Electromagnetic Induction Exemplar Solutions Class 12 - Free PDF

Strategic angle. A loop only ``sees'' the field component along its normal. Project the field onto n first; everything else is arithmetic.

1. Loop normal: n=k. Component of B along n is B_z=4B₀.
2. Flux =B_z·(area)=4B₀· L²=4B₀ L² Wb.

Why this matters. The same trick handles every constant-field flux problem: identify n, read off the parallel component, and multiply by area.

Alternative – via |B|cosθ. The angle between B and k satisfies cosθ=B·k|B|=4√29, so _B=|B|Acosθ=B₀√29· L²· 4√29=4B₀L² Wb – exactly the same answer, just a longer route. Choice (D) is the trap that drops the cosθ factor.

Unit cross-check. [B][A]=T·m²= Wb. Coefficient is dimensionless, so 4B₀L² carries units of Wb as required.

Concept linkage. The very same projection trick reappears in Chapter 1 (electric-flux through a tilted area) and in Chapter 4 (force on a current-carrying loop – only the perpendicular component of B contributes to the torque). Flux is fundamentally a ``component-along-normal × area'' object.

_B=4B₀ L² Wb.

Strategic angle. Use the fact that flux depends only on the boundary, not on the chosen surface. Pick the easiest surface – two squares meeting along AD.

1. Square in xy-plane bounded by ABCD: normal k, area L², gives B₀ L².
2. Square in yz-plane bounded by ADEF: normal i, area L², gives B₀ L².
3. Sum: 2B₀ L² Wb.

Why this matters. Flux through any cap with the same boundary is the same – a direct consequence of ∇·B=0.

Alternative – shrink the boundary. You can replace the L-shaped loop with the flat triangle that has the same edge sequence projected onto each face. Because B=B₀(i+k) has equal x- and z-components, the two square-projections each carry B₀L², giving the same total 2B₀L² – a nice consistency check.

Sign discipline. The corner order A fixes both normals to point outward along +k and +i respectively. Had the problem said FEDCBA, both normals would flip and the answer would be -2B₀L². Marks are often lost on this.

Concept linkage. The independence of flux on the chosen cap is exactly why we can apply Faraday's law to any convenient surface – critical when handling solenoids and toroids later in this chapter.

_B=2B₀ L² Wb.

Strategic angle. Spin-symmetry test – if rotating the source leaves it looking identical, the field pattern is time-independent and no e.m.f. can appear.

1. The bar magnet is cylindrically symmetric about its axis. Rotation about that axis is a symmetry of the system.
2. A symmetry of the source cannot change the flux through any fixed circuit – so dΦ/dt=0 and ε=0.

Why this matters. A real homopolar generator needs the wire itself to move through the field, not just the magnet to spin; this question is its negative counterpart.

Alternative – motional emf check. Try to compute a motional e.m.f. ∫(v×B) along the axis wire: v=0 for points on the axis (no translation), and along the radial brush contact the magnet's surface also moves with velocity tangential, but the closed loop is stationary in the lab. The conductor (wire + ammeter) does not move at all, so the line integral vanishes element-by-element.

Common pitfall. Students often confuse this setup with the Faraday disc (rotating conducting disc in a fixed field), where a real DC e.m.f. does appear. The difference: in the Faraday disc the conductor rotates through the field. Here the conductor is fixed and only the field source spins – but the field pattern itself is rotation-invariant, so nothing changes.

Concept linkage. ``A field that is invariant under a continuous symmetry of its source cannot induce an e.m.f.'' generalises the ``dΦ/dt=0'' criterion to arbitrary geometries.

No current; symmetry forbids any change in flux.

Strategic angle. Decouple the two causes of flux change – current variation vs. geometric motion – and test each against the data.

1. ``B has current iff A moves'' ⇒ flux change is purely geometric, not current-driven ⇒ I_A is constant.
2. ``Current in A is counter-clockwise (given)''. So I_A is constant and counter-clockwise.

Why this matters. This is exactly the principle of a transformer's primary at DC – moving a DC-energised coil mimics an AC primary for the secondary.

Alternative – mutual-inductance view. If _B=M(t) I_A, then _B=-d_Bdt= -dMdtI_A-MdI_Adt. The data (_B exists ⇔ A moves) tells us only the first term contributes: dM/dt≠ 0 (because separation is changing) but dI_A/dt=0 (because I_A is steady). This is exactly the algebraic statement of step 1.

Common pitfall – mistaking _B for I_A. Some students see ``current in B is in some direction'' and conclude that I_A must be in that same direction. The current in B is the induced one and its sign is set by Lenz's law – it has no direct correspondence with I_A. The question states I_A's direction outright; that's what fixes (A) vs. (D).

Concept linkage. The same logic underlies wireless-charging pads – a DC-energised coil produces no e.m.f. in the receiver unless the geometry (separation, alignment, or current) changes.

Constant counter-clockwise current in A.

Strategic angle. Two pieces of evidence: (i) only motion drives B's current, so I_A is steady; (ii) Lenz's law converts the given direction in B at t=0 into a direction in A.

1. ``At rest ⇒ no current in B'' fixes I_A=const.
2. Sign analysis via Lenz's law on the instantaneous geometry of Fig. 6.3 reverses the sense from B to A ⇒ clockwise in A.

Why this matters. A rotating DC-energised coil is the simplest AC generator – the same mechanism powers commercial alternators.

Alternative – flux-sign book-keeping. At t=0 let n_B point out of the page. If I_B is counter-clockwise (right-hand rule) then the induced field at B's centre is along +n_B. By Lenz's law the induced field opposes the change, so the flux through B from A must be decreasing along +n_B. The only way this happens with I_A steady is that A's axis is rotating away from coincidence with n_B, with I_A producing field along -n_B at the instant shown – which (right-hand rule applied to A) means I_A runs clockwise.

