Question

surface charge density on a ring of radius a and width d is sigma it rotate with frequency f about its own axis . assume that charge is only on outer surface.the magnetic field induction at centre is ? when d<<a

Answer 1

The answer is in the image shown below

Answer 2

Answer: The magnetic field induction at the center is $B = \dfrac{\mu_{0}d\sigma \omega}{2}$.

Explanation:

Given that,

Radius of the ring = a

Width of the ring = d

Charge density = $\sigma$

The area of the ring is

$A = \pi[(a+d)^{2}-a^{2}]$

$A = \pi(2ad)$

Here, d is the very small compared to radius of the ring so, neglect of the higher power of d

$A = 2\pi ad$

Now, The magnetic moment produces at the center of the ring is

$M = \dfrac{q}{T}\times\pi a^{2}$

$M = \dfrac{2\pi ad\sigma\omega}{2\pi}\times \pi a^{2}$

$M = \pi a^{3}d\sigma\omega$

The magnetic field induction at center is

$B = \dfrac{\mu_{0}}{4\pi}\times\dfrac{2M}{a^{3}}$

$B = \dfrac{\mu_{0}d\sigma \omega}{2}$

Hence, The magnetic field induction at the center is $B = \dfrac{\mu_{0}d\sigma \omega}{2}$.