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A conducting rod of length $L$ rotates with angular speed $\omega$ in a uniform magnetic field of induction $B$ which is perpendicular to its motion. The induced emf developed between the two ends of the rod is
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The correct answer is:
$\frac{B L^2 \omega}{2}$
Linear velocity of the rod $v=r \omega=\frac{L \omega}{2}$
$\therefore \quad$ Induced emf
$$
e=B v L=B \times \frac{L \omega}{2} L=\frac{1}{2} B \omega L^2
$$
$\therefore \quad$ Induced emf
$$
e=B v L=B \times \frac{L \omega}{2} L=\frac{1}{2} B \omega L^2
$$
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