Numerical cross-check. If _B(t)=M₀ I_Acos(ω t) with M₀>0 and I_A>0, then _B=-_B=+M₀ I_A(ω t). At t=0⁺ this is positive in our convention, so the current in B flows in the positive sense (counter-clockwise as drawn) – matching the data with I_A steady.

Concept linkage. This is literally how a household ceiling fan generates back-EMF when momentarily disconnected – a DC-energised rotor in a stator coil acts as an instantaneous AC source.

Constant clockwise current in A.

Strategic angle. Memorise the ratio A/l for a solenoid; the question is then pure proportionality.

1. L=₀ N²A/l, so the ``increase'' directions are A↑ or l↓.
2. Only option (B) does both unambiguously.

Why this matters. Real inductors increase L by packing more turns into a shorter, fatter coil – exactly the (B) recipe.

Alternative – per-unit-length form. With n=N/l (turns per metre), B=₀ nI and the flux linkage is NΦ=N· BA=₀N²lAI, so L=₀N²Al. The same formula, derived from Φ=LI, makes the A/l dependence obvious.

Numerical cross-check. For N=1000, A=10^-3 m², l=0.5 m: L=4π10^-7· 10⁶· 10^-30.5 ≈ 2.5 mH. Halving l to 0.25 m doubles it to 5.0 mH; doubling A to 210^-3 m² doubles it again to 10 mH – consistent with (B).

Common pitfall. Many students try option (A) thinking ``bigger means more inductance''. But a longer coil with the same N reduces n=N/l and so weakens B, dropping Φ per turn faster than the extra length helps. The dependence is on A/l, not on volume Al.

Concept linkage. The energy stored is U=12 LI²=12₀B²· Al, so the energy density u=B²/(2₀) is set by B alone (Section 6.10 of NCERT). L and volume together encode the same physics.

l↓, A↑ ⇒ L↑.

Strategic angle. Pin down the two energy sources – an externally driven current ((A) and (D)) and induced eddy currents ((B)). Static spatial variation alone does no work, ruling out (C).

1. External current ⇒ I²R heating: (A), (D).
2. Time-varying B ⇒ eddy currents ⇒ heat: (B).
3. Static field (even if non-uniform in space) ⇒ no e.m.f. ⇒ (C) is false.

Why this matters. Eddy-current heating is the principle of induction cooktops – a time-varying field heats the pan without contact.

Alternative – energy bookkeeping. Joule heat per unit volume is j²/σ where j is the current density and σ the conductivity. So heating requires j≠ 0 somewhere inside the plate. The only ways to make j≠ 0 are: connect a battery (DC) or supply oscillation (AC) – options (A) and (D); or drive eddy currents with ∂B/∂ t≠ 0 – option (B). A purely spatial B(r) that does not change in time leaves ∂B/∂ t=0 everywhere, hence E=0 in the plate's rest frame and j=σE=0.

Why (C) is wrong, in detail. ``Space-varying but constant in time'' is the field of a permanent magnet held still next to a stationary plate. No part of the plate sees a changing flux, so ∮E=0 around every small loop. No e.m.f., no eddy currents, no heating.

Concept linkage. Eddy currents are not just heaters – they're also brakes. In electromagnetic braking of trains a copper plate moves through a static field; the motion of the conductor turns the static field into a time-varying one in the conductor's frame, dissipating kinetic energy as heat. (C) becomes a heating mechanism the moment the plate is allowed to move.

(A), (B), (D).

Strategic angle. Total dΦ/dt has two pieces – ``B changes'' and ``loop moves''. As long as at least one is non-zero, ε is non-zero.

1. (A) B changes only: yes.
2. (B) both change: yes.
3. (C) loop moves only: yes (motional e.m.f.).
4. (D) nothing changes: no.

Why this matters. A generator (case C) and a transformer (case A) are two sides of the same Faraday-law coin.

Alternative – the Leibniz decomposition. The total time-derivative of Φ=_SBA for a moving surface S(t) obeys dΦdt= _S∂B∂ tA - _{∂ S}(v×B). The first term is the ``transformer'' piece (option A); the second is the ``motional'' piece (option C); option B has both. Option D zeros both terms.

Why (D) really fails. ``Spatially varying but time-independent'' B has ∂B/∂ t=0. A stationary loop has v=0, so both terms vanish identically. Even if B varies wildly in space, no induced e.m.f. appears – Φ is a fixed number set by the spatial integral.

Concept linkage. The two terms map onto two physical pictures we learn separately: transformers (term 1) and AC generators / linear generators / electromagnetic guns (term 2). Faraday's law unifies them.

(A), (B), (C).

Strategic angle. Test each option against the defining relation M₁₂=₁₂/I₂ and the reciprocity M₁₂=M₂₁.

1. Closer coils ⇒ more flux linkage ⇒ (A) true.
2. M independent of current ⇒ (B) false.
3. Rotation can decrease coupling ⇒ (C) not universally true.
4. Reciprocity ⇒ (D) true.

Why this matters. Reciprocity (D) lets us choose the easier of the two flux integrals when computing M.

Alternative – vector-potential view. M₁₂ can be written as Neumann's double-line integral M₁₂=₀4π_C1C2 d₁₂|r₁-r₂|. The integrand is manifestly symmetric under 1↔ 2, which proves M₁₂=M₂₁ on the spot (option D). And distance |r₁-r₂| appears in the denominator, so reducing it (bringing coils closer) increases M (option A).

