The figure shows a non-conducting ring of radius R carrying a charge q. In a circular region of radius r, a uniform magnetic field B perpendicular to the plane of the ring varies at a constant rate $\frac{d B}{d t} = β$ . The torque acting on the ring is

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$\frac{1}{2} q r^{2} β$

$\frac{1}{2} q R^{2} β$

$q r^{2} β$

zero

answer is A.

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Detailed Solution

$\begin{array}{rcl} E l & = & \frac{d ϕ}{d t} = S \frac{d B}{d t} \\ ∴ E (2 π r) & = & (π r^{2}) (β) \\ E & = & \frac{r^{2}}{2 R} β \Rightarrow F = q E \\ and τ & = & F R = q E R = \frac{1}{2} q r^{2} β \end{array}$

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