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Error Notes — Electromagnetic Induction

Aamirah Fathima  ·  Complete chapter, 80 questions  ·  marked +4 / −1 / 0  ·  14 October 2026

295out of 320
+30above target
92.2%percentage
75correct
5wrong
0left blank
Target met, and met on the part of the chapter she had never been tested on. The paper was set at 265 with a danger line of 235. She scored 295, thirty marks clear of target, with nothing left blank — the first full paper in this series with no unattempted question.

Two sections came back perfect. Self-inductance and mutual inductance, 24 out of 24. That block had never appeared in either of the two earlier EMI papers or the 30-question retest — it was entirely new ground when the paper was set, and she did not drop a single mark on it. Motional emf, 18 out of 18.

All five losses are in the two sections she has already been tested on twice before. That is the finding worth acting on: the gap is not in what she has just learnt, it is in the material she has seen most often.

Section by section

SectionQ✓✗MarksAccuracy
Flux, Faraday's Law and Lenz's Law2218468/8882%
Motional Emf, Charge and Energy1818072/72100%
Self-Inductance and Mutual Inductance2424096/96100%
Eddy Currents and the A.C. Generator1615159/6494%
Total80755295/32094%
DifficultyQuestionsCorrectRate
Easy171694%
Moderate403792%
Hard232296%

She scored 96% on the hard questions and 94% on the easy ones. When the rate on hard items is the higher of the two, the losses are not about difficulty — they are about three specific habits, set out below.

Where the twenty marks went

1  ·  One Lenz rule applied to both directions  —  Q16

This is the most important single thing in the paper, and the proof is in the pair of questions, not in either one alone. Q16 and Q17 ask the opposite question and she gave the same answer to both.

The pair, side by side

QuestionThe fieldHer answerCorrect
Q16into the page, increasingclockwiseanticlockwise
Q17into the page, decreasingclockwiseclockwise ✓

She got Q17 right and Q16 wrong, but she wrote the same word for both. That means the direction was not worked out from the physics either time — one rule was recalled and applied to both cases. Her diagnostic file has carried direction reversal as a confirmed recurring error since September; this is it, caught cleanly.

Q16Moderatewrong  −1
B is directed into the page and is increasing× × × uniform field B × × ×A loop in a changing fieldIn the situation shown, the induced current in the loop flows
  1. (A) first clockwise then anticlockwise
  2. (B) clockwise  ✗ her answer
  3. (C) anticlockwise  ✓
  4. (D) not at all
Why C: The flux into the page is increasing, so the induced current must set up a field out of the page inside the loop.
The fix, and it is one sentence long. Never answer a Lenz question from the field direction. Answer it in two steps: first ask is the flux through the loop growing or shrinking? — then the induced current must set up a field that opposes that change inside the loop. Growing into the page → induced field out of the page → anticlockwise. Shrinking into the page → induced field into the page → clockwise. Say the word “growing” or “shrinking” out loud before choosing.

2  ·  Reading the height of a graph instead of its slope  —  Q11 and Q12

These two questions share one figure and one mistake, so they cost eight marks for a single idea. In Q11 she chose region I as the steepest; in Q12 she gave region I's emf as 0.8 V. 0.8 is the flux reached at the end of region I, not the rate at which it rose. Region I climbs 0.8 Wb in 2 s, a slope of 0.4; region III falls 0.8 Wb in 1 s, a slope of 0.8 — which is why III is the steepest and I is not.

Worth noting that she answered Q13 correctly: region II is flat, so the emf is zero. The idea that a flat graph means no emf is secure. What failed was comparing two slopes that both look like big rises.

Q11Moderatewrong  −1
Φ / Wbt / s01234560.40.8IIIIIIIVFlux through a coil against timeFrom the graph, the induced emf in the single-turn coil is greatest in region
  1. (A) I  ✗ her answer
  2. (B) IV
  3. (C) II
  4. (D) III  ✓
Why D: The emf is the slope of the graph, and region III is the steepest.|slope| = 0.8/1 = 0.8 Wb s⁻¹
Q12Hardwrong  −1
Φ / Wbt / s01234560.40.8IIIIIIIVFlux through a coil against timeIn region I of the same graph the induced emf has magnitude
  1. (A) 0.4 V  ✓
  2. (B) 0.8 V  ✗ her answer
  3. (C) 0.2 V
  4. (D) 1.6 V
Why A: Read the slope of the straight line in region I.0.8 Wb ÷ 2 s = 0.4 V

3  ·  The angle measured from the plane instead of the normal  —  Q3

Q3Moderatewrong  −1
A coil is held in a uniform magnetic field with its plane containing the direction of the field. The flux through the coil is
  1. (A) equal to BA  ✗ her answer
  2. (B) half the maximum value
  3. (C) maximum
  4. (D) zero  ✓
Why D: With the plane containing B, the normal is perpendicular to B, so cos 90° = 0.
Same root as Q1, which she got right. In Q1 the plane was at 30° to the field and she correctly used 60° from the normal. One question later, with the plane containing the field, she took the flux as maximum instead of zero. The rule to hold: θ in Φ = BA cosθ is always measured from the normal, so plane parallel to B → normal perpendicular → cos 90° → zero.