Why (B) is firmly false. M is a geometric coupling coefficient, not a function of I. This is the same statement as ``capacitance does not depend on the voltage stored''.

Why (C) is conditional. Starting from 90^∘ misalignment (M≈ 0) and rotating toward coaxial alignment does increase M. The blanket statement ``rotation increases M'' is too broad – it depends on the starting orientation. So (C) is not universally true and is counted wrong.

Numerical cross-check. For two coaxial circular coils of radii a and b (b≪ a) separated by z: M=₀π a²b²2(a²+z²)^3/2. Independent of I. Decreases as z grows – consistent with (A).

Concept linkage. Reciprocity (D) is the same principle that gives equal mutual capacitance in a network – a deep result from linearity of Maxwell's equations.

(A), (D).

Strategic angle. Write Φ=B_⊥ A and force the product to be constant (or zero).

1. B_⊥=0 identically ⇒ Φ=0 ⇒ (B).
2. B_⊥(t)A(t)=const ⇒ dΦ/dt=0 ⇒ (C).

Why this matters. A vanishing time-derivative of a product is not the same as either factor being constant – a subtlety that eliminates (A) and (D).

Alternative – product-rule cross-check. dΦdt=B_⊥ A+B_⊥A. For this to vanish in case (C), the two terms must cancel: B_⊥/B_⊥= -A/A, i.e. B_⊥ must decrease at the same logarithmic rate the area grows. Calling r(t) the loop radius, B_⊥B_⊥=-2rr. So B_⊥∝ 1/r² keeps Φ constant – a very specific tuning.

Why (D) really fails. With B constant and perpendicular, Φ=B(π r²). As r grows, dΦdt=2π Brr≠ 0 unless r=0 (loop is not expanding). The problem states it is expanding, so (D) generates an e.m.f. – it cannot be one of the zero-emf cases.

Concept linkage. This setup is the inverse of a homopolar generator: instead of the e.m.f. driving current, here we force a special B(r,t) to keep the e.m.f. at zero. The same idea is used in superconducting flux pumps to maintain constant flux against a changing area.

(B), (C).

Strategic angle. Separate the topology of the circuit (open vs. closed) from the time-derivative of the flux. Only the latter drives an e.m.f.

1. Closing the switch changes resistance from ∞ to a finite value but not _B.
2. dΦ/dt=0 ⇒ ε=0 ⇒ no current.

Alternative – circuit analogy. Think of the magnet's flux ₀ as a fixed reservoir. Faraday's law acts like a generator that needs Φ to change; if the reservoir level is constant, the generator outputs zero volts no matter whether you've connected a load (closed circuit) or left it open. Closing S only attaches the load.

Common pitfall. ``Closing the switch should send a current into the loop because flux now passes through the closed loop.'' Wrong on two counts: (i) flux through the wire-bounded surface existed before too (Faraday's law cares about dΦ/dt, not about loop ``closedness''); (ii) the magnet has not moved, so Φ never changes. Currents need a time-varying flux, not just any flux.

Concept linkage. If immediately after closing S you started pulling the magnet away (or pushed it in), then dΦ/dt≠ 0 and a current would appear. This is exactly Faraday's original 1831 experiment.

No current flows on closing the switch.

Strategic angle. Apply Lenz's law to the change in flux caused by stretching. The induced response always opposes the change, here a decrease in linked flux.

1. Gaps open between turns ⇒ flux Φ leaks through the gaps and drops.
2. Induced e.m.f. opposes this fall ⇒ aids the source ⇒ current I rises.
3. Also n=N/l falls ⇒ L=₀ n²V falls ⇒ inductive reactance falls ⇒ more current.

Quantitative flavour. Take a solenoid with N turns, length l₀, radius r. Initial n₀=N/l₀ and ₀=₀ n₀ I π r² N links the coil. After stretching to length l₁>l₀, n₁=N/l₁0, ₁<₀="" _ind="-dΦ/dt" (i.="" at="" during="" e.m.f.="" induced="" is="" same="" so="" stretching="" the="">0\) – adding to the source e.m.f. and momentarily boosting the current.

Why the ``r=ρ/A so current falls'' argument is not what NCERT wants. A real solenoid's wire length changes only marginally on stretching (the helix simply un-tilts); the dominant electromagnetic effect is the flux leakage through the newly opened gaps. The exemplar's official answer therefore rests on Lenz's law, not on a small ohmic correction.

Concept linkage. The same logic explains why a coil with ferromagnetic core (large _r) carries less DC than the air-core case during transient core insertion: in both problems we follow the flux change and apply Lenz's law.

Current increases (flux leakage ⇒ Lenz-aiding e.m.f.).

Strategic angle. Track the flux: any change in _r re-routes flux lines, and Lenz's law fixes the sign of the response.

1. Iron core enters ⇒ _r rises ⇒ flux Φ rises.
2. Lenz: induced e.m.f. opposes the rise ⇒ opposes the source ⇒ I decreases.
3. Quantitatively L has grown by factor _r, so the back-e.m.f. is large during insertion and the steady current settles at a smaller value than before.

Quantitative back-e.m.f. With L=_r0 n²V, inserting the core in time _ins gives dLdt∼(_r-1)L_0ins. The back-e.m.f. is then approximately _back=IdLdt∼ I(_r-1)L_0ins. For _r∼ 5000 (soft iron), L₀∼ 1 mH, I∼ 0.1 A and _ins∼ 1 s, we get _back∼ 0.5 V – enough to drop the current substantially.

