Question

A non conducting disc of radius r has a sigma

Answer 1

Explanation:

surface charge density σ = q/πr²

Now let us take a small cross section of length dR whose radius varies from R to R + dR

The charge on this annular region is ,

dq = 2πRσdR

= 2πR x q/πr² x dR

= 2q/r² x RdR

This charge dq passes through a line once in a time, T = 2π/ω

So charge crossing per unit time i.e. current dI = dq/T

= dqω/2π

=2q/r² x RdR x ω/2π

The area enclosed by this current = πR²

Hence, magnetic moment due to current loop ,

dM = πR² dI

=πR² x  2q/r² x RdR x ω/2π

= ωq/r²  x R³dR

Taking integration on both the sides where limit of R varies from 0 to R we get,

M = ωq/r²  x

=> M = ωq/r²  x r⁴/4

=> M = ωqr²/4

Hence the moment of inertia will be ωqr²/4