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A circular loop of radius $r$ carrying a current $i$ is held at the centre of another circular loop of radius $R (> > r)$ carrying a current $I$ .The larger loop is stationary and the smaller loop can rotate about a fixed spindle along its diametre . The plane of the smaller loop makes an angle $30 °$ with that of the larger loop. If the smaller loop is held fixed in this position by applying a single force at a point on its periphery. If the minimum magnitude of this force is $\frac{μ_{0} πiIr}{k R}, f i n d t h e v a l u e o f k .$

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Detailed Solution

The magnetic field at the centre of the smaller loop, $B = \frac{μ_{0} I}{2 R}$ .

According to right hand screw rule, it is along as shown in the given figure. Since smaller loop is placed at its centre and assuming that field is uniform all over it, the torque exerted by magnitude field.

$ζ = M B \sin θ$

$= (i {πr}^{2}) \times \frac{μ_{0} I}{2 R} \sin 30 °$

$= \frac{μ_{0} {πiIr}^{2}}{4 R}$

If $F$ is the force at the periphery then, to keep the loop at its position

$F \times r = \frac{μ_{0} {πiIr}^{2}}{4 R}$

or $F = \frac{μ_{0} πiIr}{4 R}$

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