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A positive electric charge $q$ is distributed over a circular ring of radius $a$ . It is placed in a horizontal plane and is rotated about its axis at a uniform angular speed $ω$ . A horizontal magnetic field $B$ exists in the space. The torque acting on the ring due to the magnetic force is :

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$\frac{1}{2 π} q ω a^{2} B$

$\frac{1}{π} q ω a^{2} B$

$q ω a^{2} B$

$\frac{1}{2} q ω a^{2} B$

answer is B.

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Detailed Solution

$i = \frac{q}{t} = \frac{q ω}{2 π}$

Now, $τ = M B \sin θ = i A B \sin 90 ° = \frac{q ω}{2 π} \times π a^{2} B = \frac{q ω a^{2} B}{2}$

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