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}

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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 …

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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) …

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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

Flux through the growing loop

One emf, from one derivative

Current

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