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 rotating about its midpoint, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Parameters**:
- Length of the rod, \( L \)
- Mass of the rod, \( M \)
- Charge distributed uniformly over the rod, \( q \)
- Angular velocity of rotation, \( \omega \)

2. **Charge per Unit Length**:
- The charge per unit length \( \lambda \) is given by:
\[ \lambda = \frac{q}{L} \]

3. **Consider a Small Element**:
- Take a small element of the rod at a distance \( x \) from the midpoint. The length of this element is \( dx \).
- The distance from the midpoint to this element is \( x \).

4. **Calculate the Current Element**:
- The current \( I \) associated with the charge moving in a circular path can be expressed as:
\[ I = \frac{dq}{dt} \]
- The time period \( dt \) for one complete revolution is given by:
\[ dt = \frac{2\pi}{\omega} \]
- Therefore, the current \( I \) can be expressed as:
\[ I = \frac{dq}{\frac{2\pi}{\omega}} = \frac{\omega \cdot dq}{2\pi} \]

5. **Charge in the Small Element**:
- The charge \( dq \) in the small element \( dx \) is:
\[ dq = \lambda \cdot dx = \frac{q}{L} \cdot dx \]

6. **Substituting for Current**:
- Substitute \( dq \) into the expression for current:
\[ I = \frac{\omega}{2\pi} \cdot \frac{q}{L} \cdot dx \]

7. **Magnetic Moment of the Small Element**:
- The magnetic moment \( d\mu \) of the small element is given by:
\[ d\mu = A \cdot I \]
- The area \( A \) for the circular path of the small element is \( \pi x^2 \). Therefore:
\[ d\mu = \pi x^2 \cdot \left(\frac{\omega q}{2\pi L} \cdot dx\right) = \frac{\omega q}{2L} x^2 \cdot dx \]

8. **Integrate to Find Total Magnetic Moment**:
- Integrate \( d\mu \) from \( -\frac{L}{2} \) to \( \frac{L}{2} \):
\[ \mu = \int_{-\frac{L}{2}}^{\frac{L}{2}} \frac{\omega q}{2L} x^2 \, dx \]
- The integral of \( x^2 \) is:
\[ \int x^2 \, dx = \frac{x^3}{3} \]
- Evaluating the limits:
\[ \mu = \frac{\omega q}{2L} \left[ \frac{x^3}{3} \right]_{-\frac{L}{2}}^{\frac{L}{2}} = \frac{\omega q}{2L} \left( \frac{(\frac{L}{2})^3}{3} - \left(-\frac{L}{2}\right)^3 \frac{1}{3} \right) = \frac{\omega q}{2L} \cdot \frac{L^3}{12} = \frac{q \omega L^2}{24} \]

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