4  ·  Q74 — the one mark that is genuinely unclear on the script

Q74 carries a single straight red slash with no hook, running from the question text down to the left margin and ending level with option (B). Every other tick on the paper has a clear V whose point sits on the chosen option. Read by the usual rule — lowest point of the mark — this is (B), and I have scored it that way. Read as a strike through (A), it is correct.

So the score is 295 as marked, or 299 if that one mark was meant as (A). Worth asking her which she intended, and worth telling her to close her ticks properly — on a real OMR sheet an unclear mark is simply lost.

Q74Easyambiguous mark
For a coil of N turns and area A rotating with angular velocity ω in a field B, the instantaneous emf is
  1. (A) NBAω sin ωt  ✓
  2. (B) NBAω cos ωt  ✗ her answer
  3. (C) NBω sin ωt
  4. (D) NBA sin ωt
Why A: Differentiating Φ = NBA cos ωt gives the sine form.ε = NBAω sin ωt
One caveat on the score, the same as last time. This paper was built before the specification was corrected. Its four assertion–reason questions are all keyed to the same option and she got all four; its two “how many of the following” questions are both keyed “Only three” and she got both. That is six questions, 24 marks, that a reflex answers. The effect here is much smaller than on the Biology papers, and it does not change the shape of the result — but the Thermal Properties paper, which varies those keys across all four options, will give the cleaner reading.

Against the last physics paper

PaperScore%BlanksWeakest block
AC · EM Waves · Ray Optics (120Q, 10 Oct)360/48075.0%4Ray Optics, 68%
Electromagnetic Induction (80Q, today)295/32092.2%0Flux, Faraday and Lenz, 82%

Seventeen percentage points in four days, and the blanks are gone. On the October paper four questions were abandoned, three of them consecutive; here she attempted all eighty. That change is worth more than the score.

The complete paper, question by question

Every one of the eighty questions is reproduced below in the order she sat them, with the option she ticked and the correct one marked on each, and the working underneath. The five she lost are the ones with a red edge; the one unresolved mark has an amber edge.

