Two coils have a mutual inductance 0.005 H . The current changes in the first coil according to equation I=I_0 sin ω t , where I_0=10 ~A and ω= 100 π radian / sec . The maximum value of e.m.f. in the second coil is

Correct answer: (B) 5 π. e =M di/dt=0.005 × d/dt (I_0 sin ω t ) =0.0005 × I_0 ω cos ω t ∴ e_max =0.005 × 10 × 100 π=5 π [∵ cos ω t=1]

Q91 / 120 · VITEEE 2018 · Physics

Two coils have a mutual inductance $0.005 \mathrm{H}$. The current changes in the first coil according to equation $\mathrm{I}=\mathrm{I}_{0} \sin \omega \mathrm{t}$, where $\mathrm{I}_{0}=10 \mathrm{~A}$ and $\omega=$ $100 \pi$ radian $/ \mathrm{sec}$. The maximum value of e.m.f. in the second coil is

A
$2 \pi$
Correct answer
B
$5 \pi$
C
$\pi$
D
$4 \pi$

Solution

$$ \begin{array}{l} \begin{aligned} \mathrm{e} &=\mathrm{M} \frac{\mathrm{di}}{\mathrm{dt}}=0.005 \times \frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{I}_{0} \sin \omega \mathrm{t}\right) \\ &=0.0005 \times \mathrm{I}_{0} \omega \cos \omega \mathrm{t} \\ \therefore \mathrm{e}_{\max } &=0.005 \times 10 \times 100 \pi=5 \pi \quad[\because \cos \omega \mathrm{t}=1] \end{aligned} \end{array} $$

Hence, the correct answer is option: 'B'.

Solution

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