Question

A copper rod AB of length L is pivoted at end A and rotates at constant angular velocity $\omega$ at right angles to a uniform magnetic field of induction B. The e.m.f developed between the mid point of C of the rod and nd B is

• A. $\frac{B\omega L^2}{4}$
• B. $\frac{B\omega L^2}{2}$
• C. $\frac{3B\omega L^2}{4}$
• D. $\frac{B\omega L^2}{8}$

Answer 1

The correct option is D $\frac{B\omega L^2}{8}$

EMF induced in the element is
d$ϵ$ = Bdx v
= Bdx (xw)
P d by centre and the end point is,
$dV={\int }_{\frac{l}{2}}^{l}Bxwdx=Bw{\frac{x^2}{2}|}_{\frac{l}{2}}^l=\frac{Bw}{2}[l^2-\frac{l^2}{4}]=\frac{3Bw{l}^2}{8}$