Question 223: Electromagnetic Induction

Question

A simple pendulum with bob of mass m and conducting wire of length $L$ swings under gravity through an angle $	heta$. The component of earth's magnetic field in the direction perpendicular to swing is B . Maximum e.m.f. induced across the pendulum is ( $\mathrm{g}=$ acceleration due to gravity)

Accepted Answer

At the extreme position, the bob of the pendulum at a height ' $h$ ' from the mean position.
$$ h=L-L \cos 	heta=L(1-\cos 	heta)=2 L \sin ^2\left(\frac{	heta}{2}ight) $$
When the bob reaches the mean position, it loses potential energy and gains equal kinetic energy.
$$ \begin{array}{ll} 	herefore & \frac{1}{2} \mathrm{mv}^2=\mathrm{mgh}=\mathrm{mg} 	imes 2 \mathrm{Lsin}^2\left(\frac{	heta}{2}ight) \ 	herefore & \mathrm{v}=2 \sin \left(\frac{	heta}{2}ight) \sqrt{\mathrm{gL}} \end{array} $$
Induced emf (using Instantaneous max velocity) At lowest point, entire wire of length $L$ is moving with speed v .
So, maximum emf is:
$$ \begin{aligned} & \varepsilon_{\max }=B \cdot L \cdot v=B \cdot L \cdot 2 \sqrt{g L} \sin \left(\frac{	heta}{2}ight) \ 	herefore \quad & \varepsilon_{\max }=2 B L \sqrt{g L} \sin \left(\frac{	heta}{2}ight) \end{aligned} $$

Answer

$2 \mathrm{BL}(\sqrt{\mathrm{gL}})\left(\sin \frac{	heta}{2}ight)$

Answer

$\mathrm{BL}(\sqrt{\mathrm{gL}})\left(\sin \frac{	heta}{2}ight)$

Answer

$\mathrm{BL}(\sqrt{\mathrm{gL}})^2\left(\sin \frac{	heta}{2}ight)$

Answer

$2 \mathrm{BL}(\sqrt{\mathrm{gL}})\left(\sin ^2 \frac{	heta}{2}ight)$

Electromagnetic Induction

A simple pendulum with bob of mass m and conducting wire of length L swings under gravity through an angle \theta. The component of earth's magnetic field in the direction perpendicular to swing is B . Maximum e.m.f. induced across the pendulum is ( \mathrm{g}= acceleration due to gravity)

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