A disc of radius R rotates with constant angular velocity `omega` about its own axis. Surface charge density of this disc varies as `sigma = alphar^(2)`, where r is the distance from the centre of disc. Determine the magnetic field intensity at the centre of disc.

Options

A. `mu_(0)alphaomegaR^(3)`
B. `(mu_(0)alphaomegaR^(3))/(6)`
C. `(mu_(0)alphaomegaR^(3))/(8)`
D. `(mu_(0)alphaomegaR^(3))/(3)`

Answer

To determine the magnetic field intensity at the center of a rotating disc with a given surface charge density, we can follow these steps:

### Step-by-Step Solution

1. **Understand the Problem**: We have a disc of radius \( R \) rotating with a constant angular velocity \( \omega \). The surface charge density varies as \( \sigma = \alpha r^2 \), where \( r \) is the distance from the center of the disc.

2. **Consider a Differential Element**: To find the magnetic field at the center, we can consider a thin ring of radius \( r \) and thickness \( dr \) on the disc. The area of this ring is \( dA = 2\pi r \, dr \).

3. **Calculate the Charge on the Ring**: The charge \( dq \) on this ring can be calculated using the surface charge density:
\[ dq = \sigma \cdot dA = \alpha r^2 \cdot (2\pi r \, dr) = 2\pi \alpha r^3 \, dr \]

4. **Determine the Current Due to Rotation**: The ring rotates with angular velocity \( \omega \), which means it behaves like a current loop. The current \( i \) due to the charge \( dq \) can be expressed as:
\[ i = \frac{dq}{T} \]
where \( T \) is the time period for one complete rotation. The time period \( T \) is given by:
\[ T = \frac{2\pi}{\omega} \]
Therefore,
\[ i = \frac{dq}{T} = \frac{2\pi \alpha r^3 \, dr}{\frac{2\pi}{\omega}} = \alpha \omega r^3 \, dr \]

5. **Calculate the Magnetic Field Contribution from the Ring**: The magnetic field \( dB \) at the center of the disc due to the current \( i \) in the ring is given by the formula:
\[ dB = \frac{\mu_0 i}{2r} \]
Substituting for \( i \):
\[ dB = \frac{\mu_0 (\alpha \omega r^3 \, dr)}{2r} = \frac{\mu_0 \alpha \omega r^2}{2} \, dr \]

6. **Integrate to Find Total Magnetic Field**: To find the total magnetic field \( B \) at the center of the disc, integrate \( dB \) from \( r = 0 \) to \( r = R \):
\[ B = \int_0^R dB = \int_0^R \frac{\mu_0 \alpha \omega r^2}{2} \, dr \]
This integral evaluates to:
\[ B = \frac{\mu_0 \alpha \omega}{2} \int_0^R r^2 \, dr = \frac{\mu_0 \alpha \omega}{2} \left[\frac{r^3}{3}\right]_0^R = \frac{\mu_0 \alpha \omega}{2} \cdot \frac{R^3}{3} \]
Simplifying gives:
\[ B = \frac{\mu_0 \alpha \omega R^3}{6} \]

7. **Final Result**: The magnetic field intensity at the center of the disc is:
\[ B = \frac{\mu_0 \alpha \omega R^3}{6} \]