Key:correctwrongmark unclear

Flux, Faraday's Law and Lenz's LawQ1–Q22  ·  18 of 22 correct  ·  68/88
Q1Easycorrect  +4
A plane coil of area 0.025 m2 is held so that its plane makes an angle of 30° with a uniform field of 0.4 T. The flux through it is
  1. (A) 1.0 × 10−2 Wb
  2. (B) zero
  3. (C) 8.7 × 10−3 Wb
  4. (D) 5.0 × 10−3 Wb  ✓ her answer
Why D: Φ = BA cosθ, and θ is measured from the normal. A plane at 30° to the field puts the normal at 60°.Φ = 0.4 × 0.025 × cos 60° = 5.0 × 10⁻³ Wb
Q2Easycorrect  +4
The weber is equivalent to
  1. (A) tesla per square metre
  2. (B) volt second  ✓ her answer
  3. (C) joule per ampere squared
  4. (D) ampere metre
Why B: From ε = −dΦ/dt, one weber is one volt times one second.
Q3Moderatewrong  −1
A coil is held in a uniform magnetic field with its plane containing the direction of the field. The flux through the coil is
  1. (A) equal to BA  ✗ her answer
  2. (B) half the maximum value
  3. (C) maximum
  4. (D) zero  ✓
Why D: With the plane containing B, the normal is perpendicular to B, so cos 90° = 0.
Q4Easycorrect  +4
NSmoved towards the coilcoilGgalvanometer deflectsFaraday's experimentIn the experiment shown, the galvanometer shows a deflection
  1. (A) only if the coil carries a current of its own
  2. (B) as long as the magnet is anywhere near the coil
  3. (C) only while the magnet is moving  ✓ her answer
  4. (D) only when the magnet is at rest inside the coil
Why C: An emf exists only while the flux is changing; a stationary magnet changes nothing.
Q5Easycorrect  +4
The flux through a single-turn coil falls from 0.25 Wb to 0.05 Wb in 0.1 s. The average induced emf is
  1. (A) 2.5 V
  2. (B) 20 V
  3. (C) 0.2 V
  4. (D) 2 V  ✓ her answer
Why D: Faraday's law with N = 1.ε = ΔΦ/Δt = 0.20/0.1 = 2 V
Q6Moderatecorrect  +4
A coil of 250 turns experiences a flux change of 4 × 10−3 Wb through each turn in 0.02 s. The induced emf is
  1. (A) 500 V
  2. (B) 0.05 V
  3. (C) 50 V  ✓ her answer
  4. (D) 5 V
Why C: Multiply the rate of change of flux by the number of turns.ε = N ΔΦ/Δt = 250 × 4×10⁻³ / 0.02 = 50 V
Q7Moderatecorrect  +4
The flux through a single-turn coil varies as Φ = 5t2 + 3t + 2, with Φ in weber and t in second. The induced emf at t = 2 s is
  1. (A) 43 V
  2. (B) 28 V
  3. (C) 20 V
  4. (D) 23 V  ✓ her answer
Why D: Differentiate and substitute.dΦ/dt = 10t + 3; at t = 2, ε = 23 V
Q8Moderatecorrect  +4
A coil of 40 turns and area 0.25 m2 lies with its plane perpendicular to a field that is increasing steadily at 0.02 T s−1. The induced emf is
  1. (A) 0.2 V  ✓ her answer
  2. (B) 2 V
  3. (C) 0.5 V
  4. (D) 0.005 V
Why A: The plane is perpendicular to B, so the flux is NBA and only B changes.ε = NA dB/dt = 40 × 0.25 × 0.02 = 0.2 V
Q9Hardcorrect  +4
A circular loop of radius 0.1 m lies perpendicular to a field that falls from 0.5 T to 0.1 T in 0.2 s. The induced emf is nearly
  1. (A) 6.3 × 10−2 V  ✓ her answer
  2. (B) 4.0 × 10−1 V
  3. (C) 1.3 × 10−1 V
  4. (D) 3.1 × 10−2 V
Why A: Work out the area first, then multiply by the rate of change of B.A = πr² = 0.0314 m²; ε = 0.0314 × 0.4/0.2 = 6.3×10⁻² V
Q10Hardcorrect  +4
A coil of area 0.06 m2 in a field of 0.25 T is turned in 0.1 s from the position in which its plane is perpendicular to the field to the position in which its plane contains the field. The average emf is
  1. (A) 0.30 V
  2. (B) 0.15 V  ✓ her answer
  3. (C) 0.015 V
  4. (D) 1.5 V
Why B: The flux falls from BA to zero.ΔΦ = 0.25 × 0.06 = 0.015 Wb; ε = 0.015/0.1 = 0.15 V
Q11Moderatewrong  −1
Φ / Wbt / s01234560.40.8IIIIIIIVFlux through a coil against timeFrom the graph, the induced emf in the single-turn coil is greatest in region
  1. (A) I  ✗ her answer
  2. (B) IV
  3. (C) II
  4. (D) III  ✓
Why D: The emf is the slope of the graph, and region III is the steepest.|slope| = 0.8/1 = 0.8 Wb s⁻¹
Q12Hardwrong  −1
Φ / Wbt / s01234560.40.8IIIIIIIVFlux through a coil against timeIn region I of the same graph the induced emf has magnitude
  1. (A) 0.4 V  ✓
  2. (B) 0.8 V  ✗ her answer
  3. (C) 0.2 V
  4. (D) 1.6 V
Why A: Read the slope of the straight line in region I.0.8 Wb ÷ 2 s = 0.4 V
Q13Moderatecorrect  +4
Φ / Wbt / s01234560.40.8IIIIIIIVFlux through a coil against timeIn region II of the same graph the induced emf is
  1. (A) zero  ✓ her answer
  2. (B) changing steadily
  3. (C) 0.4 V
  4. (D) 0.8 V
Why A: The flux is constant in region II, so there is no emf at all.
Q14Moderatecorrect  +4
Match Column I with Column II.
Column IColumn II
AMagnetic fluxIampere
BInduced emfIIvolt
CRate of change of fluxIIIweber
DInduced currentIVweber per second
  1. (A) A-III, B-IV, C-II, D-I
  2. (B) A-II, B-III, C-IV, D-I
  3. (C) A-I, B-II, C-IV, D-III
  4. (D) A-III, B-II, C-IV, D-I  ✓ her answer
Why D: Flux is in weber, emf in volt, rate of change of flux in weber per second, current in ampere.
Q15Easycorrect  +4
Lenz's law states that the induced emf is set up in such a direction that it
  1. (A) is always clockwise when seen from above
  2. (B) assists the change of flux that produces it
