Question

A copper rod of length $l$ rotates about its end with angular velocity $\omega$ in a uniform magnetic field B. The emf developed between the ends of the rod if the field is normal to the plane of rotation is:

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

Answer 1

Answer: (B)

$\frac{1}{2}B\omega l^2$


$dE=B\omega xdxE={\int }_0^{L}B\omega xdx=B\omega \frac{x^2}{2}=\frac{1}{2}B\omega l^2$