A simple pendulum with a bob of mass m and a | vedclass

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(A) The vertical height $h$ reached by the bob when it makes an angle $	heta$ with the vertical is given by $h = L(1 - \cos 	heta)$.Using the conser

Vedclass · Accepted Answer

$2BL\sin \left( {\frac{	heta }{2}} ight){(gL)^{1/2}}$

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(A) The vertical height $h$ reached by the bob when it makes an angle $	heta$ with the vertical is given by $h = L(1 - \cos 	heta)$.Using the conservation of energy, the maximum velocity $v$ at the equilibrium position (lowest point) is given by $v^2 = 2gh$.Substituting $h = L(1 - \cos 	heta) = L(2 \sin^2(	heta/2))$, we get $v^2 = 2gL(2 \sin^2(	heta/2)) = 4gL \sin^2(	heta/2)$.Taking the square root, $v = 2\sqrt{gL} \sin(	heta/2)$.The motional electromotive force $(EMF)$ induced across a conductor of length $L$ moving with velocity $v$ in a magnetic field $B$ is $V = BvL$.Substituting the value of $v$, the maximum potential difference is $V_{\max} = B 	imes (2\sqrt{gL} \sin(	heta/2)) 	imes L = 2BL \sin(	heta/2) (gL)^{1/2}$.

$A$ simple pendulum with a bob of mass $m$ and a conducting wire of length $L$ swings under gravity through an angle $2\theta$. The Earth's magnetic field component in the direction perpendicular to the swing is $B$. The maximum potential difference induced across the pendulum is

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