A bar magnet has a magnetic moment equal to 5 × 10 ^ - 5 Wbm . It is suspended in a magnetic field which has a magnetic induction B = 8π × 10 ^ - 4 T . The magnet vibrates with a period of vibration equal to 15 s . The

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Correct answer: D. 4.54 × 10 ^ 5 kgm^ 2

Here, magnetic moment is given in weber-meter, which is the unit of

{\mu}_{0}M

.

∴ {\mu}_{0}M = 5 \times {10}^{- 5}

Wb-m

M = \dfrac{5 \times {10}^{- 5}}{{\mu}_{0}}Am^{2}

Also,

B = 8\pi \times {10}^{- 4} = {\mu}_{0}H

∴ H = \dfrac{8\pi \times {10}^{- 4}}{{\mu}_{0}}

T = 2\pi\sqrt{\dfrac{I}{MH}}

I = \dfrac{MHT^{2}}{4{\pi}^{2}} = \dfrac{5 \times {10}^{- 5} \times 8\pi \times {10}^{- 4} \times {15}^{2}}{4{\pi}^{2}{\mu}_{0}^{2}}

= \dfrac{5 \times {10}^{- 5} \times 8\pi \times {10}^{- 4} \times 225}{4{\pi}^{2}{\left(4\pi \times {10}^{- 7}\right)}^{2}}

I = \dfrac{2250 \times {10}^{- 9}}{\pi ( 16{\pi}^{2} ) \times {10}^{- 14}} = 4.54 \times {10}^{5}kgm^{2}

A bar magnet has a magnetic moment equal to 5 × 10 ^ - 5 Wbm . It...

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