Question

A rod of length L is rotating with an angular velocity $\omega$ in a uniform magnetic field $B_0$ as shown in figure. Potential difference between the ends of rod A and B is
(Take $OB=\frac{L}{4}$)

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

Answer 1

The correct option is D $\frac{B\omega L^2}{4}$

$V_{AB}=V_{AO}+V_{OB}=V_{AO}-V_{BO}{V}_{AO}=\frac{b\omega l^2}{2}=\frac{B\omega }{2}(\frac{3L}{4}{)}^2=\frac{9B\omega L^2}{32}$
$V_{BO}=\frac{Bw}{2}(\frac{L}{4}{)}^2=\frac{B\omega L^2}{32}$
$\Rightarrow V_{AB}=V_{AO}-V_{BO}=\frac{B\omega L^2}{4}$