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

Question

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

Infinity Learn · Accepted Answer

τ$$1=MBsin$$θWhere θ  is the angle between  M→ and B→.When the ring is rotated, new angle between  M→ and B→  is $$900$$−θ .∴  τ$$2=MBsin900$$−θ$$=MBcos$$θ∴  τ$$12+$$τ$$22=MB2$$⇒$$M=$$τ$$12+$$τ$$22B=0.32+0.420.25Amp$$−$$m2$$⇒$$M=2$$ Amp−$$m2$$

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