A magnetic needle having moment of inertia of 40 g cm2 has time period of 3 s in earth's horizontal field of 3.6 x 10-5 Wb/m2. Its magnetic moment will be

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0.5 A m2

5 A m2

0.250 A m2

5 x 102 A m2

answer is A.

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Detailed Solution

T=2πIMBHI=40 g cm2=400 x 10-8 kg m2∴3=2π400 x 10-836 x 10-6 x M⇒1M=94π2 x 364⇒M=0.5 A m2

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