Apart from physically moving a coil or a magnet, state two other ways in which the magnetic flux linked with a

Exercise · Q16

Q. Apart from physically moving a coil or a magnet, state two other ways in which the magnetic flux linked with a stationary coil can be changed so as to induce an emf in it, and give one everyday or laboratory example of each.

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Flux $\Phi_B=BA\cos\theta$ depends on three quantities, and motion of a conductor through the field is only one way to change one of them ( $\theta$ , effectively). The other two are:

(1) Changing the magnetic field $B$ itself. A coil can be left completely stationary while the field passing through it varies with time -- for instance, if a nearby electromagnet's current is switched on, off, or varied continuously. Since $\Phi_B(t)=B(t)A$ changes purely through $B(t)$ , an emf $\mathcal{E}=-NA\,dB/dt$ is induced in the stationary coil. This is exactly the mechanism by which a TRANSFORMER's secondary coil has an emf induced in it, by the changing current (and hence changing field) in the primary coil, without either coil physically moving at all. …

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

Showing the 12 most recent of 15 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. In electromagnetic induction, the induced e.m.f. is independent of which of the following?$
(a) Change in flux
(b) Time
(c) Number of turns
(d) Resistance of the coil
›Reveal solutionSolution

By Faraday's law, $\varepsilon=-N\dfrac{d\Phi}{dt}$ — resistance appears only later, when you use $\varepsilon$ to find the induced current, not in the e.m.f. itself.

Faraday's law of electromagnetic induction states that the induced e.m.f. in a coil of $N$ turns is $\varepsilon=-N\dfrac{d\Phi}{dt}$ , where $\Phi$ is the magnetic flux linked with one turn. This expression explicitly contains the change in flux, the time over which it changes (through the derivative), and the number of turns $N$ — so the e.m.f. genuinely depends on all three. The resistance $R$ of the coil only comes into …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The flux on a coil of 50 turns changes from 0.3 Wb to 0.5 Wb in 8 seconds. The induced e.m.f. in the coil is$
(a) 10 V
(b) 0.6 V
(c) $-12$ V
(d) $-1.25$ V
›Reveal solutionSolution
Faraday's law of electromagnetic induction, $\varepsilon=-N\,d\Phi/dt$ , directly gives the induced EMF from the given rate of change of flux linkage.

Given data

Number of turns, $N=50$

Initial flux, $\Phi_1 = 0.3$ Wb

Final flux, $\Phi_2 = 0.5$ Wb

Time interval, $\Delta t = 8$ s

Faraday's law

$\varepsilon = -N\,\frac{\Delta\Phi}{\Delta t}$

Substituting

$\Delta\Phi = \Phi_2-\Phi_1 = 0.5 - 0.3 = 0.2\text{ Wb}$

…
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CBSE 2026Set SEM31 markMCQ
$Q. The expression for the magnetic flux in a circuit having resistance 7 Ω is φ = 6t² - 5t + 4. In time t = 1 second, the induced current will be$
(a) 1·2 A
(b) 0·8 A
(c) 0·5 A
(d) 1 A
›Reveal solutionSolution

Induced emf is the time-rate of change of flux: ε = −dφ/dt = −(12t − 5). At t = 1 s, |ε| = 7 V, and I = ε/R = 7/7 = 1 A. Option (d).

Step 1 — Faraday's law (NCERT/CBSE Class 12 Physics, Electromagnetic Induction): ε = −dφ/dt.

Step 2 — differentiate φ = 6t² − 5t + 4: …

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CBSE 2025Set 55/5/11 markMCQ
$t$
›Reveal solutionSolution
The induced emf is the negative rate of change of flux, $e = -\frac{d\phi}{dt}$ . Differentiating $\phi = 8t^2 + 5t + 7$ gives $e = -(16t + 5)$ . At $t = 4$ s, this evaluates to $e = -69$ V, so the magnitude is 69 V, which corresponds to option (D).

The core idea here is Faraday’s law of electromagnetic induction: whenever the magnetic flux linked with a coil changes with time, an emf is induced in the coil. The law states that the induced emf is equal to the negative rate of change of flux. The negative sign (Lenz’s law) tells us the direction of the induced emf — it opposes the change causing it. But for magnitude questions like this one, we often just take the absolute value.

The flux is given as a simple polynomial in time: $\phi = 8t^2 + 5t + 7$ . Notice that the constant term $7$ Wb represents a steady flux that doesn’t change with time — it contributes nothing to the induced emf. Only the terms that depend on $t$ matter.

Let’s work through it step by step.

Write down Faraday’s law The instantaneous induced emf $e$ is given by:

$e = -\frac{d\phi}{dt}$
Here, $\phi$ is in weber (Wb) and $t$ in seconds, so $e$ comes out in volts (V).

Differentiate the flux expression

$\phi = 8t^2 + 5t + 7$
Differentiate term by term:

Derivative of $8t^2$ is $16t$

Derivative of $5t$ is $5$

Derivative of $7$ (constant) is $0$ So:

$\frac{d\phi}{dt} = 16t + 5$

Apply the negative sign

$e = -(16t + 5)$

Substitute $t = 4$ s

$e = -(16 \times 4 + 5) = -(64 + 5) = -69 \text{ V}$

Interpret the result …
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CBSE 2025Set ANNUAL1 mark
$Q. The magnitude of the induced e.m.f. is equal to the time rate of change of ____________ through the circuit.$
›Reveal solutionSolution

This is Faraday's law of electromagnetic induction: the induced EMF's magnitude equals the rate at which magnetic flux through the circuit changes.

