Motional emf numericals Q27–Q38: 12 / 12. The rod-on-rails chain (emf → current → force → power → heat), terminal velocity, rotating rod, disc and aircraft were all correct. The three-line rule is working on calculations.
Eddy currents Q63–Q68: 6 / 6 — only the statement question and the Foucault name were missed.
Flux and Faraday's law Q1–Q13: 13 / 13. No slips in the basic block at all.
A.c. generator Q71–Q80: 9 / 10.
So this is not a chapter problem. It is one topic — the sense of the induced current — plus a handful of factor slips.
3. The rule that fixes half the errors
Three lines on every direction question, before looking at the options:
1. Which way does the flux point? 2. Is it growing or dying? 3. The induced current opposes the CHANGE — not the original current.
Then convert to a pole with the diagram above. Q14, 16, 18, 21, 22, 24 and 26 all fall to these three lines.
Lenz's Law and the Sense of the Induced Current — 5 wrong · 3 blank
Q14WrongCDirection reversalCoil face and current sense
The south pole of a bar magnet moves towards a coil. Seen from the magnet side, the induced current and the near face are?
Her answerAnticlockwise, near face becomes a south pole
Correct answerClockwise, near face becomes a south pole
WhyThe near face must become S to repel the approaching S pole — that half she had right. But a face is a SOUTH pole when the current runs CLOCKWISE as seen from that face (anticlockwise = north).
Where it went wrong · fixPole correct, sense reversed. Rule to memorise: Clockwise = South, Anticlockwise = North, always judged from the side you are looking at.
Q16WrongCDirection reversalIncreasing flux out of the page
A loop lies in a field out of the page whose magnitude is increasing. Induced current and force?
Her answerAnticlockwise, loop pushed away
Correct answerClockwise, loop pushed away
WhyFlux out of the page is growing, so the induced current must make flux INTO the page inside the loop → clockwise. The loop and source repel, so it is pushed away.
Where it went wrong · fixShe got the force right and the sense wrong. Write the three lines every time: flux direction → growing or dying → current that opposes it.
Q18BlankTBlank — A–RA–R: current switched off nearby
A: Switching off a nearby current induces a current in a closed loop. R: The induced current flows in the SAME sense as the original, to maintain the dying flux.
Her answerLeft blank
Correct answerBoth true, R explains A
WhyWhen flux is dying, Lenz's law means the induced current tries to keep it — so it flows in the same sense as the original current. When flux is growing, it flows opposite.
Where it went wrong · fixBlank on an A–R she could have solved with the three-step method. Opposing the CHANGE is not the same as opposing the current.
Q21WrongCDirection reversalEntry and exit of a field region
A loop is pulled at constant speed into and later out of a uniform field region. The senses of the induced current during entry and exit are?
Her answerOpposite only if the speed changes
Correct answerOpposite to each other
WhyEntering, flux grows; leaving, flux dies. Opposite changes give opposite induced currents, whatever the speed. Speed changes only the size of the current, never its sense.
Where it went wrong · fixSpeed controls magnitude; the direction of the flux CHANGE controls sense.
Q22WrongSArrangement / orderMatch: flux change → current sense
Match: flux into page increasing / decreasing; magnet's S pole receding; loop at rest in a steady field.
Her answerA–IV, B–II, C–III, D–I
Correct answerA–II, B–IV, C–III, D–I
WhyInto-page flux increasing → anticlockwise (II). Into-page flux decreasing → clockwise (IV). S pole receding → near face becomes N to attract it (III). Loop at rest in a steady field → no induced current (I).
Where it went wrong · fixRows C and D were right; A and B were swapped — the same right-content-wrong-arrangement pattern as Botany Q107 in the ILTS. Solve all four rows in the margin first.
Q23BlankTBlank — statement typeStatements on Lenz's law
A: Lenz's law follows from conservation of energy. B: The induced current opposes the change whether flux rises or falls. C: An induced emf can exist across an open conductor. D: If the induced effect aided the change, energy would come from nothing.
