Electro Magnetic Induction And Alternating Currents Question 351

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. Maximum potential difference induced across the pendulum is [MP PET 2005]

Options:

A) $ 2BL\sin ( \frac{\theta }{2} ){{(gL)}^{1/2}} $

B) $ BL\sin ( \frac{\theta }{2} )(gL) $

C) $ BL\sin ( \frac{\theta }{2} ){{(gL)}^{3/2}} $

D) $ BL\sin ( \frac{\theta }{2} ){{(gL)}^{2}} $

Show Answer

Answer:

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( 2{{\sin }^{2}}\frac{\theta }{2} ) $

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}}. $



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