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

Taking an elemental ring of radius r and thickness dr, we obtain magnetic field at the centre of the ring.
`dB=(mu_(0))/(2r)dI=(mu_(0))/(2r) . (dq)/T=(mu_(0))/(2r) . (Q.(piR^(2)).2pirdr)/((2r)/(omega))=(mu_(0)Qomega)/(2piR^(2))dr`
So, net magnetic field at the centre of disk,
`B=int_(0)^(R)dB=(mu_(0)omegaQ)/(2piR^(2))int_(0)^(R)dr=(mu_(0)omegaQ)/(2piR)` i.e. `B prop 1/R`