  3. (C) is independent of the direction of the change
  4. (D) opposes the change of flux that produces it  ✓ her answer
Why D: The induced effect always opposes the change that causes it.
Q16Moderatewrong  −1
B is directed into the page and is increasing× × × uniform field B × × ×A loop in a changing fieldIn the situation shown, the induced current in the loop flows
  1. (A) first clockwise then anticlockwise
  2. (B) clockwise  ✗ her answer
  3. (C) anticlockwise  ✓
  4. (D) not at all
Why C: The flux into the page is increasing, so the induced current must set up a field out of the page inside the loop.
Q17Moderatecorrect  +4
A field directed into the page through a loop is steadily decreasing. The induced current in the loop is
  1. (A) zero
  2. (B) clockwise  ✓ her answer
  3. (C) alternating
  4. (D) anticlockwise
Why B: A decreasing inward flux is opposed by an induced current that adds to it, so the current is clockwise.
Q18Moderatecorrect  +4
The north pole of a bar magnet is pushed towards a coil. Seen from the side of the magnet, the induced current in the coil flows
  1. (A) anticlockwise, making the near face a south pole
  2. (B) clockwise, making the near face a south pole
  3. (C) anticlockwise, making the near face a north pole  ✓ her answer
  4. (D) clockwise, making the near face a north pole
Why C: The coil must repel the approaching north pole, so the near face must itself be a north pole.
Q19Moderatecorrect  +4
The north pole of a bar magnet is now withdrawn from the same coil. Seen from the side of the magnet, the induced current flows
  1. (A) clockwise, making the near face a north pole
  2. (B) anticlockwise, making the near face a south pole
  3. (C) anticlockwise, making the near face a north pole
  4. (D) clockwise, making the near face a south pole  ✓ her answer
Why D: Now the coil must attract the retreating pole, so the near face becomes a south pole.
Q20Hardcorrect  +4
A bar magnet is dropped down a long vertical copper pipe. Compared with the same magnet dropped outside the pipe, it
  1. (A) falls in exactly the same time
  2. (B) does not fall at all
  3. (C) falls faster
  4. (D) falls more slowly  ✓ her answer
Why D: The eddy currents induced in the pipe oppose the motion of the magnet, so it is retarded.
Q21Hardcorrect  +4
Assertion (A): Lenz's law is a statement of the conservation of energy.
Reason (R): If the induced effect assisted the change producing it, energy would be created out of nothing.
  1. (A) Both A and R are true and R is the correct explanation of A  ✓ her answer
  2. (B) Both A and R are true but R is NOT the correct explanation of A
  3. (C) A is true but R is false
  4. (D) A is false but R is true
Why A: An induced effect that helped the change would grow without limit, which would create energy from nothing.
Q22Hardcorrect  +4
How many of the following statements are correct?
A. An emf is induced only while the flux is changing.
B. The induced emf depends on how fast the flux changes, not on the value of the flux.
C. Lenz's law fixes the direction of the induced current.
D. An emf is induced in a stationary coil held in a steady magnetic field.
  1. (A) Only one
  2. (B) Only two
  3. (C) Only three  ✓ her answer
  4. (D) All four
Why C: A, B and C are correct. No emf is induced in a stationary coil in a steady field.
Motional Emf, Charge and EnergyQ23–Q40  ·  18 of 18 correct  ·  72/72
Q23Easycorrect  +4
Rrodvlthe field B is directed into the pageA rod sliding on railsThe rod has length 0.5 m and slides at 4 m s−1 in a field of 0.3 T. The emf between its ends is
  1. (A) 2.4 V
  2. (B) 0.15 V
  3. (C) 1.2 V
  4. (D) 0.6 V  ✓ her answer
Why D: A rod moving at right angles to B.ε = Blv = 0.3 × 0.5 × 4 = 0.6 V
Q24Moderatecorrect  +4
Rrodvlthe field B is directed into the pageA rod sliding on railsFor that rod, if the rails are joined through R = 3 Ω and all other resistance is negligible, the current is
  1. (A) 0.1 A
  2. (B) 0.2 A  ✓ her answer
  3. (C) 0.6 A
  4. (D) 2.0 A
Why B: Ohm's law with the emf just found.I = ε/R = 0.6/3 = 0.2 A
Q25Moderatecorrect  +4
For the same rod and current, the magnitude of the magnetic force that opposes its motion is
  1. (A) 0.015 N
  2. (B) 0.06 N
  3. (C) 0.3 N
  4. (D) 0.03 N  ✓ her answer
Why D: The current-carrying rod in the field feels a force that opposes its motion.F = BIl = 0.3 × 0.2 × 0.5 = 0.03 N
Q26Hardcorrect  +4
The power that the external agent must supply to keep that rod moving steadily is
  1. (A) 0.48 W
  2. (B) 0.06 W
  3. (C) 0.24 W
  4. (D) 0.12 W  ✓ her answer
Why D: The agent must balance that retarding force at the given speed.P = Fv = 0.03 × 4 = 0.12 W, and ε²/R = 0.36/3 = 0.12 W
Q27Hardcorrect  +4
For the same arrangement, the rate at which heat appears in the resistance is
  1. (A) half the power supplied by the agent
  2. (B) twice the power supplied by the agent
  3. (C) equal to the power supplied by the agent  ✓ her answer
  4. (D) zero, because the rod moves steadily
