A magnetic field $\mathbf{B} = B_0 \sin(\omega t)\,\hat{k}$ covers a large region where a wire $AB$ slides smoothly over two parallel conductors separated by a distance $d$. The wires are in the $x$-$y$ plane. The wire $AB$ (of length $d$) has resistance $R$ and the parallel wires have negligible resistance. If $AB$ is moving with velocity $v$, what is the current in the circuit? What is the force needed to keep the wire moving at constant velocity?

A magnetic field mathbfB = B₀ sin(ω t)hatk covers a large region where a wire AB slides smoothly over two para

NCERT Exemplar · Q18

\mathbf{B} = B_0 \sin(\omega t)\,\hat{k}

Kerala DhseSubjective· 5mImportance★★★★★

72% · 36/50 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Electromagnetic Induction

Electromagnetic Induction

Electromagnetic induction is the phenomenon in which a changing magnetic flux through a circuit produces an electromotive force (emf) — and hence a current, if the circuit is closed. It is the single idea behind generators, transformers, inductors, and the entire AC power grid.

The Central Discovery

Michael Faraday found (1831) that a current is induced in a coil not when a magnet sits still near it, but only while the magnet moves — that is, only while the magnetic flux linked with the coil is changing. A steady magnet, however strong, induces nothing.

Magnetic Flux

The key quantity is magnetic flux $\Phi_B$ through a surface of area $A$ in a field $\vec{B}$ :

\Phi_B = \vec{B} \cdot \vec{A} = BA\cos\theta

where $\theta$ is the angle between $\vec{B}$ and the area's normal. Its SI unit is the weber (Wb), where $1\ \text{Wb} = 1\ \text{T}\cdot\text{m}^2$ .

Flux can change in three distinct ways, and any of them induces an emf:

the field strength $B$ changes,
the area $A$ of the loop changes,
the orientation $\theta$ changes (a coil rotating in a field — the basis of the generator).

Faraday's Law

The induced emf equals the negative rate of change of flux. For a coil of $N$ turns:

\mathcal{E} = -N\,\frac{d\Phi_B}{dt}

The faster the flux changes, the larger the emf. This is why a magnet dropped quickly through a coil gives a bigger deflection than one moved slowly.

Lenz's Law — the Minus Sign

The negative sign expresses Lenz's law: the induced current flows in the direction that opposes the change producing it. Push a magnet's north pole toward a coil, and the coil's near face becomes a north pole to repel it; pull it away, and the face becomes a south pole to attract it. This is simply energy conservation — you must do work against this opposition, and that work becomes the electrical energy of the induced current.

Motional emf

A special, very useful case: a conducting rod of length $l$ moving with speed $v$ perpendicular to a field $B$ sweeps out area and develops an emf

\mathcal{E} = Blv

Here the emf arises because the free charges in the rod experience a magnetic force $q\vec{v}\times\vec{B}$ , which drives them along the rod. …

Why this formula?

Electromagnetic Induction

Electromagnetic induction is the effect discovered by Faraday: a changing magnetic flux through a circuit drives an induced EMF (and hence a current). The key word is changing — a steady field, however strong, induces nothing.

Magnetic flux

Flux measures how many field lines thread a surface bounded by the loop:

\Phi_B = \int \vec{B} \cdot d\vec{A} = BA\cos\theta

It can change three ways: by changing $B$ , by changing the area $A$ , or by rotating the loop (changing $\theta$ ).

Faraday's law

The induced EMF equals the rate of change of flux:

\mathcal{E} = -\frac{d\Phi_B}{dt}

For a coil of $N$ turns, $\mathcal{E} = -N\,\dfrac{d\Phi_B}{dt}$ . The EMF depends on how fast the flux changes, not on the flux itself — a slow change gives a small EMF, a rapid change a large one.