Her answerLeft blank
Correct answerA, B, C and D (all four)
WhyAll four are standard NCERT statements. C is the one usually doubted: an open rod still develops an emf across its ends — only the current needs a closed circuit.
Where it went wrong · fixLeft blank. If you cannot name the false word in a statement, the statement is true.
Q24BlankTBlank — not attemptedRing below a solenoid, current switched off
A metallic ring lies coaxially below a solenoid whose steady current is suddenly switched OFF. The ring is?
Her answerLeft blank
Correct answerAttracted towards the solenoid
WhyFlux through the ring is dying, so the induced current tries to maintain it — the ring's current runs in the same sense as the solenoid's, and parallel currents attract.
Where it went wrong · fixPair with the switch-ON case: flux growing → opposite currents → repelled. Learn the two as a pair.
Q26WrongCDirection reversalA–R: why the induced current always opposes
A: The induced current always opposes the change that produces it. R: If it aided the change, the current and hence the energy would grow without work being done.
Her answerA is true but R is false
Correct answerBoth true, R is the correct explanation of A
WhyR is exactly the energy argument behind Lenz's law: an aiding current would be self-amplifying and create energy from nothing. So R is true and it explains A.
Where it went wrong · fixJudge R on its own before deciding. Here R is a true physical statement that also happens to be the reason for A.
Motional emf — 5 wrong · 0 blank
Q39WrongSFactor / scaling slipCharge through the circuit
Rod on rails moves 2 m. B = 0.5 T, l = 0.4 m, R = 0.8 Ω. Charge that flows?
Her answer0.25 C
Correct answer0.5 C
Whyq = ΔΦ/R = B·l·Δx / R = 0.5 × 0.4 × 2 / 0.8 = 0.5 C.
Where it went wrong · fixExactly half the right answer — the familiar factor-of-2 slip. Write the formula fully before substituting.
Q42WrongCConcept gapWhen is the motional emf zero?
The emf of a rod moving in a magnetic field is zero when?
Her answerThe rod is perpendicular to both the field and the velocity
Correct answerIts velocity is parallel to the field
Whyε = B l v sin θ, where θ is the angle between v and B. The emf vanishes when v ∥ B (θ = 0). A rod perpendicular to both v and B is the case of MAXIMUM emf.
Where it went wrong · fixShe picked the maximum-emf configuration. The emf dies when the rod sweeps no field lines.
Q43WrongSFactor / scaling slipVelocity at 30° to the field
v makes 30° with B = 0.4 T, rod perpendicular to both. l = 0.5 m, v = 12 m/s. emf?
Her answer2.08 V
Correct answer1.2 V
Whyε = B l v sin θ = 0.4 × 0.5 × 12 × sin 30° = 0.4 × 0.5 × 12 × 0.5 = 1.2 V.
Where it went wrong · fix2.08 V is the cos 30° answer. Only the component of v perpendicular to B produces emf, so it is always sin θ.
Q45WrongSFactor / scaling slipRotating rod: axis to mid-point
A 1 m rod rotates about one end with angular velocity ω. PD between the axis and the MID-POINT?
Her answerBω/4
Correct answerBω/8
WhyFor a rod rotating about one end, ε = ½Bωl². Up to the mid-point, l = ½: ε = ½Bω(½)² = Bω/8.
Where it went wrong · fixBω/4 is half of the full-length value — she scaled the length linearly. The length is SQUARED, so half the length gives a quarter.
Q46WrongNConcept gapStatements about motional emf
A: emf ∝ length for a rod moving perpendicular to B. B: emf ∝ l² for a rod rotating about one end. C: The magnetic force on the induced current always opposes the motion. D: The emf depends on the external resistance.
