A charge `Q` is uniformly distributed over the surface of non - conducting disc of radius `R`. The disc rotates about an axis perpendicular to its plane and passing through its centre with an angular to its plane and passing through its centre of the disc. If we keep both the amount of charge placed on the disc and its angular velocity to be constant and vary the radius of the disc then the variation of the magnetic induction at the centre of the disc will br represented by the figure:

Options

A.
B.
C.
D.

Answer

A

Explanation

To calculate magnetic field at the centre of disc we need to select one ring segment of radius x and thickness dx on the surface of disc. Charge on this ring segment can be written as follows:
`dq = Q/(pi R^(2)) 2 pi x d x = (2Q)/R^(2) x d x `
Its equivalent current due to rotation can be written as follows:
` di = (dq)/T = ((2Q)/R^(2)xdx )/((2pi)/omega) = (Q omegaxdx)/(pi R^(2))`
Magnetic field at the centre of disc due to ring segment only can be written as follows:
`dB = (mu_(0)(di))/(2x) = (mu_(0) Q omega x dx )/(2x xx pi R^(2)) = (mu_(0) Q omega)/(2 pi R^(2)) dx `
All such ring segments will produce magnetic field in the same direction at the centre of disc. Hence, net magnetic field at the centre can be calculated by integrating the above result.
`rArr" " B = int dB = (mu_(0)Qomega)/(2 pi R^(2)) underset(0)overset(R) int dx = (mu_(0)Q omega)/(2 pi R) `
Magnetic field at the centre of disc is inversely proportional to its radius for a given charge magnitude. Thus, option (a) is correct.