Question

A 100 turns closely wound circular coil of radius 10 cm carries a current of 3.2 A. A uniform horizontal magnetic field of strength 2T is turned on. The axis of the coil is initially in the direction of the field. The coil rotates through an angle of 90${}^{\circ }$under the influence of the magnetic field. What is the angular speed acquired by the coil when it has rotated by ${90}^o$?

Answer 1

Given $n=100,i=3.2A,r=10cm=0.1m$

as $\tau =I\alpha$

where I=moment of inertia

and $\alpha$ = angular acceleration

but $\tau =MBsin\theta$

$\therefore I\alpha =MBsin\theta$

$Iw\frac{dw}{d\theta }=MBsin\theta$

$I{\int }_0^{w}wdw=MB{\int }_{0^{\circ }}^{{90}^{\circ }}sin\theta d\theta$

$I\frac{w^2}{2}=MB$

$w=\sqrt{\frac{2MB}{I}}$

M=magnetic moment of coil

$=niA=100\ast 3.2\ast 3.14\ast (0.1{)}^2$

$=10A{m}^2$

$I=0.1kg{m}^2$ (Given)

$\therefore w=\sqrt{\frac{2\ast 10\ast 2}{0.1}}=20rad/sec$