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A uniform current carrying ring of mass m and radius R is connected by a massless string as shown in figure. A uniform magnetic field B₀ exists in the region to keep the ring in horizontal position, then the current in the ring is (l = length of string)

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$\frac{mg}{{πRB}_{0}}$

$\frac{mg}{{RB}_{0}}$

$\frac{mg}{3 {πRB}_{0}}$

$\frac{mgl}{{πR}^{2} B_{0}}$

answer is A.

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Detailed Solution

Torque due to magnetic field $τ_{mag} = {MB}_{0} = {IπR}^{2} B_{0}$ …….(i)

Torque due to weight about the point where string is connected

$τ_{weight} = mgR$ …….(ii)

If ring remains horizontal, then $τ_{mag} = τ_{weight}$

${IπR}^{2} B_{0} = mgR \Rightarrow I = \frac{mg}{{πRB}_{0}}$

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