A flat disc of radius R charged uniformly on its surface at a surface charge density `sigma`. About its central axis of rotation it rotates at an angular speed `omega`. Find the magnetic moment of disc due to rotation of charges.

Explanation

Let us first select a circular segment of radius and thickness dx on the surface of discs shown in the figure.

Charge on this circular segment can be written as:
` dq = [(2 pi x)dx] sigma = 2 pi sigma x dx `
Now, the equivalent current when it rotates with angular speed a can be written as follows:
` i = (dq)/(2 pi // omega) = ((2 pi sigma x d x) omega)/(2 pi) = omega sigma x d x `
Magnetic field due to this equivalent circular loop at a point on its axis at a distance d can be written as follows:
`dB = (mu_(0)ix^(2))/(2(d^(2) + x^(2))^(3//2)) = (mu_(0)(omega sigma xdx)x^(2))/(2(d^(2) + x^(2))^(3//2)) = (mu_(0)omega sigma)/2 (x^(3)dx)/((d^(2) + x^(2))^(3//2))`
Let
` t = d^(2) + x^(2) rArr dt = 2xdx`
`dB = (mu_(0)omega sigma)/2 (x^(3)dx)/((d^(2) + x^(2))^(3//2)) = (mu_(0)omegasigma)/4 (x^(2)(2xdx))/((d^(2) + x^(2))^(3//2))`
`rArr" " dB = (mu_(0)omega sigma)/4 ((t - d^(2))(dt))/(t^(3//2)) `
`rArr" " dB = (mu_(0)omegasigma)/4 [t^(-1//2) - d^(2) t^(-3//2)] dt`
The magnetic field due to all the circular segments will be in the same direction, hence we can integrate it directly.
`rArr" " B = intdB = (mu_(0)omegasigma)/4 underset(t=d^(2))overset(t=d^(2)+R^(2))int [t^(-1//2) - d^(2)t^(-3//2)] dt`
In the above step we have used limits of integration as per new variablet. Variable x is from 0 to R and hence, according to `t = d^(2) + x^(2)` we can understand the limits of t.
`rArr B = (mu_(0)omegasigma)/4 [2 sqrtt + 2 d^(2)/sqrtt]_(d^(2))^(d^(2)+R^(2))`
`rArr B = (mu_(0)omegasigma)/4 [{2sqrt(d^(2)+R^(2))+2 d^(2)/(sqrt(d^(2) + R^(2)))} - {2sqrtd^(2) + 2 d^(2)/sqrtd^(2)}]`
`rArr B = (mu_(0)omegasigma)/4[{2 sqrt(d^(2) + R^(2)) + 2 d^(2)/(sqrt(d^(2) + R^(2)) - 2d - 2d) }]`
`rArr B = (mu_(0)omegasigma)/2 [(2d^(2) + R^(2))/sqrt(d^(2) + R^(2)) - 2d]`