Question

A circular coil of mass $m$, radius $R$ and $N$ turns carries current $I$. Plane of the coil is perpendicular to the magnetic meridian. The coil is suspended by a thread of negligible torsion at a place where angle of dip is ${37}^{\circ }$ and earth's magnetic field is $B$. If the coil is rotated by a small angle with thread as the axis and released, it starts oscillating. Time period of such oscillations is

• A. $2\pi \sqrt{\frac{m}{3N\pi IB}}$
• B. $2\pi \sqrt{\frac{3m}{8N\pi IB}}$
• C. $2\pi \sqrt{\frac{2m}{NIB}}$
• D. $2\pi \sqrt{\frac{5m}{8N\pi IB}}$

Answer 1

The correct option is D $2\pi \sqrt{\frac{5m}{8N\pi IB}}$

$\tau =M{B}_{H}sin\theta =MB\cos {37}^{\circ }\theta$
$[\theta <<1\Rightarrow \sin \theta =\theta ]$

As, $M=NIA$
$\tau =\left(NI\pi R^2\right)B\times \left(\frac{4}{5}\right)\theta$ .......(1)
&
${\tau }_{\text{restoring}}=-I_{MOI}\alpha =-\frac{m{R}^2}{2}\alpha$ .......(2)

From (1) & (2):

$\alpha =-\frac{8}{5}\frac{N\pi IB}{m}\theta \Rightarrow \omega =\sqrt{\frac{8N\pi IB}{5m}}$
$\Rightarrow T=2\pi \sqrt{\frac{5m}{8N\pi IB}}$