Why C: The rod moves steadily, so its kinetic energy does not change and all the work done appears as heat.
Q28Moderatecorrect  +4
A rod of length l moves with speed v perpendicular to a field B. If the speed is doubled, the emf and the heat produced per second become
  1. (A) twice and four times  ✓ her answer
  2. (B) twice and unchanged
  3. (C) four times and twice
  4. (D) twice and twice
Why A: ε ∝ v, and the power is ε²/R.ε → 2ε and P → 4P
Q29Moderatecorrect  +4
A straight conductor moves in a magnetic field with its velocity parallel to the field. The emf induced across its ends is
  1. (A) zero  ✓ her answer
  2. (B) Blv sin 45°
  3. (C) maximum
  4. (D) Blv
Why A: With v parallel to B the force qv × B on the charges is zero.
Q30Hardcorrect  +4
A rod moves so that its velocity makes an angle θ with a uniform field B, the rod being perpendicular to both. The emf across it is
  1. (A) Blv
  2. (B) Blv sin θ  ✓ her answer
  3. (C) Blv tan θ
  4. (D) Blv cos θ
Why B: Only the component of v perpendicular to B drives the charges along the rod.ε = Blv sin θ
Q31Moderatecorrect  +4
A rod of length 0.4 m rotates about one end in a plane perpendicular to a field of 0.4 T, at 15 rad s−1. The emf between its ends is
  1. (A) 0.96 V
  2. (B) 0.24 V
  3. (C) 2.40 V
  4. (D) 0.48 V  ✓ her answer
Why D: Use the formula for a rod rotating about one end.ε = ½Bωl² = 0.5 × 0.4 × 15 × 0.16 = 0.48 V
Q32Hardcorrect  +4
A rod of length 1 m rotates about one end at 120 revolutions per minute in a field of 0.5 T perpendicular to its plane of rotation. The emf developed is about
  1. (A) 3.14 V  ✓ her answer
  2. (B) 0.79 V
  3. (C) 1.57 V
  4. (D) 6.28 V
Why A: Convert the rate of rotation to rad s⁻¹ first.ω = 2π × 2 = 12.57 rad s⁻¹; ε = ½ × 0.5 × 12.57 × 1 = 3.14 V
Q33Moderatecorrect  +4
For a rod of length l rotating about one end with angular velocity ω in a field B, the emf is
  1. (A) Bωl
  2. (B) 2Bωl2
  3. (C) ½Bωl2  ✓ her answer
  4. (D) Bωl2
Why C: Each element at distance r moves at ωr; integrating Bωr dr from 0 to l gives ½Bωl².
Q34Hardcorrect  +4
An aeroplane with a wingspan of 25 m flies horizontally at 300 m s−1 where the vertical component of the earth's field is 5 × 10−5 T. The emf between its wing tips is
  1. (A) 0.0375 V
  2. (B) 3.75 V
  3. (C) 37.5 V
  4. (D) 0.375 V  ✓ her answer
Why D: Only the vertical component of the earth's field matters for horizontal flight.ε = Blv = 5×10⁻⁵ × 25 × 300 = 0.375 V
Q35Moderatecorrect  +4
The charge that flows round a circuit of resistance R when the flux linked with it changes by ΔΦ is
  1. (A) ΔΦ / R  ✓ her answer
  2. (B) ΔΦ / (RΔt)
  3. (C) ΔΦ × Δt / R
  4. (D) ΔΦ × R
Why A: q = ∫I dt = ∫(ΔΦ/Δt)/R dt, and the time cancels.q = ΔΦ/R
Q36Moderatecorrect  +4
A flux change of 0.6 Wb takes place in a single-turn circuit of resistance 4 Ω. The charge that flows is
  1. (A) 2.4 C
  2. (B) 0.15 C  ✓ her answer
  3. (C) 0.6 C
  4. (D) 0.067 C
Why B: Apply that result.q = 0.6/4 = 0.15 C
Q37Hardcorrect  +4
Two identical coils are linked by the same total change of flux, one slowly and one quickly. Compared with the slow change, the quick change produces
  1. (A) a smaller emf and the same charge
  2. (B) a larger emf and a larger charge
  3. (C) a larger emf and the same charge  ✓ her answer
  4. (D) the same emf and a larger charge
Why C: The charge depends only on the total flux change and the resistance, but the emf depends on how fast it happens.
Q38Moderatecorrect  +4
The electrical energy produced by a rod sliding on rails comes from
  1. (A) the thermal energy of the rod
  2. (B) the mechanical work done by the agent that keeps the rod moving  ✓ her answer
  3. (C) the resistance of the circuit
  4. (D) the magnetic field, which is weakened in the process
Why B: The work done against the retarding magnetic force is the source of the electrical energy.
Q39Hardcorrect  +4
Assertion (A): A motional emf can be understood without using Faraday's law at all.
Reason (R): The magnetic force on the free charges in the moving conductor drives them towards one end until an electric field builds up to balance it.
  1. (A) Both A and R are true and R is the correct explanation of A  ✓ her answer
  2. (B) Both A and R are true but R is NOT the correct explanation of A
  3. (C) A is true but R is false
  4. (D) A is false but R is true
Why A: The separation of charge by the magnetic force gives Blv directly, without invoking the flux rule.
Q40Hardcorrect  +4
Which of the following statements are correct?
A. The motional emf of a rod is Blv when B, l and v are mutually perpendicular.
B. The force opposing the motion of the rod is B2l2v / R.
C. The work done by the external agent appears entirely as heat in the resistance.
D. The induced charge depends on how quickly the flux changes.
  1. (A) 1, 3 and 4 only
  2. (B) 1 and 4 only
  3. (C) 2, 3 and 4 only
  4. (D) 1, 2 and 3 only  ✓ her answer
Why D: Statements 1, 2 and 3 are right. The induced charge depends only on the total flux change, not on its speed.
Self-Inductance and Mutual InductanceQ41–Q64  ·  24 of 24 correct  ·  96/96
