Question

# A simple pendulum with bob of mass m and conducting wire of length L swings under gravity through an angle θ. The component of earth's magnetic field, in a direction perpendicular to swing is B. Then

A
The maximum velocity of pendulum is 2gLsin(θ2)
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B
The maximum velocity of pendulum is gLsin(θ2)
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C
The maximum potential difference induced across the pendulum is BLgLsin(θ2)
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D
The maximum potential difference induced across the pendulum is 2BLgLsin(θ2)
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Solution

## The correct option is D The maximum potential difference induced across the pendulum is 2BL√gLsin(θ2) From the figure, h=L(1−cosθ) The maximum emf will be induced when the velocity of the pendulum is maximum, i.e. when the pendulum is at the lowest position. The maximum velocity of the pendulum is, v2max=u2+2gh =02+2gL(1−cosθ) [∵u=0 ] =2gL×2sin2(θ2) ∴vmax=2√gLsin(θ2) So, option (A) is correct. Now, the maximum induced potential difference is, Emax=BvmaxL =B×2√gLsin(θ2)×L ∴ Emax=2BL√gLsin(θ2) Thus, option (D) is also correct.

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