Common pitfall. ``Flux Φ=LI increases, so I must increase too.'' Wrong: the inserted core makes L much larger, and the source e.m.f. cannot maintain the previous current against the larger inductance. The induced e.m.f. opposing the source forces I down.

Concept linkage. Iron-cored chokes in power supplies use exactly this effect to limit current; without the core, the same coil would pass a larger current and saturate the load.

Current decreases (Lenz-opposing e.m.f. during _r↑).

Strategic angle. Treat the ring as the secondary of a transformer and apply Lenz's law to read off the direction, then use the parallel/anti-parallel loop force rule.

1. Flux growth ⇒ induced current opposes ⇒ anti-parallel to solenoid.
2. Anti-parallel coaxial loops repel ⇒ ring jumps up.

Alternative – pole picture. At the top of the solenoid, the field lines emerge upward (say); this is a magnetic N-pole. By Lenz's law, the induced current in the ring must produce a flux opposing the growing upward flux. The ring therefore presents a N-pole on its bottom face – and two N-poles facing each other repel. Result: the ring is launched upward, exactly as observed.

Energy budget. Where does the ring's kinetic energy come from? From the source e.m.f. driving the solenoid current. Power ε I_sol goes partly to I_sol²R heating, partly to building the magnetic-field energy, and partly to mechanical work on the ring. Conservation of energy is rigorously maintained.

Concept linkage. The jumping ring is a stock demonstration in Indian physics labs. A more violent version – a copper ring on an iron core with an AC coil – can shoot the ring several metres. Same Lenz mechanism, just at 50 Hz.

Repulsion from induced anti-parallel ring current.

Strategic angle. Mirror image of 6.14: ``flux decreasing'' is opposed by an induced current that supports the existing flux.

1. Flux falling ⇒ induced ring current parallel to solenoid.
2. Parallel coaxial currents attract ⇒ ring pulled down.

Alternative – pole picture. While energised, the top of the solenoid presented (say) a N-pole. Switching the solenoid off does not change the ring's pole instantly, but during the decay the induced current in the ring tries to preserve the existing flux through itself. The bottom face of the ring therefore behaves like a S-pole and the top face of the solenoid (now decaying) still has residual N-character. S meets N ⇒ attraction ⇒ ring snaps downward.

Why the effect is brief. The attractive impulse lasts only while dI_sol/dt≠ 0. For a typical mechanical switch this is microseconds, and the ring acquires a small downward momentum but is then in free fall under gravity. The net observed behaviour: a small ``thud'' onto the cardboard.

Concept linkage. Compare with 6.14: switching on gave a sustained jump because the current rise lasted milliseconds and the magnetic pressure built up; switching off gives a much shorter pulse because the inductor's stored energy drops faster (an arc may form across the switch, draining the energy quickly).

Ring is attracted downward.

Strategic angle. Compare flux change: a moving magnet drives dΦ/dt in the pipe; a moving piece of iron does not.

1. Magnet falling ⇒ eddy currents in pipe ⇒ Lenz-law retarding force ⇒ slower descent.
2. Iron bar falling ⇒ no net flux change ⇒ no eddy currents ⇒ near-free fall.

Terminal-velocity estimate. At terminal velocity v_t the magnetic braking force equals weight: F_br(v_t)=mg. A simple scaling argument gives F_br≈α B_eff²vR_pipe, where B_eff is the field at the pipe wall and R_pipe is the eddy-current loop resistance. So v_t≈mgR_pipeα B_eff². For a strong NdFeB magnet (B_eff∼ 0.5 T) in a copper pipe of 1 cm wall, v_t is typically a few cm/s – demonstrably slow to the eye.

Why an unmagnetised iron bar still feels some drag. Iron is ferromagnetic, so it does become weakly magnetised by stray fields and by the Earth's field. But the moment is tiny compared to a permanent magnet's, so the flux change as it falls through the pipe is also tiny – the eddy-current braking force is negligible, and gravity dominates.

Concept linkage. Electromagnetic braking on rails – some high-speed trains use a strong magnet hovering over the iron rail; eddy currents in the rail provide contactless braking. The pipe-and-magnet demo is the laboratory analogue.

Magnet is slowed by eddy-current braking; iron bar is not.

Strategic angle. Differentiate the cosine, plug in the three phases, attach directions via the right-hand rule.

1. Φ=π a²B₀cosω t ⇒ ε=π a²B₀ω t.
2. At the three quarter-periods sinω t=1,0,-1 ⇒ I=+I₀,0,-I₀ with directions as above.

Direction check via Lenz at t=π/(2ω). At this instant B=B₀cos(π/2)k=0, but Ḃ=-B₀(π/2)k=-B₀ωk – the flux through the loop is decreasing along +k. Lenz's law says the induced current must produce a flux supporting +k. By the right-hand rule, that current flows counter-clockwise as viewed from +k – at (a,0,0) this points along +j, matching the sign found above.

Numerical sanity check. Pick B₀=0.1 T, a=0.05 m, ω=100 rad/s, R=10 Ω: I₀=π(0.05)²(0.1)(100)/10=7.85× 10^-4 A ≈ 0.78 mA. A meter would register this comfortably.

Concept linkage. This is a textbook AC dynamo problem – a sinusoidal field through a fixed loop produces a sinusoidal current that is π/2 out of phase with the field (current is maximum when B is maximum, which is when B is zero). The same phase relation governs every transformer secondary.

I₀=π a²B₀ω/R; signs follow sinω t.

Strategic angle. ``Two caps on one boundary'' = closed surface; apply ∮BA=0.

1. S₁∪ S₂ closes the surface; net outward flux is zero.
2. Sign-flipping one surface to match the boundary gives equal fluxes through S₁ and S₂.

