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`

Explanation

To solve the problem of a thin annular disc with outer radius \( R \) and inner radius \( \frac{R}{2} \) that rotates with a constant angular velocity \( \omega \) and has a uniformly distributed charge \( Q \), we will follow these steps:

### Step 1: Calculate the Surface Charge Density
The surface charge density \( \sigma \) is given by the total charge \( Q \) divided by the area of the annular disc.

The area \( A \) of the annular disc is:
\[ A = \pi R^2 - \pi \left(\frac{R}{2}\right)^2 = \pi R^2 - \frac{\pi R^2}{4} = \frac{3\pi R^2}{4} \]

Thus, the surface charge density \( \sigma \) is:
\[ \sigma = \frac{Q}{A} = \frac{Q}{\frac{3\pi R^2}{4}} = \frac{4Q}{3\pi R^2} \]

### Step 2: Determine the Current Element
Consider a small ring element of radius \( r \) and thickness \( dr \). The area \( dA \) of this ring is:
\[ dA = 2\pi r \, dr \]

The charge \( dq \) on this ring is:
\[ dq = \sigma \, dA = \sigma \cdot 2\pi r \, dr = \frac{4Q}{3\pi R^2} \cdot 2\pi r \, dr = \frac{8Q}{3R^2} r \, dr \]

### Step 3: Calculate the Current \( di \)
The current \( di \) due to the rotating charge is given by:
\[ di = \frac{dq}{T} \]
where \( T \) is the period of rotation. The period \( T \) can be expressed in terms of angular velocity \( \omega \):
\[ T = \frac{2\pi}{\omega} \]

Thus,
\[ di = \frac{dq}{T} = \frac{8Q}{3R^2} r \, dr \cdot \frac{\omega}{2\pi} = \frac{4Q\omega}{3\pi R^2} r \, dr \]

### Step 4: Calculate the Magnetic Field at the Center
The magnetic field \( dB \) at the center of the disc due to the current element \( di \) is given by:
\[ dB = \frac{\mu_0 \, di}{2r} \]

Substituting \( di \):
\[ dB = \frac{\mu_0}{2r} \cdot \frac{4Q\omega}{3\pi R^2} r \, dr = \frac{2\mu_0 Q \omega}{3\pi R^2} \, dr \]

To find the total magnetic field \( B \) at the center, integrate \( dB \) from \( r = \frac{R}{2} \) to \( r = R \):
\[ B = \int_{R/2}^{R} dB = \int_{R/2}^{R} \frac{2\mu_0 Q \omega}{3\pi R^2} \, dr = \frac{2\mu_0 Q \omega}{3\pi R^2} \left( R - \frac{R}{2} \right) = \frac{2\mu_0 Q \omega}{3\pi R^2} \cdot \frac{R}{2} \]
\[ B = \frac{\mu_0 Q \omega}{3\pi R} \]

### Step 5: Calculate the Magnetic Moment
The magnetic moment \( m \) of the disc is given by:
\[ m = \text{Number of turns} \times \text{Current} \times \text{Area} \]

For our case, the area of the annular disc is \( A = \frac{3\pi R^2}{4} \), and the current \( I \) is given by integrating \( di \):
\[ m = 1 \cdot \int_{R/2}^{R} di \cdot \frac{3\pi R^2}{4} \]

Substituting \( di \):
\[ m = \frac{3\pi R^2}{4} \cdot \int_{R/2}^{R} \frac{4Q\omega}{3\pi R^2} r \, dr = \frac{3\pi R^2}{4} \cdot \frac{4Q\omega}{3\pi R^2} \cdot \left[ \frac{r^2}{2} \right]_{R/2}^{R} \]
\[ = \frac{3\pi R^2}{4} \cdot \frac{4Q\omega}{3\pi R^2} \cdot \left( \frac{R^2}{2} - \frac{(R/2)^2}{2} \right) = \frac{3\pi R^2}{4} \cdot \frac{4Q\omega}{3\pi R^2} \cdot \left( \frac{R^2}{2} - \frac{R^2}{8} \right) \]
\[ = \frac{3\pi R^2}{4} \cdot \frac{4Q\omega}{3\pi R^2} \cdot \frac{3R^2}{8} = \frac{Q\omega R^2}{8} \]

### Conclusion
The magnetic field intensity at the center of the disc is:
\[ B = \frac{\mu_0 Q \omega}{3\pi R} \]
And the magnetic moment of the disc is:
\[ m = \frac{5Q\omega R^2}{16} \]