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A uniform current carrying ring of mass $m$ and radius $R$ is connected by a massless string as shown. A uniform magnetic field $B_{0}$ exists in the region to keep the ring in horizontal position then the current in the ring is
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$\frac{m g}{{πRB}_{0}}$

$\frac{m g}{{RB}_{0}}$

$\frac{m g}{3 {πRB}_{0}}$

$\frac{m g}{{πR}^{2} B_{0}}$

answer is A.

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Detailed Solution

$τ_{m g}$ about the left end (from where string is connected)

$= |M \times B| = M B \sin 90 °$

or $(m g R) = (N i A) B_{0} = i ({πR}^{2}) B_{0}$

or $i = \frac{m g}{{πRB}_{0}}$

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