Question

A suspended bar magnet of moment $3A{m}^2$ is deflected through an angle of ${60}^0$ in a magnetic field of induction $4\times {10}^{-5}T$ and then released. If its angular velocity, when passing through its equilibrium position is $0.5radians/s$, the moment of inertia of the magnet is:

• A. $24\times {10}^{-5}kg{m}^2$
• B. $12\times {10}^{-5}kg{m}^2$
• C. $48\times {10}^{-5}kg{m}^2$
• D. $64\times {10}^{-5}kg{m}^2$

Answer 1

The correct option is C $48\times {10}^{-5}kg{m}^2$

Kinetic Energy = $\frac{1}{2}I{\omega }^2$
$m=3A{m}^2$
$\omega =\frac{1}{2}rad/s$
${\theta }_1={60}^0$
${\theta }_2=0^0$
$U_1=-Bmcos60=\frac{-Bm}{2}$
$U_2=-Bmcos0=-Bm$
$\left(\frac{1}{2}I{\omega }^2\right)=U_1-U_2$
$=\frac{Bm}{2}$
$\frac{1}{2}I\times {\left(\frac{1}{2}\right)}^2=\frac{4\times {10}^{-5}\times 3}{2}$
$I=48\times {10}^{-5}Kg{m}^2$