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

Face area of disc `=piR^2`
Charge per unit area of disc `=(Q)/(piR^2)`
Consider an elementary ring of disc of radius r and thickness dr, as shown in figure. Face area of elementary ring `=2pirdr`
Change on elementary ring,
`dq=(Q)/(piR^2)xx2pirdr=(2Qrdr)/(R^2)`

Time period of rotation of disc, `T=2pi/omega`
Current I associated with rotating charge `dq` is
`I=(dq)/(T)=(2Qrdr//R^2)/(2pi//omega)=(Qomegadr)/(piR^2)`
Magnetic field induction `dB` at the centre O of the elementary ring due to current I is
`dB = (mu_(0))/(4pi) (2piI)/(r) = (mu_(0)I)/(2r) = (mu_(0))/(2r) xx(Q omega r dr)/(piR^(2)) = (mu_(0)Q omega dr)/(piR^(2)) = (mu_(0)Q omega dr)/(2piR^(2))`
Total magnetic field induction at O due to entire rotating disc is
`B = int_(0)^(R) (mu_(0)Q omega dr)/(2piR^(2)) = (mu_(0)Q omega R)/(2pi R^(2)) = (mu_(0)Q omega)/(2piR)`
`:. B prop (1)/(R) ( :' Q` and `omega` are unchanges)
Hence, variation of B and R should be a rectangular hyperbola as re[resenmted in option (a).