A short bar magnet placed with its axis at 30⁰ with an external field of 800 G experiences a torque of 0.016 Nm. What is the magnetic moment of the magnet? (b) The bar magnet is replaced by a solenoid of cross - sectional area 2 × 10^–4 m² and 1000 turns, but of the same magnetic moment, then the current flowing through the solenoid is:

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0.25 Am², 3A

0.40Am², 2A

0.15 Am², 5A

0.10 Am², 2A

answer is A.

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Detailed Solution

$τ = MBsin θ; 0 .016 = M \times 800 \times 10^{- 4} \times \frac{1}{2} M = NiA = \frac{0.016 \times 2}{800 \times 10^{- 4}} = 0.4 {Am}^{2} i = \frac{M}{NA} = \frac{0.4}{1000 \times 2 \times 10^{- 4}} = 2 A$

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