Question

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 $d \ll a$ )

• A. $\pi \,\mu_{0} \,f\, \sigma d$
• B. $\mu_{0}\, f \,\sigma d$
• C. $2 \pi \,\mu_{0}\, f \,\sigma d$
• D. $\frac{\pi^{2}}{2 \mu_{0}} f \,\sigma\, d$

Answer 1

Answer: (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 $
$\vec{B}=\frac{\mu_{0} l}{2 \pi a}=\frac{\mu_{0}(\sigma 2 \pi a d f)}{2 \pi a}=\pi \mu_{0} \sigma d f$