Question

A simple pendulum with bob of mass $m$ and conducting wire of length $L$ swings under gravity through an angle $2\theta$. The earth's magnetic field component in the direction perpendicular to swing is $B$. The maximum potential difference induced across the pendulum is

• A. $2BL\sin \left(\frac{\theta }{2}\right)\cdot \sqrt{gL}$
• B. $BL\sin \left(\frac{\theta }{2}\right)\sqrt{(gl)}$
• C. $BL\sin \left(\frac{\theta }{2}\right)\cdot (gL{)}^{3/2}$
• D. $BL\sin \left(\frac{\theta }{2}\right)\cdot (gL{)}^2$

Answer 1

Answer: (B)

Using conservation of energy
$mgl(1-\cos \theta )=\frac{1}{2}l{\omega }^2$
$mgl(1-\cos \theta )=\frac{1}{2}m{l}^2{\omega }^2$
$2gl\left(2{\sin }^2\left(\frac{\theta }{2}\right)\right)=l^2{\omega }^2$
$\omega =2\sqrt{\frac{g}{l}}\sin \frac{\theta }{2}$
$\epsilon =\frac{1}{2}B\omega l^2=\frac{1}{2}\times \left(2\sqrt{\frac{g}{I}}\sin \frac{\theta }{2}\right)I^2\times B$
$=Bl\sin \frac{\theta }{2}\sqrt{gl}$