Question

A current $I=10sin\left(100 \pi t\right) \, A$ is passed in first coil, which induces a maximum EMF of $5\pi \, V$ in second coil. The mutual inductance between the coils is

• A. $5 \, mH$
• B. $10 \, mH$
• C. $15 \, mH$
• D. $20 \, mH$

Answer 1

Answer: (A)

Let current is $I=I_{0} \, sin \, \omega t,$
Where $I_{0} \, =10, \, \omega =100 \, \pi $
Then induced e.m.f. is given by
$\omega =-\frac{M d I}{d t}=-\frac{M d}{d t}\left(I_{0}\right)$
$\omega =-MI_{0}\omega cos \omega t$
$\therefore $ Maximum e.m.f. is
$\omega \underlinemax =MI_{0}\omega $
$5\pi \, = \, M \, \times \, 10 \, \times \, 100\pi $
$M \, = \, 5 \, mH$