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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 plane and lies in the equatorial plane of the sphere. Current in the coil is i. The value of B if the sphere is in equilibrium is

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$\frac{m g \tan θ}{π i R}$

$\frac{m g \cos θ}{π i R}$

$\frac{m g \sin θ}{π i R}$

$\frac{m g}{π i R}$

answer is B.

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Detailed Solution

For rotational equilibrium of sphere about point p

$(m g \sin θ) R = M (B \sin θ) o r (m g R \sin θ) = i π R^{2} B \sin θ$

$∴ B = \frac{m g}{π i R}$

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