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# A circular coil of 16 turns and radius $10\;cm$ carrying a current of $0.75 \;A$ 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 $2.0 m/s$ . What is the moment of inertia of the coil about its axis of rotation ?

$\begin{array}{1 1} 1.2 \times 10^{-4}\;kg\; m^2 \\ 1.2 \times 10^{4}\;kg\; m^2 \\ 1.2 \times 10^{-2}\;kg\; m^2 \\ 1.2 \times 10^{3}\;kg\; m^2 \end{array}$

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A)
Solution :
given $n=16$
$r= 10\;cm=0.1 \;m$
$I= 0.75 \;A$
$B= 5 \times 10^{-2} T$
$f= \large\frac{2}{s}$
$M= nIA$
$\qquad= 16 \times 0.75 \times \pi (0.1)^2$
$\qquad= 0.377\;J/T$
Frequency $f= \large\frac{1}{2\pi} \sqrt {\large\frac{M \times B}{I}}$
(ie) $f^2 =\large\frac{1}{4 \pi ^2} \frac{MB}{I}$
$\therefore I= \large\frac{1}{4 \pi ^2} \frac{MB}{f^2}$
$\qquad = \large\frac{ 0.377 \times 5 \times 10^{-2}}{4 \times (3.14)^2 \times 2 \times 2 }$
$\qquad= 1.2 \times 10^{-4}\;kg\; m^2$
Hence the normal of inertia of the coil $1.2 \times 10^{-4}\;kg\; m^2$
Answer : $1.2 \times 10^{-4}\;kg\; m^2$