Q41Easycorrect  +4
The SI unit of inductance is the
  1. (A) tesla
  2. (B) volt second squared
  3. (C) henry  ✓ her answer
  4. (D) weber
Why C: Inductance is measured in henry.
Q42Easycorrect  +4
One henry is equivalent to
  1. (A) one weber per second
  2. (B) one weber per ampere  ✓ her answer
  3. (C) one ampere per weber
  4. (D) one volt per ampere
Why B: From NΦ = LI, one henry is one weber per ampere.
Q43Easycorrect  +4
Ilength l, N turns, cross-section Aa changing current in the coil induces an emf in the same coilSelf-inductance of a solenoidThe self-inductance of the solenoid shown is given by
  1. (A) μ0NA / l
  2. (B) μ0N2l / A
  3. (C) μ0NAl
  4. (D) μ0N2A / l  ✓ her answer
Why D: For a long solenoid the flux linkage is μ₀N²AI/l, so L = μ₀N²A/l.
Q44Moderatecorrect  +4
An air-cored solenoid has 500 turns wound over a length of 0.3 m and a cross-section of 4 × 10−4 m2. Its self-inductance is about
  1. (A) 8.4 × 10−4 H
  2. (B) 4.2 × 10−3 H
  3. (C) 4.2 × 10−4 H  ✓ her answer
  4. (D) 2.1 × 10−4 H
Why C: Substitute into L = μ₀N²A/l.L = 4π×10⁻⁷ × 500² × 4×10⁻⁴ / 0.3 = 4.2×10⁻⁴ H
Q45Moderatecorrect  +4
If the number of turns of a solenoid is doubled, everything else being unchanged, its self-inductance becomes
  1. (A) half
  2. (B) unchanged
  3. (C) four times  ✓ her answer
  4. (D) double
Why C: L is proportional to N².2² = 4
Q46Moderatecorrect  +4
The self-inductance of a coil depends on
  1. (A) the emf applied to it
  2. (B) its geometry and the medium inside it  ✓ her answer
  3. (C) the resistance of its wire
  4. (D) the current flowing through it
Why B: L is fixed by the number of turns, the size and shape of the coil and the material inside it.
Q47Moderatecorrect  +4
Inserting a soft iron core into an air-cored coil makes its self-inductance
  1. (A) exactly half
  2. (B) unchanged
  3. (C) very much larger  ✓ her answer
  4. (D) very much smaller
Why C: Iron has a large relative permeability, so the flux linkage for a given current rises steeply.
Q48Easycorrect  +4
The emf induced in a coil of self-inductance L when the current through it changes is
  1. (A) −L dI/dt  ✓ her answer
  2. (B) −½LI2
  3. (C) −L I
  4. (D) −I dL/dt
Why A: The induced emf opposes the change of current that produces it.ε = −L dI/dt
Q49Moderatecorrect  +4
The current in a coil of self-inductance 0.5 H changes at the rate of 4 A s−1. The induced emf is
  1. (A) 8 V
  2. (B) 0.5 V
  3. (C) 0.125 V
  4. (D) 2 V  ✓ her answer
Why D: Multiply the inductance by the rate of change of current.ε = 0.5 × 4 = 2 V
Q50Moderatecorrect  +4
The current in a 2 H coil falls steadily from 3 A to 1 A in 0.4 s. The magnitude of the induced emf is
  1. (A) 10 V  ✓ her answer
  2. (B) 2.5 V
  3. (C) 20 V
  4. (D) 5 V
Why A: Work out the rate of change first.dI/dt = 2/0.4 = 5 A s⁻¹; ε = 2 × 5 = 10 V
Q51Moderatecorrect  +4
The flux linkage of a coil of self-inductance 0.3 H carrying a current of 4 A is
  1. (A) 0.3 Wb
  2. (B) 1.2 Wb  ✓ her answer
  3. (C) 2.4 Wb
  4. (D) 0.075 Wb
Why B: Flux linkage is LI.NΦ = 0.3 × 4 = 1.2 Wb
Q52Hardcorrect  +4
Assertion (A): Self-inductance is sometimes called the inertia of electricity.
Reason (R): An inductor opposes any change in the current through it, just as mass opposes any change in velocity.
  1. (A) Both A and R are true and R is the correct explanation of A  ✓ her answer
  2. (B) Both A and R are true but R is NOT the correct explanation of A
  3. (C) A is true but R is false
  4. (D) A is false but R is true
Why A: An inductor resists any change of current exactly as mass resists any change of velocity.
Q53Easycorrect  +4
The energy stored in an inductor L carrying a steady current I is
  1. (A) LI
  2. (B) ½LI2  ✓ her answer
  3. (C) ½L2I
  4. (D) LI2
Why B: The work done against the back emf while the current grows is stored as ½LI².
Q54Moderatecorrect  +4
The energy stored in a coil of self-inductance 0.2 H carrying a current of 5 A is
  1. (A) 5.0 J
  2. (B) 2.5 J  ✓ her answer
  3. (C) 25 J
  4. (D) 0.5 J
Why B: Substitute into ½LI².U = ½ × 0.2 × 25 = 2.5 J
Q55Moderatecorrect  +4
If the current in an inductor is doubled, the energy stored becomes
  1. (A) unchanged
  2. (B) half as great
  3. (C) four times as great  ✓ her answer
  4. (D) twice as great
Why C: The energy goes as the square of the current.2² = 4
Q56Hardcorrect  +4
The work done against the back emf while the current in an inductor is being built up is stored
  1. (A) in the magnetic field of the inductor  ✓ her answer
  2. (B) as chemical energy in the source
  3. (C) in the electric field between the turns
  4. (D) as heat in the resistance of the coil
Why A: The energy resides in the magnetic field that the current has set up.
Q57Easycorrect  +4
primary, N₁ turnssecondary, N₂ turnsI₁Ga changing current in the primary induces an emf in the secondaryTwo coupled coilsIn the arrangement shown, an emf appears in the secondary coil
  1. (A) only if the secondary carries a current of its own
