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 2θ . 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) 2BLsin(θ2)(gL)1/2

B) BLsin(θ2)(gL)

C) BLsin(θ2)(gL)3/2

D) BLsin(θ2)(gL)2

Show Answer

Answer:

Correct Answer: A

Solution:

Þ h=L(1cosθ) …(i)

Maximum velocity at equilibrium is given by

v2=2gh=2g,L(1cosθ) =2g L(2sin2θ2)

Þ v=2gLsinθ2

Thus, max. potential difference Vmax=BvL =B×2gLsinθ2L =2BLsinθ2(gL)1/2.



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