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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 magnetic field of magnitude $5 \times 10^{- 2} T$ .The coil is free to turn about an axis perpendicular to the field. when it is turned slightly and released, it oscillates about its position of stable equilibrium with a frequency of 2Hz. What is the moment of inertia of the coil about its axis of rotation ?

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$1.2 \times 10^{- 4} kg - m^{2}$

$1.4 \times 10^{- 6} kg - m^{2}$

$1.2 \times 10^{- 6} kg - m^{2}$

$1.4 \times 10^{- 5} kg - m^{2}$

answer is C.

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Detailed Solution

Magnetic dipole moment of the coil $M = NiA = 16 \times 0.75 \times 3.14 \times 10^{- 2} A M^{2} \Rightarrow M = 0.3768 A M^{2}$
We know, $T = 2 π \sqrt{\frac{I}{BM}} \Rightarrow I = 1.2 \times 10^{- 4} kg - m^{2}$

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