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 earth?s magnetic field component in the direction perpendicular to swing is B. Maximum potential difference induced across the pendulum is                           [MP PET 2005]

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

Correct Answer: A

Solution :

                                                     Þ \[h=L(1-\cos \theta )\]                         ??.(i)                    Maximum velocity at equilibrium is given by                    \ \[{{v}^{2}}=2gh=2g\,L(1-\cos \theta )\]\[=2g\ L\left( 2{{\sin }^{2}}\frac{\theta }{2} \right)\]                    Þ \[v=2\sqrt{gL}\sin \frac{\theta }{2}\]                    Thus, max. potential difference                    \[{{V}_{\max }}=BvL\]\[=B\times 2\sqrt{gL}\sin \frac{\theta }{2}L\]\[=2BL\sin \frac{\theta }{2}{{(gL)}^{1/2}}.\]