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Q.

The pole strength of a bar magnet is $48 A - m$ and the distance between its pole is $25 c m .$ The torque, by which it can be held at an angle of $30 °$ with the uniform magnetic field of strength $0.15 w b / m^{2},$ will be

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a

$0.90 N - m$

b

$1.80 N - m$

c

$2.70 N - m$

d

$3.6 N - m$

answer is A.

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Detailed Solution

$τ = M B \sin 30 ° = 48 \times \frac{25}{100} \times 0.15 \times \frac{1}{2} = 0.90 N - m$

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