Question

A conducting rod of length 2l is rotating with constant angular speed $\omega$ about its perpendicular bisector. A uniform magnetic field $\overrightarrow{B}$ exists parallel to the axis of rotation. The e.m.f. induced between two ends of the rod is

• A. $B\omega l^2$
• B. $\frac{1}{2}B\omega l^2$
• C. $\frac{1}{8}B\omega l^2$
• D. $Zero$

Answer 1

The correct option is D $Zero$

Potential difference between
O and A is $V_0-V_A=\frac{1}{2}B{l}^2\omega$
O and B is $V_0-V_B=\frac{1}{2}B{l}^2\omega$
so $V_A-V_B=0$