Question

A copper rod of length $L$ rotates at an angular velocity $\omega$ in a uniform magnetic field $B$ as shown. What is the induced emf across the ends?

• A. $\frac{1}{2} \omega B L^{2}$
• B. $\frac{1}{3} \omega B L^{2}$
• C. $\frac{1}{2} \omega B L^{3}$
• D. $\frac{1}{4} \omega B L^{2}$

Answer 1

Answer: (A)

Induced emf for element on the rod
$d e=B v d L \sin 90^{\circ}=B v d L$
Net emf, $e=\int d e=\int\limits_{0}^{L} B v d L=\int\limits_{0}^{L} B r \omega d L=\left(\frac{B r^{2} \omega}{2}\right)_{0}^{L}$