A dielectric circular disc of radius R carries a uniform surface charge density `sigma`. If it rotates about its exis with angular velocity `omega`, the magnetic field at the cente of disc is :

Options

A. `(mu_(0)sigma omega R^(2))/(2pi)`
B. `(mu_(0)sigma omega R)/(2)`
C. `(mu_(0)sigma omega R^(2))/(4)`
D. `(mu_(0)sigma omega R^(2))/(2sqrt(2))`

Answer

B

Explanation

Let us assumed a ring like element of the disc of radius r and thickness dr. If `sigma` is the surface charge density, then the charge on the element.
`dq=sigma(rpi r) dr` ...(i)
Current di associated with rotating charge dq is
`di=(Dq)/(T)=(dq omega)/(2 pi) ` ( `:.T=(2pi)/(omega)` ) ...(ii)

From Eqs. (i) and (ii), we get
`rArr di = sigma omega r dr `
As, the magnetic field dB at centre due to the element,
`dB=(mu_(0)di)/(2r)=(mu_(0)sigma omega r dr)/(2r) rArr dB=(mu_(0)sigma omega)/(2) dr`
`rArr B_("net")=(mu_(0)sigma omega)/(2)underset(0)overset(R)int dr rArr B_("net")=(mu_(0)sigma omega R)/(2)`
Hence, the correct option is (b).