question_answer

A simple pendulum with bob of mass m and conducting wire of length L swings under gravity through an angle\[2\theta \]. The earths magnetic field component in the direction perpendicular to swing is B. Maximum potential difference induced across the pendulum is:

A)  \[2BL\sin \left( \frac{\theta }{2} \right){{(gL)}^{1/2}}\]

B)  \[BL\sin \left( \frac{\theta }{2} \right)(gL)\]

C)  \[BL\sin \left( \frac{\theta }{2} \right){{(gL)}^{3/2}}\]

D)  \[BL\sin \left( \frac{\theta }{2} \right){{(gL)}^{2}}\]

E)  \[BL\sin \left( \frac{\theta }{2} \right){{(gL)}^{5/2}}\]

Correct Answer: A

Solution :

\[h=L-L\cos \theta \] \[\Rightarrow \]               \[h=L(1-\cos \theta )\]                ...(1) \[\therefore \]  \[{{v}^{2}}=2gh=2gL(1-\cos \theta )\]                 \[=2g\,L\left( 2{{\sin }^{2}}\frac{\theta }{2} \right)\] \[\Rightarrow \]               \[v=2\sqrt{gL}\sin \frac{\theta }{2}\] Thus, maximum potential difference                 \[{{V}_{\max }}=BvL\]                 \[=B\times 2\sqrt{gL}\sin \frac{\theta }{2}L\]                 \[=2BL\sin \frac{\theta }{L}{{(gL)}^{1/2}}\]