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

Answer 1

Answer: (B)

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