Her answerA, C and D only
Correct answerA, B and C only
WhyD is false — resistance changes the CURRENT, never the emf. B is true because ε = ½Bωl².
Where it went wrong · fixShe rejected B (the ½Bωl² statement) and accepted D. The same ½Bωl² formula was mis-scaled in Q45, so both errors are one gap.
Induced Charge and the Flux–Time Graph — 1 wrong · 0 blank
Q55WrongSFactor / scaling slipCurrent in region I
Flux–time graph, 50-turn coil, R = 5 Ω. The current during region I?
Her answer0.12 A
Correct answer6 A
WhyIn region I the emf is N·ΔΦ/Δt = 50 × 0.6/1 = 30 V (as answered correctly in Q52). Then I = ε/R = 30/5 = 6 A.
Where it went wrong · fix0.12 A is 0.6 ÷ 5 — the FLUX divided by resistance instead of the emf. She had already found the 30 V three questions earlier; carry the answer forward.
Eddy Currents and the a.c. Generator — 2 wrong · 2 blank
Q69BlankTBlank — statement typeStatements on eddy currents
A: They oppose the motion producing them. B: They dissipate energy as heat. C: They can be reduced by lamination. D: They are produced only in magnetic materials.
Her answerLeft blank
Correct answerA, B and C only
WhyD is false — eddy currents appear in any conductor, including copper and aluminium, which are not magnetic. The swinging copper plate in Q64 is the proof.
Where it went wrong · fixLeft blank although Q63–Q68 were all correct. The knowledge was there; the statement format stopped her.
Q70WrongNNCERT factAnother name for eddy currents
Eddy currents are also known as?
Her answerAmpere currents
Correct answerFoucault currents
WhyEddy currents are called Foucault currents, after Léon Foucault.
Where it went wrong · fixPure recall. Card: eddy = Foucault. Ampère, Lenz and Faraday all have laws, not currents.
Q74BlankTBlank — formula not attemptedInstantaneous emf
Peak emf 40 V. Instantaneous emf when the coil has turned 30° past the position of zero emf?
Her answerLeft blank
Correct answer20 V
Whyε = ε₀ sin θ = 40 sin 30° = 40 × 0.5 = 20 V. The zero-emf position is θ = 0 in this formula.
Where it went wrong · fixOne-line substitution left blank. Note the contrast with Q43, where sin 30° was needed and cos was used — the same fact, both ways.
Q75WrongSFactor / scaling slipPeak to opposite peak
An a.c. generator runs at 25 Hz. Time from one peak to the next opposite peak?
Her answer10 ms
Correct answer20 ms
WhyT = 1/f = 40 ms. A peak and the next opposite peak are half a cycle apart: T/2 = 20 ms.
Where it went wrong · fix10 ms is T/4, the peak-to-zero time. Sketch the sine curve and count: peak → zero = T/4, peak → opposite peak = T/2.
Fix list before 4 October
Saturday 26 Sep repair hour: Lenz's law only. Redraw all eight figures and write the three lines for each.
Re-test 1 (30 Q, Sat 26 Sep): direction and sense questions only, no numericals. Pass = 27/30 with at most 2 wrong.
Never leave a statement question blank (Q23, Q69). If you cannot name the false word, the statement is true.
Factor card: q = BlΔx/R · rotating rod ε = ½Bωl² (length SQUARED, so half the rod gives a quarter) · ε = Blv sin θ, always sin · peak → opposite peak = T/2, peak → zero = T/4 · ε = ε₀ sin θ.
Carry your own answer forward (Q55): the 30 V found in Q52 was the input for Q55.
Recall card: eddy currents = Foucault currents.
Error tags: C concept · N NCERT/recall · S silly or factor slip · T not attempted. All numerical answers verified by calculation. Six strokes on the sheet sat between two options (Q27, 39, 41, 45, 46, 70) and were read as marking the option whose letter the stroke starts on; if Q70 and Q45 were intended differently the score becomes 245.