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}}. $
BETA
