A positive charge q is distributed over a circular ring of radius $a$ . It is placed in a horizontal plane and is rotated about it 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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$q ω a^{2} B$

$\frac{1}{2} q ω^{2} a B$

$\frac{1}{2} q ω a^{2} B$

$\frac{1}{2} q ω a B$

answer is A.

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Detailed Solution

Using torque,

$τ = M \times B = IA \times B = \frac{q}{t} \times {πa}^{2} \times B$

$\Rightarrow τ = \frac{q ω}{2 π} \times {πa}^{2} \times B [∵ ω = \frac{2 π}{t} or, t = \frac{2 π}{ω}]$

$∴ τ = \frac{1}{2} {qωa}^{2} B$

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