A thin annular disc of outer radius R and `R/2` inner radius has charge Q uniformly distributed over its surface. The disc rotates about its axis with a constant angular velocity `omega`. Choose the correct option(s):

Options

A. The magnetic field intensity at the centre of the disc is `(mu_(0)Qomega)/(3piR)`
B. The magnetic field intensity at the centre of the disc is `(mu_(0)Qomega)/(4R)`
C. The magnetic moment of the disc is `(5Q omega R^(2))/16`
D. The magnetic moment of the disc is `(3Qomega R^(2))/4`

Answer

A, C

Explanation

Charge per unit area of the disc. `sigma=Q/(pi(R^(2)-(R^(2))/4))=(4Q)/(3piR^(2))`
Let us consider a ring element of inner and outer radii x and x+dx
Charge on this element `dq=sigma(2pix dx)=(8Q)/(3R^(2))x dx`
If a charged circular ring rotates about its axis, it can be treated like a current carrying loop. To calculate the equivalent current, take a point on the circumference of the ring, imagine a surface perpendicular to the ring at this point, and think about how much charge is crossing this surface per unit time. The entire charge on the ring will pass through this surface in the time it takes the ring to make one complete rotation.
So equilavent current `di=(dq)/(((2pi)/(omega)))=(omegadq)/(2 pi)=((4 Q omega)/(3 piR^(2)))x dxx`
Therefore, magnetic field at the centre of the loop due to this ring element,
`dB=(mu_(0)(di))/(2x)=((2mu_(0)Qomega)/(3piR^(2)))dx`
So net magnetic field `B=intdB=((2mu_(0)Qomega)/(3piR^(2)))int_(R//2)^(R)dx=(mu_(0)Q omega)/(3piR)`
No magnetic moment of the ring element
`dm=(di)(pix^(2))=((4Qomega)/(3R^(2)))x^(3)dx`
So, net magnetic moment of the rotating disc
`m=intdx=((4Qomega)/(3R^(2)))int_(R//2)^(R)x^(3)dx=(5Qomega R^(2))/4`