A nonconducing disc of radius R is rotaing about axis passing through its cente and perpendicual its plance with an agular velocity `omega` . The magnt moment of this disc is .

Options

A. `(1)/(4)qomegaR^(2)`
B. `(1)/(2)qomegaR`
C. `qomegaR`
D. `(1)/(2)qomegaR^(2)`

Answer

To find the magnetic moment of a non-conducting disc of radius \( R \) rotating with an angular velocity \( \omega \), we can follow these steps:

### Step 1: Define the Problem
We have a non-conducting disc of radius \( R \) rotating about an axis through its center and perpendicular to its plane. The disc has a total charge \( Q \) uniformly distributed over its surface.

### Step 2: Consider an Element of the Disc
Take a small ring element of radius \( r \) and thickness \( dr \) within the disc. The area \( dA \) of this ring is given by:
\[ dA = 2\pi r \, dr \]

### Step 3: Calculate the Charge in the Element
The charge density \( \sigma \) (charge per unit area) on the disc is:
\[ \sigma = \frac{Q}{\pi R^2} \]
Thus, the charge \( dq \) in the ring element is:
\[ dq = \sigma \cdot dA = \frac{Q}{\pi R^2} \cdot 2\pi r \, dr = \frac{2Q}{R^2} r \, dr \]

### Step 4: Determine the Current Due to Rotation
The current \( I \) associated with the charge \( dq \) moving in a circular path with angular velocity \( \omega \) is given by:
\[ I = \frac{dq}{T} \]
where \( T \) is the time period of rotation. The time period \( T \) is:
\[ T = \frac{2\pi}{\omega} \]
Thus, the current \( I \) becomes:
\[ I = \frac{dq}{T} = \frac{dq \cdot \omega}{2\pi} \]

### Step 5: Substitute for \( dq \)
Substituting \( dq \) into the equation for current:
\[ I = \frac{\frac{2Q}{R^2} r \, dr \cdot \omega}{2\pi} = \frac{Q \omega r \, dr}{\pi R^2} \]

### Step 6: Calculate the Magnetic Moment of the Element
The magnetic moment \( d\mu \) of the ring element is given by:
\[ d\mu = I \cdot A \]
where \( A \) is the area of the ring. The magnetic moment becomes:
\[ d\mu = \left(\frac{Q \omega r \, dr}{\pi R^2}\right) \cdot (2\pi r) = \frac{2Q \omega r^2 \, dr}{R^2} \]

### Step 7: Integrate to Find Total Magnetic Moment
Now, we integrate \( d\mu \) from \( 0 \) to \( R \):
\[ \mu = \int_0^R d\mu = \int_0^R \frac{2Q \omega r^2 \, dr}{R^2} \]
Calculating the integral:
\[ \mu = \frac{2Q \omega}{R^2} \int_0^R r^2 \, dr = \frac{2Q \omega}{R^2} \left[\frac{r^3}{3}\right]_0^R = \frac{2Q \omega R^3}{3R^2} = \frac{2Q \omega R}{3} \]

### Final Result
Thus, the magnetic moment \( \mu \) of the disc is:
\[ \mu = \frac{Q \omega R^2}{2} \]