Question

Find the angular acceleration of the loop when it has rotated through an angle $\varphi$ from its original orientation and is rotating downward at an angular speed $\omega$.

Answer 1

We need to find Net torque that act on the loop that has rotated through an angle $\varphi$ with speed $\omega$.

let current Induced in the loop is $I$.

$I=\frac{B{L}^2\omega }{R}$

each side of the loop has mass $\frac{m}{4}$

the flux in the loop=$BA\cos \varphi$

Torque due to gravity is,

$T_g$=$\sum r\times mg$

$T_g$=$\frac{L}{2}\frac{M}{4}g\sin (90-\varphi )+\frac{L}{2}\frac{M}{4}g\sin (90-\varphi )+\frac{L}{2}\frac{M}{4}g\sin (90-\varphi )$

$T_g$=$\frac{mgL}{2}\cos \varphi$ in clockwise direction

Torque due to magnetic feild.

$T_B$=$|T\times B|=I\times A\times B\times \sin \varphi$

=$\frac{B^2{L}^2\omega }{R}.L^2\times {\sin }^2\varphi$

=$\frac{B^2{L}^4\omega }{R}{\sin }^2\varphi$ in anti clockwise.

total torque=$\frac{mgL}{2}\cos \varphi -\frac{B^2{L}^4\omega }{R}{\sin }^2\varphi$ ($\alpha$ is angular accelaration)

$I\alpha =\frac{mgL}{2}\cos \varphi -\frac{B^2{L}^4\omega }{R}{\sin }^2\varphi$

$\alpha$=$\frac{\frac{mgL}{2}\cos \varphi -\frac{B^2{L}^4\omega }{R}{\sin }^2\varphi }{\frac{5m{L}^2}{12}}$

$\alpha$=$\frac{12}{5m}\left[\frac{mg}{2L}\cos \varphi -\frac{B^2{L}^4\omega }{R}{\sin }^2\varphi \right]$