Question

A circular coil of $16$ turns and radius $10$ cm carrying a current of $0.75A$ rests with its plane normal to an external field of magnitude $5.0\times {10}^{-2}T$. The coil is free to turn about an axis in its plane perpendicular to the filed direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of $2.0/s$. What is the moment of inertia of the coil about its axis of rotation?

• A. $1.2\times {10}^{4}g-c{m}^2$
• B. $3\times {10}^{4}kg-m^2$
• C. $0.3\times {10}^{4}kg-m^2$
• D. $1.2\times {10}^{4}kg-m^2$

Answer 1

The correct option is D $1.2\times {10}^{4}kg-m^2$

Given that,
Number of turns in the circular coil, $N=16$
Radius of the coil, $r=10$ cm $=0.1m$
Current in the coil, $I=0.75A$
Magnetic field strength, $B=5.0\times {10}^{2}T$
Frequency of oscillations of the coil, $v=2.0s^{-1}$
Magnetic moment, $M=NIA$
$M=NI\pi r^2=\left(16\right)\left(0.75\right)\pi (0.1{)}^2=0.3768J{T}^{-1}$
The frequency is given by,
$\nu =\frac{1}{2\pi }\sqrt{\frac{MB}{I}}$
$I=\frac{MB}{4{\pi }^2{\nu }^2}$
$I=\frac{\left(0.3768\right)(5\times {10}^2)}{4{\pi }^2{2}^2}$
$I=1.19\times {10}^4=1.2\times {10}^{4}Kg-m^2$