Question

A rod of length 60 cm rotates with a uniform angular velocity 20 rad s^–1 about its perpendicular bisector, in a uniform magnetic field 0.5 T. The direction of magnetic field is parallel to the axis of rotation. The potential difference between the two ends of the rod is ____V.

Answer 1

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 \(\frac{B\omega(L/2)^2}{2} = \frac{B\omega L^2}{8}\)

Also,

By symmetry in motion

∴ V_A − V_C = V_B − V_C

V_A − V_B = 0