Q.

A current carrying circular ring, when placed in a uniform magnetic field of 0.25 T with one diameter perpendicular to the field  experiences a torque of 0.3 N-m, when the ring is rotated about this diameter through an angle of 90o, it experiences a torques of 0.4 N-m. Then magnetic dipole moment of the ring is

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a

2 A−m2

b

4 A−m2

c

2.5 A−m2

d

1.25 A−m2

answer is A.

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Detailed Solution

τ1=MBsinθWhere θ  is the angle between  M→ and B→.When the ring is rotated, new angle between  M→ and B→  is 900−θ .∴  τ2=MBsin900−θ=MBcosθ∴  τ12+τ22=MB2⇒M=τ12+τ22B=0.32+0.420.25Amp−m2⇒M=2 Amp−m2
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