An insulating rod of length I carries a charge q distrubuted uniformly on it. The rod is pivoted at its mid-point and is rotated at a frequency f about a fixed axis perpendicular to the the rod and passing through the pivot . The magnetic moment of the rod system is

Options

A. `1/12 piqfl^2`
B. ` piqfl^2`
C. `1/6 piqfl^2`
D. `1/3 piqfl^2`

Explanation

To find the magnetic moment of an insulating rod of length \( L \) carrying a uniformly distributed charge \( q \), which is pivoted at its midpoint and rotated with a frequency \( f \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- The rod is of length \( L \) and is pivoted at its midpoint. Therefore, each half of the rod has a length of \( \frac{L}{2} \).
- The total charge \( q \) is uniformly distributed along the length of the rod.

2. **Charge Density**:
- The linear charge density \( \lambda \) can be defined as:
\[ \lambda = \frac{q}{L} \]

3. **Angular Frequency**:
- The angular frequency \( \omega \) is related to the frequency \( f \) by:
\[ \omega = 2\pi f \]

4. **Consider a Small Element**:
- Consider a small element of the rod at a distance \( x \) from the pivot with a small length \( dx \).
- The charge \( dq \) on this small element is given by:
\[ dq = \lambda \, dx = \frac{q}{L} \, dx \]

5. **Velocity of the Charge Element**:
- The velocity \( v \) of this charge element rotating in a circular path is:
\[ v = x \cdot \omega = x \cdot (2\pi f) \]

6. **Time Period for One Complete Revolution**:
- The time period \( T \) for the charge to complete one full circle is:
\[ T = \frac{2\pi x}{v} = \frac{2\pi x}{x \cdot \omega} = \frac{2\pi}{\omega} \]

7. **Current Due to the Charge Element**:
- The current \( I \) due to the moving charge is:
\[ I = \frac{dq}{T} = \frac{\frac{q}{L} \, dx}{\frac{2\pi}{\omega}} = \frac{q \cdot \omega}{2\pi L} \, dx \]

8. **Magnetic Moment Contribution**:
- The magnetic moment \( dM \) contributed by this small charge element is given by:
\[ dM = I \cdot A \]
- The area \( A \) swept by the charge element is:
\[ A = \pi x^2 \]
- Thus, the differential magnetic moment becomes:
\[ dM = \left(\frac{q \cdot \omega}{2\pi L} \, dx\right) \cdot (\pi x^2) = \frac{q \cdot \omega}{2L} \, x^2 \, dx \]

9. **Total Magnetic Moment**:
- To find the total magnetic moment \( M \), integrate \( dM \) from \( -\frac{L}{2} \) to \( \frac{L}{2} \):
\[ M = \int_{-\frac{L}{2}}^{\frac{L}{2}} \frac{q \cdot \omega}{2L} \, x^2 \, dx \]
- This can be simplified as:
\[ M = \frac{q \cdot \omega}{2L} \cdot \left[ \frac{x^3}{3} \right]_{-\frac{L}{2}}^{\frac{L}{2}} = \frac{q \cdot \omega}{2L} \cdot \left( \frac{(\frac{L}{2})^3}{3} - \left(-\frac{L}{2}\right)^3 \frac{1}{3} \right) \]
- The limits yield:
\[ M = \frac{q \cdot \omega}{2L} \cdot \left( \frac{L^3}{24} \right) = \frac{q \cdot L^2 \cdot \omega}{48} \]

10. **Final Expression**:
- Substituting \( \omega = 2\pi f \):
\[ M = \frac{q \cdot L^2 \cdot (2\pi f)}{48} = \frac{q \cdot L^2 \cdot \pi f}{24} \]

### Final Answer:
The magnetic moment of the rod system is:
\[ M = \frac{q \cdot L^2 \cdot \pi f}{24} \]