Faraday's law of electromagnetic induction states:

$\varepsilon = -\dfrac{d\Phi_B}{dt}$

…

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CBSE 2025Set ANNUAL1 markMCQ
$\phi = 5t^2 + 3t + 16$
(a) 6 volt
(b) 10 volt
(c) 53 volt
(d) 43 volt.
›Reveal solutionSolution

Induced emf $= -\dfrac{d\phi}{dt}$ ; differentiate the given flux and substitute t = 5 s.

Given $\phi = 5t^2 + 3t + 16$ (Weber).

$e = -\frac{d\phi}{dt} = -(10t + 3)$

At $t = 5\,s$ : $e = -(10\times5 + 3) = -53\,V$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. A magnetic field of flux density 1.0 Wbm⁻² acts normal to a 80 turn coil of 0.01m² area. If this coil is removed from the field in 0.1s. The emf induced is$
(a) 6V
(b) 7V
(c) 8V
(d) 9V.
›Reveal solutionSolution

Faraday's law gives emf = N × (rate of change of flux); plugging in the numbers gives 8 V.

Initial flux through one turn: $\Phi_i = BA = 1.0 \times 0.01 = 0.01\ \text{Wb}$

Final flux (coil removed from field): $\Phi_f = 0$

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. In electromagnetic induction, the induced e.m.f is independent of -$
(a) change of flux
(b) time
(c) number of turns of the coil
(d) resistance of the coil
›Reveal solutionSolution

Faraday's law gives emf purely from the rate of change of flux — resistance plays no part in it.

By Faraday's law of electromagnetic induction, the induced emf is $\varepsilon = -N\dfrac{d\Phi}{dt}$ , where $N$ is the number of turns and $\dfrac{d\Phi}{dt}$ is the rate of change of magnetic flux through one turn. This expression depends on the change of flux, the time over which it changes, and the number of turns — but NOT on the re …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. When a magnet is taken towards a coil, the induced emf depends on$
(a) the number of turns in the coil
(b) the speed of the magnet
(c) the resistance of the coil
(d) magnetic moment of the magnet.
›Reveal solutionSolution

By Faraday's law, induced emf = rate of change of flux linkage; for a given coil and magnet, that rate is controlled by the magnet's speed.

Faraday's law of electromagnetic induction states

$\varepsilon = -N\dfrac{d\phi}{dt}$

where $N$ is the number of turns and $d\phi/dt$ is the rate of change of magnetic flux through the coil.

When a magnet is moved towards a coil, the flux linked with the coil changes because the magnet's distance from the coil changes with time. The FASTER the magnet moves, the more rapidly the flux changes, and hence the larger the induced emf — this is the direct meaning of $d\phi/dt$ .

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. In AC generator, if the rotational velocity of armature is doubled, the induced emf will :$
(a) become half
(b) become double
(c) become zero
(d) remains unchanged
›Reveal solutionSolution

Peak emf of an AC generator is directly proportional to the angular (rotational) velocity of the armature.

For a coil of N turns, area A, rotating with angular velocity $\omega$ in a uniform field B, the instantaneous emf is $\varepsilon = NBA\omega \sin(\omega t)$ , so the peak emf is $\varepsilon_0 = NBA\omega$ . Since $\varepsilon_0 \propto \omega$ , doubling the r …

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CBSE 2023Set ANNUAL1 mark
$Q. The magnetic flux passing through a ring is increased from Q1 to Q2 at a constant rate in time t. The value of Induced Electromotive force will be ............. .$
›Reveal solutionSolution

By Faraday's law, the induced emf equals the rate of change of magnetic flux.

Faraday's law of electromagnetic induction states

$\varepsilon = -\frac{d\phi}{dt}$ …

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CBSE 2023Set ANNUAL1 mark
$Q. A wire cuts across a flux of 0.2 \times 10⁻² weber in 0.12 second. What is the emf induced in the wire?$
›Reveal solutionSolution

By Faraday's law, $\varepsilon = \dfrac{\Delta\Phi}{\Delta t}$ ; substituting the given values gives about $1.67\times10^{-2}$ V.

Change in magnetic flux, $\Delta\Phi = 0.2\times10^{-2}\ \text{Wb} = 2\times10^{-3}\ \text{Wb}$ .

Time taken, $\Delta t = 0.12\ \text{s}$ .

By Faraday's law of electromagnetic induction, the magnitude of induced emf is: …

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Q.Apart from physically moving a coil or a magnet, state two other ways in which the magnetic flux linked with a stationary coil can be changed so as to induce an emf in it, and give one everyday or laboratory example of each.

Concept understanding — Faraday's Law of Electromagnetic Induction

Q.Apart from physically moving a coil or a magnet, state two other ways in which the magnetic flux linked with a stationary coil can be changed so as to induce an emf in it, and give one everyday or laboratory example of each.

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