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Question: In the figure shown a coil of single turn is wound on a sphere of radius r and mass m. The plane of ...

In the figure shown a coil of single turn is wound on a sphere of radius r and mass m. The plane of the coil is parallel to the inclined plane and lies in the equatorial plane of the sphere. If sphere is in rotational equilibrium the value of B is (current in the coil is i)

A

mgπir\frac{\text{mg}}{\pi\text{ir}}

B

mgsinθπi\frac{\text{mgsin}\theta}{\pi i}

C

mgr sinθπi\frac{\text{mgr sin}\theta}{\pi i}

D

None

Answer

mgπir\frac{\text{mg}}{\pi\text{ir}}

Explanation

Solution

The gravitational torque must be counter balanced by the magnetic torque about O, for equilibrium of the sphere. The gravitational force = τgr\tau_{gr}=mg X r sin θ

⇒ τgr\tau_{gr} = mgr sinθ

The magnetic torque τm\tau_{m} = μ→XB→\overset{\rightarrow}{\mu}X\overset{\rightarrow}{B}

Where the magnetic moment of the coil

= μ = (iπr2)

⇒ τm\tau_{m} = πir2 Sin θ

πir2 B Sinθ = mgr Sin θ⇒ B= mgπir\frac{mg}{\pi ir}