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Physics · Electromagnetic Induction

JEE Main 2024 — 27 January, Shift 1 — Question 50

Two coils have mutual inductance 0.002 H . The current changes in the first coil according to the relation i=i0sin⁡ωt\mathrm{i}=\mathrm{i}_{0} \sin \omega \mathrm{t}, where i0=5 A\mathrm{i}_{0}=5 \mathrm{~A} and ω=50π\omega=50 \pi rad/s\mathrm{rad} / \mathrm{s}. The maximum value of emf in the second coil is παV\frac{\pi}{\alpha} \mathrm{V}. The value of α\alpha is \qquad

Answer: 2

Numerical answer — enter this value.

Step-by-step solution

The induced emf in the second coil is

E=−Mdidt.\mathcal{E} = -M \frac{di}{dt}.

Given:

M=0.002 H,i=i0sin⁡(ωt),i0=5 A,ω=50π rad/s.M = 0.002\,\text{H}, \qquad i = i_0 \sin(\omega t), \qquad i_0 = 5\,\text{A}, \qquad \omega = 50\pi\,\text{rad/s}.

Differentiate the current:

didt=ddt(i0sin⁡(ωt))=i0ωcos⁡(ωt).\frac{di}{dt} = \frac{d}{dt}\left(i_0 \sin(\omega t)\right) = i_0 \omega \cos(\omega t).

Thus,

E=−Mi0ωcos⁡(ωt).\mathcal{E} = -M i_0 \omega \cos(\omega t).

Maximum emf:

Emax⁡=Mi0ω.\mathcal{E}_{\max} = M i_0 \omega.

Substitute the values:

Emax⁡=(2×10−3)(5)(50π)=0.5π V.\mathcal{E}_{\max} = (2 \times 10^{-3})(5)(50\pi) = 0.5\pi\ \text{V}.

Given:

Emax⁡=πα,\mathcal{E}_{\max} = \frac{\pi}{\alpha},

so

0.5π=πα⇒α=2.0.5\pi = \frac{\pi}{\alpha} \quad\Rightarrow\quad \alpha = 2. α=2\boxed{\alpha = 2}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
27 January, Shift 1
Subject
Physics
Chapter
Electromagnetic Induction
Topic
Self-Inductance and Mutual Inductance and Energy Density

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