  2. (B) only if the two coils touch each other
  3. (C) only while the current in the primary is changing  ✓ her answer
  4. (D) as long as a steady current flows in the primary
Why C: A steady current gives a steady flux and so no emf; only a changing current induces one.
Q58Moderatecorrect  +4
The emf induced in the secondary when the current in the primary changes at the rate dI1/dt is
  1. (A) −I1 dM/dt
  2. (B) −½MI12
  3. (C) −M I1
  4. (D) −M dI1/dt  ✓ her answer
Why D: The mutual inductance relates the emf in one coil to the rate of change of current in the other.ε₂ = −M dI₁/dt
Q59Moderatecorrect  +4
An emf of 0.6 V appears in a coil when the current in a neighbouring coil changes at 3 A s−1. Their mutual inductance is
  1. (A) 0.6 H
  2. (B) 1.8 H
  3. (C) 0.2 H  ✓ her answer
  4. (D) 5.0 H
Why C: Rearrange that relation.M = ε/(dI/dt) = 0.6/3 = 0.2 H
Q60Hardcorrect  +4
Two long coaxial solenoids of the same length l and cross-section A carry N1 and N2 turns. Their mutual inductance is
  1. (A) μ0N1N2l / A
  2. (B) μ0N1N2A / l  ✓ her answer
  3. (C) μ0(N1 + N2)A / l
  4. (D) μ0N12N22A / l
Why B: The flux of the inner solenoid links every turn of the outer one.M = μ₀N₁N₂A/l
Q61Hardcorrect  +4
For any two coils, the mutual inductance M12 of coil 1 with respect to coil 2 and M21 of coil 2 with respect to coil 1 are related by
  1. (A) there is no general relation between them
  2. (B) M12 = 1/M21
  3. (C) M12 = M21  ✓ her answer
  4. (D) M12 = 2M21
Why C: Mutual inductance is a property of the pair, so it is the same whichever coil is taken as the primary.
Q62Moderatecorrect  +4
The mutual inductance of a pair of coils is increased by
  1. (A) turning one of them through 90°
  2. (B) moving them further apart
  3. (C) winding them on a common soft iron core  ✓ her answer
  4. (D) reducing the number of turns on either coil
Why C: A common iron core carries the flux of one coil almost entirely into the other.
Q63Moderatecorrect  +4
ItI₀τcurrent grows towards its steady valueGrowth of current when an inductor is switched onThe time constant marked τ on the graph is equal to
  1. (A) 1 / LR
  2. (B) L / R  ✓ her answer
  3. (C) LR
  4. (D) R / L
Why B: The current reaches a fraction 1 − 1/e of its final value in one time constant.τ = L/R
Q64Hardcorrect  +4
A coil of inductance 0.4 H and resistance 20 Ω is connected to a cell. The time constant of the circuit is
  1. (A) 50 s
  2. (B) 8 s
  3. (C) 0.05 s
  4. (D) 0.02 s  ✓ her answer
Why D: Substitute.τ = 0.4/20 = 0.02 s
Eddy Currents and the A.C. GeneratorQ65–Q80  ·  15 of 16 correct  ·  59/64
Q65Easycorrect  +4
Eddy currents are the circulating currents induced in
  1. (A) the body of a bulk piece of conductor placed in a changing field  ✓ her answer
  2. (B) an electrolyte carrying a direct current
  3. (C) a thin wire wound into a coil
  4. (D) a semiconductor at high temperature
Why A: Eddy currents swirl through the body of a solid conductor, not along a wire.
Q66Easycorrect  +4
Eddy currents are also known as
  1. (A) Faraday currents
  2. (B) displacement currents
  3. (C) Foucault currents  ✓ her answer
  4. (D) Lenz currents
Why C: They are named after Foucault.
Q67Easycorrect  +4
solid plateslotted plateboth swing through the same magnetic fieldEddy currents in a swinging plateOf the two plates shown, the one that comes to rest sooner is the
  1. (A) solid plate, because it has the greater resistance
  2. (B) solid plate, because large eddy currents can circulate in it  ✓ her answer
  3. (C) slotted plate, because it is lighter
  4. (D) slotted plate, because the slots guide the current
Why B: The solid plate allows large closed loops of current, so more energy is dissipated and it is damped more strongly.
Q68Moderatecorrect  +4
The cores of transformers and other electrical machines are laminated in order to
  1. (A) increase the resistance to eddy currents and so reduce the heat lost  ✓ her answer
  2. (B) increase the magnetic flux
  3. (C) prevent the windings from touching the core
  4. (D) make the core lighter
Why A: Laminations, insulated from one another, break the current paths and so cut the heat loss.
Q69Moderatecorrect  +4
Electromagnetic damping is used in a moving-coil galvanometer to
  1. (A) protect the coil from overheating
  2. (B) bring the coil to rest quickly without oscillating  ✓ her answer
  3. (C) increase the deflection
  4. (D) make the scale linear
Why B: The eddy currents induced in the metal frame damp the motion, so the pointer settles quickly.
Q70Moderatecorrect  +4
In an induction furnace the metal to be melted is
  1. (A) heated by a flame beneath the crucible
  2. (B) placed inside a coil carrying a high-frequency current, so that eddy currents heat it  ✓ her answer
  3. (C) connected directly to a high-voltage supply
  4. (D) bombarded with electrons in a vacuum
Why B: The high-frequency field induces strong eddy currents in the charge itself, and the heat melts it.
Q71Hardcorrect  +4
