A simple pendulum with bob of mas ' $m$ ' and conducting wire of length ' $L$ ' swings under gravity through…

A simple pendulum with bob of mas ' $m$ ' 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 ' $\mathrm{B}$ '. Maximum e.m.f. induced across the pendulum is ( $\mathrm{g}=$ acceleration due to gravity)
  1. $\mathrm{BL} \sin \left(\frac{\theta}{2}\right)(\mathrm{gL})$
  2. $2 \mathrm{BL} \sin \left(\frac{\theta}{2}\right)(\mathrm{gL})^2$
  3. $2 \mathrm{BL} \sin \left(\frac{\theta}{2}\right)(\mathrm{gL})^{3 / 2}$
  4. $2 \mathrm{BL} \sin \left(\frac{\theta}{2}\right)(\mathrm{gL})^{1 / 2}$

Solution

We have, $\mathrm{h}=\mathrm{L}(1-\cos \theta)$ Maximum velocity at equilibrium position is given by, $\begin{aligned} & \mathrm{v}^2=2 \mathrm{gh}=2 \mathrm{gL}(1-\cos \theta)=2 \mathrm{gL}\left(2 \sin ^2\left(\frac{\theta}{2}\right)\right) \\ & \mathrm{v}=2 \sin \left(\frac{\theta}{2}\right) \sqrt{\mathrm{gL}}\end{aligned}$ Thus Maximum potential difference, $\mathrm{V}_{\max }=\mathrm{BvL}=2 \mathrm{BL} \sin \left(\frac{\theta}{2}\right) \sqrt{\mathrm{gL}}$

Asked in: MHT CET 2022 (05 Aug Shift 2)

Practice more Oscillations questions on Aicharya