Lenz's law — the minus sign …

Start your 14-day free trial to unlock the full solution →

The time-varying field and the moving wire together give an emf from a single flux derivative, $\varepsilon=-B_0 d(v\sin\omega t+\omega v t\cos\omega t)$ , so $I=\dfrac{B_0 d v}{R}(\sin\omega t+\omega t\cos\omega t)$ and the force to keep $v$ constant is $F=\dfrac{B_0^2 d^2 v}{R}\sin\omega t(\sin\omega t+\omega t\cos\omega t)$ .

Flux through the growing loop

Put the wire $AB$ (length $d$ ) at position $x=vt$ on the rails, so the enclosed area is $A=d\,x=d\,vt$ . The field $\vec B=B_0\sin(\omega t)\,\hat k$ is perpendicular to the plane, giving

\Phi = B\,A = B_0\sin(\omega t)\,(d\,vt).

Both factors depend on time: the field through the product rule and the area through the wire's motion.

One emf, from one derivative

Faraday's law is $\varepsilon=-\dfrac{d\Phi}{dt}$ . Differentiating the product,

\frac{d\Phi}{dt}=B_0 d\Big(\underbrace{v\sin\omega t}_{\text{from }dA/dt}+\underbrace{\omega v t\cos\omega t}_{\text{from }dB/dt}\Big),

\varepsilon = -B_0 d\big(v\sin\omega t + \omega v t\cos\omega t\big).

Watch out

Do not compute a "transformer emf" $-d\Phi/dt$ and then add a separate motional emf $Bvd$ . The single derivative $-d\Phi/dt$ of $\Phi=B_0\sin(\omega t)\,d\,vt$ already contains the motional part (the $\,v\sin\omega t\,$ term) and the field-change part (the $\,\omega v t\cos\omega t\,$ term). Adding a motional term again double-counts it.

Current

Only $AB$ has resistance $R$ , so (taking the magnitude) …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KEAM 2026Set eng-2026-04204 marksMCQ
$10\ cm$
›Reveal solutionSolution

Average induced emf $=\dfrac{N\Delta\Phi}{\Delta t}=\dfrac{NAB}{\Delta t}=4\pi$ V.

Area. $A=\pi r^2=\pi(0.1)^2=0.01\pi\ \text{m}^2$ .

emf. …
Start your 14-day free trial to see the full solution →
KEAM 2026Set eng-2026-04214 marksMCQ
$150\,cm^2$
›Reveal solutionSolution

Flux changes from $BA$ to $0$ ; $\varepsilon=15\,$ mV.

Initially the plane is perpendicular to $B$ , so the area vector is along $B$ and flux is maximum: $\Phi_i = BA$ . Finally the plane is along the field, so $\Phi_f=0$ . …
Start your 14-day free trial to see the full solution →
KEAM 2026Set pha-2026-0419F4 marksMCQ
$Q. If an emf of 2 V is induced in a coil in 6 s, then the change of flux (in Wb) in the coil during the time is (A) 3 (B) 24 (C) 6 (D) 18 (E) 12$
›Reveal solutionSolution

$d\Phi = \varepsilon\,dt = 2\ \mathrm{V}\times 6\ \mathrm{s} = 12\ \mathrm{Wb}$ .

Faraday's law relates the induced emf to the rate of change of flux (for a single turn):
$\varepsilon = \frac{d\Phi}{dt}.$
Hence the change of flux over the interval is …
Start your 14-day free trial to see the full solution →
KEAM 2025Set eng-2025-04254 marksMCQ
$Q. The mismatched pair regarding the induced emf is (A) Eddy current : Induction furnace (B) Transformer : Laminated core (C) Induced emf : Biot - Savart law (D) Ac generator : Electromagnetic induction (E) Coaxial coils : Mutual inductance$
›Reveal solutionSolution
Induced emf comes from Faraday's law (changing flux); the Biot–Savart law gives the magnetic field of a current element. Pairing induced emf with Biot–Savart law is incorrect.

Checking the associations:

(A) Eddy currents power an induction furnace — correct.

(B) Transformers use a laminated core to cut eddy losses — correct. …
Start your 14-day free trial to see the full solution →
KEAM 2025Set eng-2025-04264 marksMCQ
$m^2$
›Reveal solutionSolution

With the coil plane perpendicular to the field, the field is fully along the area-normal, so flux is maximum: $\phi = NBA = 2$ Wb.

