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

Answer 1

The correct option is A $2BL\sin \left(\frac{\theta }{2}\right)\sqrt{gL}$

We have $h=L(1-Cos\theta )$

Maximum velocity at equilibrium is given by,

$v^2=2gh=2gL(1-Cos\theta )$

$=2gL\left(2Si{n}^2\right(\frac{\theta }{2}\left)\right)$

$v=2\sqrt{gL}sin\left(\frac{\theta }{2}\right)$

Thus Maximum potential difference,

$V_{max}=BvL$

$=B\times 2\sqrt{gL}sin\left(\frac{\theta }{2}\right)L$

$=2BLsin\left(\frac{\theta }{2}\right)\sqrt{gL}$