Charge q is uniformly distributed on a disc of radius r. If the disc is rotated with a frequency v, then magnetic field induction at the centre will be

Options

A. `(mu_0q)/(vr)`
B. `(mu_0qv)/(r)`
C. `(mu_0v)/(qr)`
D. `(mu_0)/(qvr)`

Answer

B

Explanation

Consider a small element of the disc of radius x with `dx`, figure. Face area of the small element of disc `=2pixdx`
Charge on the elementary part of disc,
`dq=(q)/(pir^2)xx2pixdx` or `dq=(2qxdx)/(r^2)`

Magnetic field induction at O due to rotation of elementary ring with frequency v is
`dB=(mu_0)/(4pi)(2pidqv)/(x) ( :' dI=dqxxv)'
'=(mu_0)/(2x)xx(2qxdx)/(r^2)xxv=(mu_0qv)/(r^2)dx`
`:.` Total magnetic field conduction at O due to rotation of whole disc with frequency is
`B=int_(0)^(r)(mu_0qv)/(r^2)dx=(mu_0qv)/(r^2)(r-0)=(mu_0qv)/(r)`