Question

A coil having magnetic moment 15 A$m^2$ placed in a uniform magnetic field of 4 T in the horizontal direction exists such that initially, the axis of coil is in the direction of the field. If the coil is rotated by ${45}^o$ and the moment of inertia of the coil is 0.5 kg $m^2$ then the angular speed acquired by the coil is

• A. 20$rad{s}^{-1}$
• B. 10$rad{s}^{-1}$
• C. 8.34$rad{s}^{-1}$
• D. 4.5$rad{s}^{-1}$

Answer 1

The correct option is C 8.34$rad{s}^{-1}$

Since $\tau$=$I\frac{d\omega }{dt}=mB\sin \Theta$____(1)
Now, $\frac{d\omega }{dt}$ = $\frac{d\omega }{d\Theta }\frac{d\Theta }{dt}$ =$\omega \frac{d\omega }{d\Theta }$___(2)

then,from (1) and (2),

$I\omega \frac{d\omega }{d\Theta }$ = $mB\sin \Theta$;; so $I\omega d\omega$ = $mBsin\Theta d\Theta$

Integrating both sides
$I{\int }_0^{\omega }\omega d\omega$ = $mB{\int }_0^{{45}^0}\sin \Theta d\Theta$

$I$ ${\left[\frac{w^2}{2}\right]}_0^{wf}$ = $mB[-\cos \Theta {]}_{0^0}^{{45}^0}$

$\frac{1}{2}I{w}_f^2=-mB[\cos {45}^0-\cos 0^0]$

$=-mB[\frac{1}{\sqrt{2}}-1]=0.29mB$

$w_f^2=\frac{2\times 0.29mB}{I}=\frac{2\times 0.29\times 15\times 4}{0.50}$

$w^2=69.6;w_f=\sqrt{6}9.6$ = $8.34rad$$s^{-1}$