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. $3nBA\omega$
• B. $\frac{3}{2}nBA\omega$
• C. $nBA\omega$
• D. $\frac{1}{2}nBA\omega$

Answer 1

Answer: (C)

We know that flux through the coil is given by

$\varphi =n\vec{B}\cdot \vec{A}\Rightarrow \varphi =nBA\cos \theta$
$\omega =\frac{\theta }{t}\Rightarrow \theta =\omega t$...(1)
$\varphi =nBA\cos \omega t$
Therefore, emf induced in the coil is given by $\epsilon =-\frac{d}{dt}\varphi$
So, $\epsilon =-\frac{d}{dt}nBA\cos \omega t=-nBA[-\sin \omega t]\omega$
$\epsilon =nBA\omega \sin \omega t$
For $\theta =\omega t=0,\sin \omega t=0\epsilon =0$
For $\theta =\omega t=90,\sin \omega t=1{\epsilon }_0=nBA\omega \cdot 1$
$\Rightarrow \epsilon ={\epsilon }_0\sin \omega t$
Therefore ${\epsilon }_0=BAn\omega$