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 $(q)=\sigma (2\pi a)d$
$\Rightarrow i=\frac{q}{T}=\sigma (2\pi a)df$
$\vec{B}=\frac{{\mu }_{0}l}{2\pi a}=\frac{{\mu }_0(\sigma 2\pi adf)}{2\pi a}=\pi {\mu }_0\sigma df$