Why this matters. The same reasoning is what allows you to choose ``any convenient surface'' when applying Faraday's law.

Alternative – vector-potential argument. Since ∇·B=0 globally, there is a vector potential A with B=∇×A. By Stokes' theorem, _SBS= _S(∇×A)S= _{∂ S}A. The last expression depends only on the boundary ∂ S=C, not on S. So Φ is a property of C alone – exactly what we wanted to show.

Where this fails. If there were a magnetic monopole inside the volume enclosed by S₁∪ S₂, then ∮BA=₀ q_m≠ 0 and the two cap fluxes would differ by ₀ q_m. The non-observation of any flux discrepancy in electromagnetic-induction experiments is one of the most precise upper bounds on monopole abundance.

Concept linkage. This freedom is what lets us draw a Faraday-law solenoid problem with a ``cup-shaped surface'' rather than a flat disc – crucial when the flat disc cuts the solenoid axis at an awkward place.

Same answer; flux depends only on the boundary.

Strategic angle. Forget the angle; compute dA/dt directly.

1. In time dt, the slider moves v dt along the rails; area swept is dA=d· v dt.
2. ε=B dA/dt=Bdv; I=Bdv/R.

Why this matters. The slanted slider does not change ε – it just stretches the slider's physical length without adding any flux-cutting power.

Alternative – motional-emf integral. Compute ε=_P^Q(v×B) along the slanted rod. With v=vx, B=Bz, we get v×B=-vBy. Parameterise the rod from P to Q as r(s)=s(cosθx+sinθy), 0≤ s≤ d/sinθ, so d=(cosθx+ sinθy) ds. The dot product is (-vB)(sinθ) ds, integrating to (-vBsinθ)·(d/sinθ)=-vBd. Magnitude ε=vBd – same answer, θ cancels.

Why the angle cancels. The slanted rod has more wire-length contributing per unit swept area, but each unit of wire moves through less field-cutting (only the perpendicular component of velocity matters). The two effects cancel and only B, d, v survive.

Concept linkage. This Bdv formula is the universal motional-emf result – it underlies linear induction generators (used in some MAGLEV systems) and is identical to the e.m.f. in a rectangular sliding-bar setup at θ=90^∘.

I=Bdv/R.

Strategic angle. The back-e.m.f. is proportional to slope – read the three slopes, scale them by |u|/slope at t=3 s.

1. Slopes: OA=0.2, AB=-0.6, BC=0.1, beyond C =0 A/s.
2. |u| is 3× on AB, 0.5× on BC, and 0 for t>30, relative to |u(3 s)|=e.
3. Max at AB. u(7):|3e|, u(15):0.5e, u(40)=0 (signs as in the main solution).

Alternative – ratio approach. Since u∝ dI/dt and L is fixed, u(t)u(3)= m(t)m_OA on any segment. So u(7)/e=m_AB/m_OA=-0.6/0.2=-3; u(15)/e=m_BC/m_OA=0.1/0.2=0.5; u(40)/e=0/0.2=0. No need to compute L explicitly.

Why the sign matters. Back-e.m.f. opposes the change in current. On OA, I is rising, so u opposes the rise (sign convention chosen so this is ``positive e''). On AB, I is falling, so u acts to oppose the fall – its sign reverses, hence u(7)=-3e. Sign tracking is what distinguishes a competent answer from a copy of formulas.

Concept linkage. This is exactly the principle behind flyback diodes across inductors and relay coils: when the supply current is suddenly cut (AB-like steep drop), the inductor generates a huge L |dI/dt| voltage that would damage transistors – the flyback diode shunts the current safely.

|u| max on AB; |u(7)|=3e, |u(15)|=0.5e, u(40)=0.

Strategic angle. The two scenarios share a single number M. Extract M from one, plug into the other.

1. M=10^-2/2=5 mH.
2. _A=M· 1=5 mWb.

Why reciprocity is non-trivial. A naive expectation would be that flux ``transferred'' from a bigger coil to a smaller one differs from the reverse. The reciprocity theorem (M₁₂=M₂₁) says no – this remarkable fact follows directly from ∇·B=0 and the linearity of Maxwell's equations.

Unit cross-check. M in henries = Wb/A. So 5× 10^-3 Wb/A =5 mH. _A=M I_B has units H·A = Wb.

What if the coils had iron in between? A high-_r medium between the coils would multiply M by roughly _r – the same flux-amplification mechanism that gives transformers their high coupling coefficient k≈ 1.

Concept linkage. Reciprocity also implies that a single coupling coefficient k controls both ``A-to-B'' and ``B-to-A'' transfer, justifying the ideal-transformer ratio V₂/V₁=N₂/N₁ used in Chapter 7.

_A=5 mWb.

Strategic angle. Recognise the two flux-change contributions and combine via the product rule.

1. Flux Φ=B(t)A(t)=B₀sinω t· dvt.
2. dΦ/dt gives ε, then I=ε/R and the agent force F=BId as above.

Alternative decomposition. Split the e.m.f. into transformer + motional pieces: ε= -B(t)· A(t)_transformer -B(t)·A(t)_motional =-B₀(ω t)· dvt-B₀sin(ω t)· dv. Factor: ε=-B₀dv[ω tcosω t+sinω t] – same answer. The split shows clearly that both mechanisms contribute and the formula is just the Faraday law written two ways.

