Q131 / 170 · Manipal (MET) 2015 · Physics
Two coils have mutual inductance 0.005 H . The current changes in the first coil according to equation $I=I_0 \sin \omega t$, where $I_0=10 \mathrm{~A}$ and $\omega=100 \pi \mathrm{rad} / \mathrm{s}$. The maximum value of emf in the second coil is
- A$12 \pi$
- B$8 \pi$
- Correct answerC$5 \pi$
- D$2 \pi$
Solution
The instantaneous current in the AC circuit is given by $I=I_0 \sin \omega t$
$\begin{aligned}
& \frac{d I}{d t}=\frac{d}{d t}\left(I_0 \sin \omega t\right) \\
& \frac{d I}{d t}=\left(I_0 \cos \omega t\right) \omega
\end{aligned}$
Substituting $\quad I_0=10 \mathrm{~A}, \omega=100 \pi$
$\begin{aligned}
\cos (100 \pi) & =1, \text { we get } \\
\frac{d I}{d t} & =1000 \pi
\end{aligned}$
Hence, induced emf is given by $E=M \frac{d I}{d t}$
On putting $M=0.005 \mathrm{H}, \frac{d I}{d t}=1000 \pi$, we get $E=(5 \pi) \mathrm{V}$
Hence, induced emf is given by $E=M \frac{d I}{d t}$
On putting $M=0.005 \mathrm{H}, \frac{d I}{d t}=1000 \pi$, we get $E=(5 \pi) \mathrm{V}$
Hence, the correct answer is option: 'C'.