Eddy currents in a metal plate swinging through a magnetic field dissipate 20 J in 4 s. If the plate is given twice the speed, the power dissipated becomes about
  1. (A) 20 W  ✓ her answer
  2. (B) 10 W
  3. (C) 5 W
  4. (D) 40 W
Why A: The eddy current is proportional to the speed, and the power to its square.P₁ = 20/4 = 5 W; P₂ = 4 × 5 = 20 W
Q72Hardcorrect  +4
How many of the following statements are correct?
A. Eddy currents oppose the motion that produces them.
B. Eddy currents are used in the magnetic braking of trains.
C. Cutting slots in a plate reduces the eddy currents in it.
D. Eddy currents are useful in every situation and are never a nuisance.
  1. (A) Only one
  2. (B) Only two
  3. (C) Only three  ✓ her answer
  4. (D) All four
Why C: A, B and C are correct. Eddy currents waste energy in transformers and motors, so they are often a nuisance.
Q73Easycorrect  +4
NScoilthe coil is rotated in the fieldto Rslip rings and brushes feed the external circuitAn a.c. generatorThe purpose of the slip rings and brushes shown is to
  1. (A) convert the alternating current into direct current
  2. (B) increase the emf of the generator
  3. (C) carry the current from the rotating coil to the stationary external circuit  ✓ her answer
  4. (D) reverse the current every half turn
Why C: The rings turn with the coil while the brushes stay still, so the circuit is never broken.
Q74Easyambiguous mark
For a coil of N turns and area A rotating with angular velocity ω in a field B, the instantaneous emf is
  1. (A) NBAω sin ωt  ✓
  2. (B) NBAω cos ωt  ✗ her answer
  3. (C) NBω sin ωt
  4. (D) NBA sin ωt
Why A: Differentiating Φ = NBA cos ωt gives the sine form.ε = NBAω sin ωt
Q75Moderatecorrect  +4
A generator coil of 100 turns and area 0.05 m2 rotates at 50 rad s−1 in a field of 0.2 T. Its peak emf is
  1. (A) 500 V
  2. (B) 50 V  ✓ her answer
  3. (C) 0.5 V
  4. (D) 5 V
Why B: The peak value is NBAω.ε₀ = 100 × 0.2 × 0.05 × 50 = 50 V
Q76Moderatecorrect  +4
The coil of a generator makes 300 revolutions per minute. Its angular frequency is about
  1. (A) 1885 rad s−1
  2. (B) 5 rad s−1
  3. (C) 300 rad s−1
  4. (D) 31.4 rad s−1  ✓ her answer
Why D: Convert revolutions per minute to radians per second.ω = 2π × 300/60 = 31.4 rad s⁻¹
Q77Moderatecorrect  +4
The emf of an a.c. generator is at its maximum when the plane of the coil is
  1. (A) parallel to the field, the flux then being zero  ✓ her answer
  2. (B) perpendicular to the field, the flux then being greatest
  3. (C) in any position, since the emf is constant
  4. (D) at 45° to the field
Why A: When the plane is parallel to B the flux is zero but it is changing fastest, so the emf is greatest.
Q78Hardcorrect  +4
A generator has a peak emf of 40 V. The instantaneous emf when the coil has turned through 30° from the position of zero emf is
  1. (A) 34.6 V
  2. (B) 20 V  ✓ her answer
  3. (C) 40 V
  4. (D) 10 V
Why B: Use ε = ε₀ sin θ, measuring θ from the position of zero emf.ε = 40 sin 30° = 20 V
Q79Moderatecorrect  +4
εtε₀one full rotation of the coilEmf of an a.c. generatorFrom the graph, the emf of the generator
  1. (A) is zero when the flux through the coil is zero
  2. (B) reverses its direction twice in every rotation of the coil  ✓ her answer
  3. (C) never reverses its direction
  4. (D) has a constant magnitude but reverses direction twice per rotation
Why B: The sine curve crosses zero and changes sign twice in each cycle, and one cycle is one rotation.
Q80Hardcorrect  +4
Assertion (A): The speed of rotation of a generator coil is doubled, and both the peak emf and the frequency of the output are doubled.
Reason (R): The peak emf is proportional to ω and the frequency is ω divided by 2π.
  1. (A) Both A and R are true and R is the correct explanation of A  ✓ her answer
  2. (B) Both A and R are true but R is NOT the correct explanation of A
  3. (C) A is true but R is false
  4. (D) A is false but R is true
Why A: Both ε₀ = NBAω and f = ω/2π are proportional to ω, so doubling ω doubles both.

This week

  1. Twenty Lenz directions, out loud. For each one say “flux is growing / shrinking” before naming the current. The failure is a skipped step, not a missing fact, so the drill has to force the step.
  2. Six flux–time graphs, slopes only. Give her the graph and ask for the emf in every region, including two regions with the same height change over different times. That is the exact comparison that failed in Q11.
  3. Write Φ = BA cosθ with “θ from the NORMAL” beside it at the top of the rough sheet, every physics paper, until it stops being a decision.
  4. Leave inductance alone. 24 out of 24 on a block she had never been examined on. There is nothing to revise there — move the time to Ray Optics, which is still the weakest physics chapter at 68%.
Marks read directly from the scanned script. Two independent readings of all 80 answers were taken and agreed on every one; each wrong answer was then re-read individually at high magnification before being written down, and Q74 was examined at the pixel level and is reported as unresolved rather than assumed.