Surface charge density on a ring of radius $a$ and width $d$ is $\sigma$ as shown in the figure. It rotates with frequency $f$ about its own axis. Assume that the charge is only on outer surface. The magnetic field induction at centre is(Assume that $d \ll a$ )
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(a)
Surface charge density $=\sigma$
Total charge on the ring $(q)=\sigma(2 \pi a) d$
$\Rightarrow i=\frac{q}{T}=\sigma(2 \pi a) d f$
$\vec{B}=\frac{\mu_0 I}{2 \pi a}=\frac{\mu_0(\sigma 2 \pi a d f)}{2 \pi a}=\pi \mu_0 \sigma d f$
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