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

The Correct Option is C

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 \(\varepsilon_{0}=B A n \omega\)