Magnetic flux linked with an $N$ -turn coil is $\phi = N\vec{B}\cdot\vec{A} = NBA\cos\theta$ , where $\theta$ is the angle between $\vec{B}$ and the coil's normal. …
Start your 14-day free trial to see the full solution →
KEAM 2025Set eng-2025-04294 marksMCQ
$^2$
›Reveal solutionSolution

Flux $\Phi=BA\cos\theta$ . With $\Phi=4$ Wb, $B=16$ T, $A=0.5$ m $^2$ : $\cos\theta=\dfrac{4}{8}=0.5$ , so $\theta=60^\circ$ . …

Start your 14-day free trial to see the full solution →
KEAM 2025Set pha-2025-0424F4 marksMCQ
$g$
›Reveal solutionSolution

By Lenz's law the ring's induced current opposes the falling magnet's changing flux, producing a retarding force; the net acceleration is therefore less than $g$ .

As the magnet falls through the ring, the magnetic flux through the ring changes, inducing an EMF and current in it. By Lenz's law this induced current opposes the change — it repels the approaching magnet and attracts the receding one, always retarding t …
Start your 14-day free trial to see the full solution →
KEAM 2025Set pha-2025-0429F4 marksMCQ
$\phi = 5t^2 - 25t - 150$
›Reveal solutionSolution

Faraday's law: induced emf $= -d\phi/dt$ . Differentiating $\phi = 5t^2 - 25t - 150$ and evaluating at $t=2\,\mathrm{s}$ gives $+5\,\mathrm{V}$ .

By Faraday's law,
$\varepsilon = -\frac{d\phi}{dt}.$
With $\phi = 5t^2 - 25t - 150$ ,

$\frac{d\phi}{dt} = 10t - 25.$ …
Start your 14-day free trial to see the full solution →
KEAM 2024Set eng-2024-06064 marksMCQ
$As^{-1}$
›Reveal solutionSolution

Induced emf is proportional to the rate of change of current: $\varepsilon=L\,dI/dt$ .

From the first case the self-inductance is

$L=\frac{\varepsilon}{dI/dt}=\frac{12}{30}=0.4\text{ H}.$ …
Start your 14-day free trial to see the full solution →
KEAM 2023Set eng-2023-P1-A14 marksMCQ
$Q. In a current carrying coil of inductance 60 mH, the current is changed from 2.5 A in one direction to 2.5 A in the opposite direction in 0.10 sec. The average induced EMF in the coil will be: (A) 1.2 V (B) 2.4 V (C) 3.0 V (D) 1.8 V (E) 0.6 V$
›Reveal solutionSolution
emf = L(Delta_I/Delta_t) with Delta_I = 5 A gives 0.06 x 5/0.10 = 3.0 V.

Concept and Intuition

The self-induced EMF is L times the rate of change of current. Reversing the current from +2.5 A to -2.5 A is a total change of 5 A, not 2.5 A.

Step-by-Step Solution

Change in current: Delta_I = 2.5 - (-2.5) = 5 A.

emf = L Delta_I/Delta_t = 0.060 x 5 / 0.10. …
Start your 14-day free trial to see the full solution →

←Previous questionQ17 — There are two coils A and B separated by some distance. If a current o…→Next questionQ19 — A conducting wire XY of mass m and negligible resistance can slide smo…

Concept understanding — Electromagnetic Induction

Electromagnetic Induction

The Central Discovery

Magnetic Flux

Faraday's Law

Lenz's Law — the Minus Sign

Motional emf

Why this formula?

Electromagnetic Induction

Magnetic flux

Faraday's law

Lenz's law — the minus sign …

Flux through the growing loop

One emf, from one derivative

Current

Method: Finding EMF When BOTH the Field and the Loop's Area Change With Time Together

Steps

Concept and Intuition

Step-by-Step Solution

Flux through the growing loop

One emf, from one derivative

Current

🎓Unlock everything free for 14 days

Unlock everything free for 14 days