Limiting checks. (a) Constant field (ω→ 0, B=B₀): ε=-B₀dv, F=B₀²d²v/R – the standard sliding-rod result. (b) Stationary rod (v→ 0): both ε and F vanish, which is wrong! Actually with v=0, the area A= const but flux Φ=B₀sinω t· A₀ still varies in time, so ε≠ 0. The expression in the answer assumed A=dvt starts at A=0 – so the limit v→ 0 collapses the loop too. A more general start with A(0)=A₀>0 would handle this case.

Concept linkage. This problem is a microcosm of all AC-generator analysis: the secondary's e.m.f. has a ``transformer'' part (changing B) and a ``motional'' part (moving rotor). Real generators harness both.

I and F as in the main solution.

Strategic angle. The rod is a one-dimensional damped system mv=-γ v with γ=B²²/R. Exponential decay follows.

1. EOM mv+γ v=0 with γ=B²²/R.
2. Solution v=u₀ e^{-γ t/m}.
3. Power dissipated I²R=γ v²=-dK/dt confirms ₀^∞ I²R dt=12m u₀².

Why this matters. Electromagnetic braking on a rail is the direct DC analogue of viscous drag.

Time-constant interpretation. The decay time is τ=mR/(B²²). Larger R ⇒ weaker eddy current ⇒ weaker braking ⇒ longer τ. Stronger B or longer rod ⇒ faster decay. This is exactly the design knob for an electromagnetic brake – engineers choose the rail resistance and magnet strength to set τ for target stopping distance.

Energy as integral. ₀^∞ I²R dt=₀^∞B²²v²Rdt. Substituting v=u₀ e^-t/τ: ₀^∞B²²u₀²Re^-2t/τdt =B²²u₀²R·τ2 =B²²u₀²R·mR2B²² =12m u₀². Exactly the initial kinetic energy – zero leakage.

Common pitfall (time-varying B case). For part (i) with non-zero B the EOM has two damping terms and an extra forcing through B(t)x. Students often drop the B(t)x term, but it represents the transformer e.m.f. which can drive the rod even when stationary.

v=u₀ e^-B22t/(mR); lost KE = heat in R.

Strategic angle. The e.m.f. of a uniformly rotating rod and its resistance both scale with the same length _OP; the _OP²/_OP ratio leaves a single power of length in the answer.

1. ε=12Bω_OP², r=λ_OP, so I=Bω_OP/(2λ).
2. Express _OP piecewise as above to get I(θ) on each side.

Derivation of ε=12Bω L². A point on the rod at distance r from the pivot has speed v=ω r. The motional-e.m.f. contribution from an element dr is dε=Bvr (dr/r)=Bω r dr. Integrate from r=0 to r=L: ε=Bω₀^L r dr=12Bω L².

Why current diverges at θ=π/4 junctions. As θ→π/4^-, _OP=l/cosθ→ l2; at the junction the formula switches to l/sinθ=l2. The two branches give the same _OP at the corner – I is continuous, not divergent. (A pure formula like 1/cosθ diverges at π/2, but the physical rod never reaches a corner with cosθ→ 0 while still using that branch.)

Concept linkage. This is essentially a disc generator with a piecewise-linear ``rim''. A circular rim (the Faraday disc) gives constant _OP=L, constant I. The rectangular boundary modulates the contact length and so modulates the output current.

Diagram-based reasoning. The current peaks at θ=π/4,3π/4 where _OP=l2 is largest, and dips to its minimum at θ=π/2 (where _OP=l). The output is therefore a non-sinusoidal but periodic waveform with period π.

I(θ)=Bω_OP/(2λ) with _OP as given.

Strategic angle. The wire's 1/r field forces an integration across the loop; the time derivative pulls λ out front.

1. Strip integration ⇒ Φ(t)∝ I(t)ln[(x₀+x)/x₀].
2. Differentiate, divide by R.

Why a logarithm appears. Field B(r)∝ 1/r integrated across a strip gives ∫ dr/r=ln r. Whenever you see a long straight wire and a parallel rectangle, expect a logarithm in the flux – it's the signature of the 1/r falloff.

Sanity check via direction. dI/dt=λ>0 means current in the wire is increasing; flux through the rectangle increases (say) into the page; by Lenz's law the induced current in the loop flows counter-clockwise (as seen from the side where flux exits) – opposing the increase. Sign of the answer depends on convention; magnitude is fixed.

Numerical check. For l=0.1 m, λ=10 A/s, x₀=0.01 m, x=0.05 m, R=1 Ω: I_loop=4π× 10^-7· 0.1· 10 2π· 1ln(6)≈ 2.0× 10^-7· 1.79 ≈ 3.6× 10^-7 A. Tiny, but non-zero.

Concept linkage. The integral ∫ B(r) l dr here is the same one used in Chapter 4 to compute the field of a Helmholtz coil and the mutual inductance between a straight wire and a parallel loop. The result M=₀ l2πln(x₀+xx₀) is a standard textbook formula.

I_loop=₀ lλ2π Rln(x₀+xx₀).

Strategic angle. ``Charge =ΔΦ/R'' bypasses the detailed time dependence completely; we only need Φ(0) and Φ(T).

1. Φ(0) from I₀, Φ(T)=0.
2. Q=ΔΦ/R gives the boxed answer.

Why this matters. The ΔΦ/R shortcut is the standard way to read total induced charge from a ballistic-galvanometer experiment.

Derivation of Q=ΔΦ/R. Q=∫ I dt=∫(ε/R)dt=∫(-dΦ/dt)· dt/R= -ΔΦ/R. The minus sign just records the convention; magnitude is |ΔΦ|/R. Remarkably, the answer depends only on the endpoints Φ(0) and Φ(T) – not on how I(t) varies in between, not on whether the source ramps linearly or oscillates wildly.

