A simple pendulum with bob of mass$\ m\ $and conducting wire of length$\ L$ swings under gravity through an angle 2 $ heta$. The earth’s magnetic field component in the direction perpendicular to swing is B. Maximum potential difference induced across the pendulum is [image:6_image12.emf]

Question

A simple pendulum with bob of mass$\ m\ $and conducting wire of length$\ L$ swings under gravity through an angle 2 $	heta$. The earth’s magnetic field component in the direction perpendicular to swing is B. Maximum potential difference induced across the pendulum is [image:6_image12.emf]

Dr. Abubacker · Accepted Answer

(a) $h = L - L\cos	heta$
[image:6_image12.emf]
$\Rightarrow \ \ \ h = L(1 - \cos	heta)$ ….(i)
$	herefore\ \ \ \ v^{2} = 2gh - 2g\ L(1 - \cos	heta)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \  = 2g\ L\left( 2\sin^{2}\frac{	heta}{2} ight)$
$\Rightarrow \ \ \ \ \ v = 2\sqrt{gL}\sin\frac{	heta}{2}$
Thus, maximum potential difference
$V_{\max} = BvL$
$\ \ \ \ \ \ \ \ \ \  = B 	imes 2\sqrt{gL}\sin\frac{	heta}{2}L$
= $2BL\sin{\frac{	heta}{2}(gL)^{1/2}}$

Q. 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

Explanation

Reviewed by Dr. Abubacker

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