Question

A coil of cross-sectional area A having n turns is placed in a uniform magnetic field B. When it is rotated with an angular velocity $\omega$, the maximum e.m.f. induced in the coil will be :

• A. $3 \, nB A \omega$
• B. $\frac{3}{2} \, nB A \omega$
• C. $\, nB A \omega$
• D. $\frac{1}{2} \, nB A \omega$

Answer 1

Answer: (C)

$\, nB A \omega$

Answer 2

We know that flux through the coil is given by

$\phi=n \vec{B} \cdot \vec{A} \Rightarrow \phi=n B A \cos \theta$
$\omega=\frac{\theta}{t} \Rightarrow \theta=\omega t$...(1)
$\phi=n B A \cos \omega t$
Therefore, emf induced in the coil is given by $\varepsilon=-\frac{d}{d t} \phi$
So, $\varepsilon=-\frac{d}{d t} n B A \cos \omega t=-n B A[-\sin \omega t] \omega$
$\varepsilon=n B A \omega \sin \omega t$
For $\theta=\omega t=0, \sin \omega t=0\,\, \varepsilon=0$
For $\theta=\omega t=90, \sin \omega t=1\,\, \varepsilon_{0}=n B A \omega \cdot 1$
$\Rightarrow \varepsilon=\varepsilon_{0} \sin \omega t$
Therefore $\varepsilon_{0}=B A n \omega$