Why a galvanometer reads Q directly. A ballistic galvanometer integrates current pulses lasting much shorter than its mechanical period. Its first throw is proportional to the total charge that flowed – which equals ΔΦ/R. Historically this is how mutual inductance and unknown B fields were measured (the ``flip coil'' method).

Alternative – direct integration. ε=-dΦ/dt. With I(t)=I₀(1-t/T), dΦ/dt= ₀ L₁ln(·)2π·(-I₀/T), so |ε|=₀ I₀ L₁ln(·)2π T – constant. I_loop=ε/R, also constant. Total charge Q=I_loop· T=₀ I₀ L₁ln(·)2π R. Matches the shortcut.

Concept linkage. Q=ΔΦ/R is to electromagnetism what impulse =Δ p is to mechanics – a ``net effect'' result that bypasses moment-by-moment details. It's the basis for fluxmeters, search coils and the classical magnetic measurements of B fields before Hall probes existed.

Q=₀ I₀ L₁2π Rln(L₂+xx).

Strategic angle. Reduce to angular-impulse = angular-momentum change, using the induced tangential E-field on the ring.

1. Tangential E=Ba²/(2bΔ t) from Faraday.
2. Torque τ=QEb=QBa²/(2Δ t).
3. Δ t=mb²ω gives ω=QBa²/(2mb²).

Key insight: Δ t cancels. The torque depends on 1/Δ t (faster decay ⇒ stronger E), but the duration of the angular impulse is Δ t itself – multiplying through gives a result independent of how quickly the field is turned off. Total angular impulse is set by the total change in flux, not its rate.

Alternative – vector-potential angular-momentum. A clean way to see why ω doesn't depend on Δ t: the total ``canonical angular momentum'' (mechanical + field contributions) is conserved. Initially the field carries L_field=12QBa² of angular momentum around the ring (a standard result for a charged ring in a coaxial uniform B). Finally the field is gone, so this angular momentum must reside in the mechanical motion: mb²ω=12QBa².

Sign of rotation – Lenz's law. B points out of the paper and is decreasing. The induced E-field circulates in the direction that would drive a current to maintain the outward flux – counter-clockwise as seen from outside. A positive charge on the ring is pushed counter-clockwise; the ring spins that way.

Concept linkage. This is the basic principle of the betatron, the first accelerator for electrons: a slowly increasing flux through a circular guide tube accelerates electrons azimuthally via the induced E-field. Same physics, opposite sign.

ω=QBa²2mb².

Strategic angle. Inclined rod with vertical B – the effective ``coupling factor'' is cosθ (twice: once in ε and once when resolving the magnetic force along the slope).

1. Effective damping coefficient γ=B²d²cos²θ/R.
2. ODE mv=mgsinθ-γ v has terminal speed v_T=mgsinθ/γ and the standard exponential approach.

Where each cosθ comes from. The rod has length d in the incline plane. Its horizontal projection (the component perpendicular to vertical B) is dcosθ. The horizontal velocity is vcosθ. Area-sweep rate is dA/dt=(dcosθ)(vcosθ)? No – here the rod's horizontal extent stays dcosθ but the swept rate is dcosθ· vcosθ=d vcos²θ only if we account for the rod sweeping only its horizontal projection. The clearer way: ε=B(d v_horiz)θ=Bdvcosθ (the last cosθ accounts for the rod's tilt relative to horizontal). The force on this rod is BId horizontally, and only BIdcosθ projects along the slope. Net: one cosθ each in ε and in F_|, giving cos²θ in damping.

Limiting checks. θ=0 (horizontal track, B perpendicular): v_T→ 0 (no driving gravity). Sensible. θ=π/2 (vertical track, B parallel to track): cosθ→ 0, γ→ 0, no braking. Then v→ gt (free fall). Sensible.

Numerical check. For B=0.5 T, d=0.5 m, θ=30^∘, m=0.1 kg, R=0.1 Ω, g=10 m/s²: v_T=0.1· 10· 0.1· 0.5(0.25)(0.25)(0.75) =0.050.0469≈ 1.07 m/s. Decay constant τ=mR/(B²d²cos²θ)=0.1· 0.10.25· 0.25· 0.75 ≈ 0.21 s – a quick approach to terminal velocity.

Concept linkage. Inclined-rod problems are the workhorse of competitive-exam EM induction. The two-cosθ pattern crops up in dozens of JEE problems involving inclined rails.

v(t)=v_T(1-e^{-γ t/m}) with v_T,γ as above.

Strategic angle. A constant motional e.m.f. charging a capacitor through R is the classic RC charging response.

1. DC source = Bvd; series R, C; time constant τ=RC.
2. I(t)=(Bvd/R) e^-t/RC.

Limiting behaviour and physical interpretation. At t=0⁺, the uncharged capacitor acts like a short circuit; the full e.m.f. drops across R, giving I_max=Bvd/R. As the capacitor charges up, the voltage across C rises and the voltage across R falls, so I decays. At t→∞, the capacitor is fully charged to V_C=Bvd, no current flows – the rod is now in electrostatic equilibrium with the capacitor.

Energy budget. Total energy supplied by the rod (acting as the e.m.f. source): E_supplied=ε Q=Bvd· CBvd=CB²v²d². Energy stored in capacitor: E_C=12 CV_C²=12 CB²v²d². Energy dissipated in R: E_R=₀^∞ I²R dt =R₀^∞(Bvd)²R²e^-2t/RCdt =(Bvd)²R·RC2=12 CB²v²d². Sum: E_C+E_R=CB²v²d²=E_supplied. Half goes to the capacitor, half to resistor – the celebrated ``RC-charging factor-of-2 energy result''.

