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 varying surface charge density, we can follow these steps:

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

### Step 2: Define the Charge Element
Consider a small ring of radius \( r \) and thickness \( dr \) on the disc. The area of this ring is given by:
\[ dA = 2\pi r \, dr \]
The charge \( dq \) on this ring can be expressed as:
\[ dq = \sigma \cdot dA = \sigma \cdot 2\pi r \, dr = \alpha r^2 \cdot 2\pi r \, dr = 2\pi \alpha r^3 \, dr \]

### Step 3: Determine the Current Element
The current \( I \) associated with the charge \( dq \) can be found using the relationship:
\[ I = \frac{dq}{dt} \]
The time \( dt \) for one complete rotation (period) is:
\[ T = \frac{2\pi}{\omega} \]
Thus, the current \( I \) due to the charge \( dq \) is:
\[ I = \frac{dq}{T} = \frac{2\pi \alpha r^3 \, dr}{\frac{2\pi}{\omega}} = \alpha r^3 \omega \, dr \]

### Step 4: Calculate the Magnetic Field
The magnetic field \( dB \) at the center of the disc due to the current element \( I \) in the ring is given by the formula:
\[ dB = \frac{\mu_0 I}{2R} \]
Substituting for \( I \):
\[ dB = \frac{\mu_0 (\alpha r^3 \omega \, dr)}{2R} \]

### Step 5: Integrate to Find Total Magnetic Field
To find the total magnetic field \( B \) at the center of the disc, we integrate \( dB \) from \( r = 0 \) to \( r = R \):
\[ B = \int_0^R dB = \int_0^R \frac{\mu_0 \alpha r^3 \omega}{2R} \, dr \]
Calculating the integral:
\[ B = \frac{\mu_0 \alpha \omega}{2R} \int_0^R r^3 \, dr = \frac{\mu_0 \alpha \omega}{2R} \left[ \frac{r^4}{4} \right]_0^R = \frac{\mu_0 \alpha \omega}{2R} \cdot \frac{R^4}{4} = \frac{\mu_0 \alpha \omega R^3}{8} \]

### Final Result
Thus, the magnetic field intensity at the center of the disc is:
\[ B = \frac{\mu_0 \alpha \omega R^3}{8} \]

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