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Q.

A magnetic needle having moment of inertia of $40 g {cm}^{2}$ has time period of 3 s in earth's horizontal field of $3.6 x 10^{- 5} Wb / m^{2}$ . Its magnetic moment will be

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a

$0.250 A m^{2}$

b

$0.5 A m^{2}$

c

$5 A m^{2}$

d

$5 x 10^{2} A m^{2}$

answer is A.

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Detailed Solution

$T = 2 π \sqrt{\frac{I}{{MB}_{H}}}$

$I = 40 g {cm}^{2} = 400 x 10^{- 8} kg m^{2}$

$∴ 3 = 2 π \sqrt{\frac{400 x 10^{- 8}}{36 x 10^{- 6} x M}}$

$\Rightarrow \frac{1}{M} = \frac{9}{4 π^{2}} x \frac{36}{4} \Rightarrow M = 0.5 A m^{2}$

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