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A conducting rod of length $\ell$ rotates about one end with angular velocity $\omega$ in a uniform magnetic field $B$ perpendicular to the plane of rotation. The induced EMF between the two ends of the rod is:
A$B\omega \ell$
B$\tfrac{1}{2} B\omega^2 \ell$
C$B\omega \ell^2$
D$\tfrac{1}{2} B\omega \ell^2$
Answer & Solution
Correct answer: D. $\tfrac{1}{2} B\omega \ell^2$
Each element $dr$ at distance $r$ from the pivot moves with velocity $\omega r$ and contributes $dE = B(\omega r)\,dr$. Integrating from 0 to $\ell$: $e = \int_0^\ell B\omega r\,dr = \tfrac{1}{2}B\omega \ell^2$.
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