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Q.

The pole strength of a bar magnet is 48A-m and the distance between its pole is 25cm. The torque, by which it can be held at an angle of 30° with the uniform magnetic field of strength 0.15wb/m2, will be

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a

0.90N−m

b

1.80N−m

c

2.70N−m

d

3.6N−m

answer is A.

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Detailed Solution

τ=MBsin30° =48×25100×0.15×12=0.90N−m

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