A uniform rod of mass M and length L carrying a charge q uniformly distributed over its length is rotated with constant velocity w about its mid point perpendicular to the rod. Its magnetic moment is

Options

A. `q omega L^(2) //6`
B. `q omega L^(2) //12`
C. `q omega L^(2)//24`
D. `qomegaL^(2)//48`

Answer

To find the magnetic moment of a uniformly charged rod of mass \( M \) and length \( L \) rotating with a constant angular velocity \( \omega \) about its midpoint, we can follow these steps:

### Step-by-Step Solution

1. **Understanding the Charge Distribution**:
The rod has a total charge \( q \) uniformly distributed over its length \( L \). Therefore, the linear charge density \( \lambda \) can be defined as:
\[ \lambda = \frac{q}{L} \]

2. **Element of Charge**:
Consider an infinitesimal element of length \( dx \) at a distance \( x \) from the midpoint of the rod. The charge \( dq \) on this element can be expressed as:
\[ dq = \lambda \, dx = \frac{q}{L} \, dx \]

3. **Current Due to Rotation**:
As the rod rotates with angular velocity \( \omega \), the element of charge \( dq \) moves in a circular path with radius \( x \). The frequency \( f \) of rotation is given by:
\[ f = \frac{\omega}{2\pi} \]
The current \( dI \) associated with the charge element \( dq \) can be calculated as:
\[ dI = dq \cdot f = \left(\frac{q}{L} \, dx\right) \cdot \left(\frac{\omega}{2\pi}\right) = \frac{q \omega}{2\pi L} \, dx \]

4. **Magnetic Moment Contribution**:
The magnetic moment \( d\mu \) contributed by the current element \( dI \) at distance \( x \) is given by:
\[ d\mu = dI \cdot A \]
where \( A \) is the area of the circular path traced by the charge element \( dq \):
\[ A = \pi x^2 \]
Thus, we can express \( d\mu \) as:
\[ d\mu = \left(\frac{q \omega}{2\pi L} \, dx\right) \cdot \pi x^2 = \frac{q \omega}{2L} x^2 \, dx \]

5. **Integrating to Find Total Magnetic Moment**:
To find the total magnetic moment \( \mu \), we integrate \( d\mu \) from \( -\frac{L}{2} \) to \( \frac{L}{2} \):
\[ \mu = \int_{-\frac{L}{2}}^{\frac{L}{2}} \frac{q \omega}{2L} x^2 \, dx \]
The integral of \( x^2 \) is:
\[ \int x^2 \, dx = \frac{x^3}{3} \]
Evaluating the integral:
\[ \mu = \frac{q \omega}{2L} \left[ \frac{x^3}{3} \right]_{-\frac{L}{2}}^{\frac{L}{2}} = \frac{q \omega}{2L} \left( \frac{(L/2)^3}{3} - \frac{(-L/2)^3}{3} \right) = \frac{q \omega}{2L} \cdot \frac{2(L/2)^3}{3} = \frac{q \omega L^2}{24} \]

6. **Final Result**:
Therefore, the magnetic moment \( \mu \) of the rod is:
\[ \mu = \frac{q \omega L^2}{24} \]