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Question: The pole strength of a bar magnet is 48 ampere-metre and the distance between its poles is 25 cm. Th...

The pole strength of a bar magnet is 48 ampere-metre and the distance between its poles is 25 cm. The moment of the couple by which it can be placed at an angle of 30o with the uniform magnetic intensity of flux density 0.15 Newton /ampere-metre will be

A

12 Newton × metre

B

18 Newton × metre

C

0.9 Newton × metre

D

None of the above

Answer

0.9 Newton × metre

Explanation

Solution

τ=MBsin⁡θ=48×25×10−2×0.15×12\tau = M B \sin \theta = 48 \times 25 \times 10 ^ { - 2 } \times 0.15 \times \frac { 1 } { 2 }