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

JEE Main · 2024 · 27 Jan 2024 · Shift 1 · Electromagnetic Induction

Two coils have mutual inductance 0.002 H. The current changes in the first coil according to the relation

i = i_{0} sin ω t

, where

i_{0} = 5 A

and

ω = 50 π rad / s

. The maximum value of emf in the second coil is

\frac{π}{α} V

. The value of

α

is ______.

Answer: 2

Solution

Induced emf:

ε = - M \frac{d i}{d t}

, with

i = i_{0} sin ω t

, so

\frac{d i}{d t} = i_{0} ω cos ω t

ε_{ma x} = M i_{0} ω = 0.002 \times 5 \times 50 π = 0.5 π = \frac{π}{2}

Comparing with

π / α

α = 2

JEE Main · 2024 · 27 Jan 2024 · Shift 1 · Integer answer

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