What keeps the rod moving? An external agent must supply mechanical energy at rate F_extv=BId· v=BvId, which equals the instantaneous power ε I=Bvd· I delivered to the circuit. The mechanical work done by the agent over all time equals E_supplied.

Concept linkage. This is conceptually identical to ``a battery charging a capacitor through a resistor'' (Chapter 3) – the motional e.m.f. plays the role of the battery. All familiar RC results carry over.

I(t)=(Bvd/R) e^-t/RC.

Strategic angle. Replace ``capacitor'' in 6.29 with ``inductor'' and the response flips from decaying current to rising current.

1. RL circuit with DC source Bvd, time constant τ=L/R.
2. I(t)=(Bvd/R)(1-e^-Rt/L).

Why the limits are reversed compared to 6.29. For an inductor, ``no current can change instantly'' so I(0<sup>+</sup>)=0. For a capacitor, ``no voltage can change instantly'' so the cap acts like a short and I(0⁺)=ε/R. So the RL circuit rises from zero to the steady current Bvd/R, while the RC circuit decays from Bvd/R to zero. Both reach steady state exponentially with time constant τ (different formula in each case).

Steady state interpretation. At t→∞, dI/dt=0 in the inductor, so V_L=0. The full e.m.f. drops across R: I_∞=Bvd/R. The inductor stores E_L=12 LI_∞²=L B²v²d²2R² of magnetic energy.

Power balance at general t. Source power: ε I=Bvd· I(t). Resistor dissipates: I²R. Inductor stores at rate: LI dI/dt. Balance: Bvd· I=I²R+LI dI/dt. Differentiating I(t) from the boxed answer verifies this identity at every instant.

Concept linkage. RL charging mirrors RC charging with the role of voltage and current interchanged – a deep duality that runs through all linear circuits. In an LC circuit (no resistance) you'd get sustained oscillations instead of exponential decay; with R≠ 0 you get damped sinusoids.

I(t)=(Bvd/R)(1-e^-Rt/L).

Strategic angle. Set up P_diss(v) from ε=Bvl-style reasoning, then equate to mgv for the terminal speed.

1. ε=π l²B₀λ v, I=ε/R, P_diss=I²R as above.
2. Terminal: mgv=P_diss gives v=mgR/(π²l⁴B₀²λ²).

Why this matters. A non-uniform static field is enough to brake a falling loop, because the loop sees a changing B_z as it descends.

Why B₀ alone (the uniform part) doesn't brake. Plug in λ=0: B_z=B₀ everywhere, no z-dependence. Then dΦ/dt=0 as the ring falls (uniform field, ring horizontal – no flux change). No e.m.f., no eddy current, no braking. The ring falls freely under gravity. The braking is entirely due to the gradient λ.

Sign of λ. If λ>0, field grows with z; if the ring falls in the +z direction (increasing z), Φ grows – the induced current opposes the growth, exerting an upward force. If λ<0, field falls with z; ring falling in +z sees decreasing Φ, induced current tries to maintain, still gives an upward force. Both signs give the same speed magnitude (it depends on λ²).

Force-from-energy cross-check. The magnetic force on the ring is F=I(2π l)B_r where B_r is the radial component near the ring (since ∇·B=0, B_r∼ -l2∂ B_z/∂ z=-l2B₀λ). Substituting and using I=π l²B₀λ v/R, the upward force is F_br=π²l⁴B₀²λ²v/R. Setting F_br=mg gives the same terminal velocity – independent confirmation.

Concept linkage. This problem is the radial-gradient analogue of the magnet-in-pipe problem (6.16). There the gradient is along the magnet's axis; here it's in the field profile. Either way, eddy currents create contactless braking.

v_T=mgR/(π²l⁴B₀²λ²).

Strategic angle. The small coil sees B=₀ nI(t) over its area π(b/2)²; differentiating once gives ε∝ dI/dt.

1. Mutual inductance: M=₀ n Nπ b²/4.
2. ε=-M dI/dt.
3. For I=mt²+C: dI/dt=2mt, so ε=-2Mmt – a line through the origin with negative slope -2Mm.

Why this matters. Knowing M=₀ n Nπ b²/4 lets you replace the messy two-coil calculation with a one-step ε=-M dI/dt.

Quick numerical estimate. n=1000 turns/m, N=20, b=2 cm, m=5 A/s²: M=4π10^-7· 1000· 20·π· 410^-44 ≈ 7.9 μH. At t=2 s, dI/dt=2mt=20 A/s, so ε=-7.9× 10^-6· 20=-1.6× 10^-4 V =-0.16 mV. Small but measurable.

Why we use the solenoid's field, not the small coil's own. The small coil is coaxial with the long solenoid and lies at its centre. Inside a long solenoid the field is essentially uniform at ₀ nI across the cross-section. The small coil's flux is just this B times its (smaller) area times N. The small coil's own field perturbs the solenoid only via self-inductance corrections, which are negligible for our problem.

Diagram-based reasoning. For I=mt²+C: at t=0, ε=0 (since dI/dt=0 when t=0, because I has a minimum there if m>0). As |t| grows, |ε| grows linearly. The graph is a straight line through the origin, slope -2Mm. For comparison, if I rises linearly (I=kt), ε is a constant; if I varies as t³, ε is parabolic. The general rule: take one time-derivative of the current profile to get the e.m.f. profile.

Concept linkage. This is the basic operation of a current sensor / Rogowski coil: a small pick-up coil reports dI/dt of a large nearby current. By integrating its output one recovers I(t) – the principle behind non-contact current measurements in power systems.

ε=-M dI/dt with M=₀ n Nπ b²/4.