A ring of radius 'r' is uniformly charged with a charge 'q'. If the ring is rotated about its own axis with an angular frequency ' $ω$ ' then the magnetic induction field at the centre is

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$\frac{μ_{0}}{4 π} (\frac{r}{qω})$

$\frac{μ_{0}}{4 π} (\frac{ω}{qr})$

$\frac{μ_{0}}{4 π} (\frac{qω}{r})$

$\frac{μ_{0}}{4 π} (\frac{q}{rω})$

answer is A.

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Detailed Solution

$\begin{array}{l} B = \frac{μ_{0} i}{2 r} = \frac{μ_{0} q}{2 rt} \\ (∵ i = \frac{q}{t}, t = \frac{2 πr}{v}, v = r ω); B = \frac{μ_{0} q ω}{4 πr} \end{array}$

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