Q.The magnetic flux through each turn of a coil of 100 turns changes from $2\times10^{-3}\ \text{Wb}$ to $8\times10^{-3}\ \text{Wb}$ in $0.2\ \text{s}$. Calculate the magnitude of the emf induced in the coil.

Showing the 12 most recent of 15 on this concept.

1. 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 play afterward, through Ohm's law $I=\varepsilon/R$, when you calculate the induced current; the e.m.f. itself, being a cause, is set purely by the changing flux linkage and does not know about the coil's resistance.

   ✓Final answer

   (d) Resistance of the coil

2. 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

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

   $\varepsilon = -50\times\frac{0.2}{8} = -50\times 0.025 = -1.25\text{ V}$

   The negative sign (Lenz's law) indicates that the induced EMF opposes the increase in flux; its magnitude is $1.25$ V.

   ✓Final answer

   $\varepsilon = -1.25\text{ V}$ (option (d))

3. 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:

   dφ/dt = 12t − 5, so ε = −(12t − 5).

   Step 3 — at t = 1 s: |ε| = |12(1) − 5| = 7 V.

   Step 4 — induced current: I = |ε|/R = 7/7 = 1 A.

   ✓Final answer

   (d) 1 A

4. CBSE 2025Set 55/5/11 markMCQ
   Q.The magnetic flux linked with a coil changes with time $t$ as $\phi=(8t^{2}+5t+7)$, where $t$ is in seconds and $\phi$ is in Wb. The value of the emf induced in the coil at $t=4$ s is: (A) 32 V (B) 37 V (C) 64 V (D) 69 V

   ›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.

   1. 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).

   2. 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:

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

   3. Apply the negative sign

   $e = -(16t + 5)$

   4. Substitute $t = 4$ s

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

   5. Interpret the result The magnitude of the induced emf is $69$ V. The negative sign indicates the direction (opposing the increase in flux), but the question asks for “the value of the emf induced” — in multiple-choice context, they usually mean the magnitude. So the answer is $69$ V.

   Watch out

   A common mistake is to forget the constant term’s derivative is zero, or to drop the negative sign and then get confused about the sign of the answer. Also, some students mistakenly integrate instead of differentiate — remember, emf is the rate of change, not the total change.

   Tip

   For a polynomial flux $\phi = at^2 + bt + c$, the induced emf is always $e = -(2at + b)$. You can directly plug the coefficients without re-differentiating each time.

   ✓Final answer

   The induced emf at $t = 4$ s is 69 V, which corresponds to option (D).

5. 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:

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

   where $\Phi_B$ is the magnetic flux linked with the circuit. The magnitude of the induced EMF is the time rate of change of magnetic flux (the negative sign, from Lenz's law, only indicates the direction that opposes the change).

   ✓Final answer

   The magnitude of the induced e.m.f. is equal to the time rate of change of magnetic flux through the circuit.

6. CBSE 2025Set ANNUAL1 markMCQ
   Q.The magnetic flux linked with a Coil is given by $\phi = 5t^2 + 3t + 16$, where $\phi$ is in Weber and t is in seconds. The induced e.m.f. in the Coil at t = 5s will be:

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

   The magnitude of the induced emf is $53\,V$ (the negative sign, from Lenz's law, only indicates the direction opposing the change in flux).

   ✓Final answer

   53 volt — option (c).

7. 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): $\Phi_f = 0$

   Induced emf magnitude: $|\varepsilon| = N\frac{\Delta\Phi}{\Delta t} = 80 \times \frac{0.01 - 0}{0.1} = 80 \times 0.1 = 8\ \text{V}$

   ✓Final answer

   The correct option is (c) 8V.

8. 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 resistance of the coil. Resistance only affects the induced CURRENT ($I = \varepsilon/R$), not the emf itself.

   ✓Final answer

   Induced emf is independent of the resistance of the coil (option d).

9. 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 $d\phi/dt$.

   While the number of turns $N$ and the magnet's own moment do affect the total flux/emf magnitude in general, the question asks specifically what the induced emf "depends on" in this dynamic situation of relative motion — the defining, variable factor here is the rate of that motion, i.e. the speed of the magnet. (Note: resistance of the coil affects the induced CURRENT, not the induced EMF itself.)

   ✓Final answer

   The induced emf depends on the speed of the magnet. Choice (b).

10. 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 rotational velocity doubles the peak (and rms) induced emf. (Note the frequency of the emf also doubles.)

    ✓Final answer

    Option (b): the induced emf becomes double.

11. 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

    $\varepsilon = -\frac{d\phi}{dt}$

    Since the flux changes at a constant rate from $Q_1$ to $Q_2$ in time $t$, the rate of change is $\dfrac{Q_2-Q_1}{t}$. Taking the magnitude,

    $|\varepsilon| = \frac{Q_2 - Q_1}{t}$

    ✓Final answer

    Induced emf $= \dfrac{Q_2 - Q_1}{t}$.

12. CBSE 2023Set ANNUAL1 mark
    Q.A wire cuts across a flux of 0.2 × 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:

    $\varepsilon = \left|\frac{\Delta\Phi}{\Delta t}\right| = \frac{2\times10^{-3}}{0.12} = 1.667\times10^{-2}\ \text{V}$

    ✓Final answer

    The induced emf is $\varepsilon \approx 1.67\times10^{-2}$ V $\approx 16.7$ mV.