Question

A circular coil of magnetic moment $0.355$J $T^{-1}$ 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 field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of $2$Hz. The moment of inertia of the coil about its axis of rotation is?

• A. $1.13\times {10}^{-1}$kg $m^2$
• B. $1.13\times {10}^{-2}$kg $m^2$
• C. $1.13\times {10}^{-3}$kg $m^2$
• D. $1.13\times {10}^{-4}$kg $m^2$

Answer 1

Answer: (D)

$1.13\times {10}^{-4}$kg $m^2$

Here, $M=0.355$J $T^{-1}$
$B=0.5\times {10}^{-2}$T, v$=2$Hz
As $v=\frac{1}{2\pi }\sqrt{\frac{MB}{I}}$
$\therefore v^2=\frac{1}{4{\pi }^2}\frac{MB}{I}$
$\Rightarrow I=\frac{MB}{4{\pi }^2{v}^2}=\frac{0.355\times 5\times {10}^{-2}}{4\times (3.14{)}^2\times 2^2}=\frac{1775\times {10}^{-2}}{157.75}$
$=1.13\times {10}^{-4}$kg $m^2$