Question

A simple pendulum with bob of mass $m$ and conducting wire of length $L$ swings under gravity through an angle $\theta$. The earth's magnetic field component in the direction perpendicular to swing is $B$. Maximum potential difference induced across the pendulum is

• A. $2 B L \sin \frac{\theta}{2}(g L)^{1 / 2}$
• B. $B L \sin \left(\frac{\theta}{2}\right)(g L)$
• C. $B L \sin \left(\frac{\theta}{2}\right)(g L)^{3 / 2}$
• D. $B L \sin \left(\frac{\theta}{2}\right)(g L)^{2}$

Answer 1

Answer: (A)

$\because h=L-L \cos \theta$
$\Rightarrow h=L(1-\cos \theta)$
$\therefore v^{2}=2 g h=2 g L(1-\cos \theta)$
$=2 g L\left(2 \sin ^{2} \theta / 2\right)$
$\Rightarrow \quad v=2 \sqrt{g L} \sin \theta / 2$
Thus, maximum potential difference
$ V_{\max } =B v L=B \times 2 \sqrt{g L}(\sin \theta / 2) L $
$=2 B L(\sin \theta / 2)(g L)^{1 / 2} $