Question

A rod of length $l$ rotates with an uniform angular velocity $\omega$ about its perpendicular bisector. A uniform magnetic field $B$ exists parallel to the axis of rotation. The potential difference between the two ends of the rod is

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

Answer 1

The correct option is A Zero


We know that motional e.m.f. for a rotating rod is $\frac{B\omega l^2}{2}$
Therefore, induced e.m.f. in AC is $\frac{B\omega (L/2{)}^2}{2}$ = $\frac{B\omega L^2}{8}$
And, induced e.m.f. in BC is $\frac{B\omega (L/2{)}^2}{2}$ = $\frac{B\omega L^2}{8}$
Also,
By symmetry in motion
$\therefore V_A-V_C=V_B-V_C$
$V_A-V_B=0$