cm and 20 turns is now located with its centre 10.0 cm from R. The current in R is kept constant at 2.0 A .

Deduce the direction of the current induced in (c)(ii).

Exam points

Relate sinusoidal emf to the negative gradient of flux: the two vary a quarter-cycle out of phase, rather than in phase or exactly opposite.
For a rotating rectangular coil, emf is zero at maximum flux magnitude and greatest when the field lies in the coil’s plane.
Check the rotation axis: rotating a coil about an axis parallel to B can leave its flux unchanged and produce zero emf.
Match a quarter-turn about the specified in-plane axis to a quarter-cycle of emf; use the actual initial field direction to choose its phase.
Use T = 2π/ω for the emf period. If the field is initially parallel to the coil surface, begin the sinusoid at positive or negative maximum.
For constant resistance, P = ε²/R gives two power peaks per revolution; the emf and rotation period is twice the power-graph period.
Find average power from energy divided by time, using the area under the supplied power–time graph; do not divide peak power by √2.

[Maximum number: 2]

A coil is rotating in a region of magnetic field with angular speed $12.56 \mathrm{rad} \mathrm{s}^{-1}$ . At t=0, the field is parallel to the surface of the coil.

Draw, on the axes, a graph to show the variation with time of the induced emf in the loop. (No numbers are required on the vertical axis.)
emf

Exam points

For an unchanged generator, peak emf is proportional to rotation frequency, while period is inversely proportional; doubling f gives 2E₀ and T/2.
When drawing the doubled-frequency voltage, double its amplitude and halve its period on the same axes; show at least one complete cycle.
Scale non-integer speed changes consistently: at 3ω/4, emf amplitude is 3/4 of its original value and period becomes 4T/3.
Calculate each rotation frequency from revolutions divided by elapsed time before comparing emf graphs; do not compare turn counts alone.
At fixed resistance, peak power scales as peak emf squared: reducing rotation frequency lowers the power peaks and lengthens their repetition time.

[Maximum number: 2]

A rod, R , lies perpendicular to a uniform magnetic field B of strength 0.50 T directed into the plane of the page. R is connected to a circuit and the electric current $I_{\mathrm{R}}$ is 2.0 A .

The coil is made to rotate such that an emf is induced. The graph shows the variation with time of the emf induced in the coil.

The frequency of rotation is doubled. Draw on the graph the variation with time of the new emf induced.

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Question (ii)

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