Question

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 $I_{0}=10 A$ and $\omega=100\, \pi\, rad / s$. The maximum value of emf in the second coil is

• A. $12 \,\pi$
• B. $8 \,\pi$
• C. $5 \,\pi$
• D. $2 \,\pi$

Answer 1

Answer: (C)

The instantaneous current in the $AC$ circuit is given by $I=I_{0} \sin \omega t$
$\frac{d I}{d t}=\frac{d}{d t}\left(I_{0} \sin \omega t\right)$
$\frac{d I}{d t}=\left(I_{0} \cos \omega t\right) \omega$
Substituting $I_{0}=10 A , w =100 \pi$
$\cos \left(100 x\right)=1$ we get
$\frac{d I}{d t}=1000 \,x$
Hence, induced emf is given by
$B=M \frac{d I}{d t}$
On putting $M=0.005\, H , \frac{d I}{d t}=1000\, \pi$, we get
$B=(5 \pi) V$