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. `(piqfl^2)/12`
B. `(piqfl^2)/2`
C. `(piqfl^2)/6`
D. `(piqfl^2)/3`

Explanation

To find the magnetic moment of the insulating rod system, we can follow these steps:

### Step 1: Understand the Charge Distribution
The insulating rod of length \( L \) carries a total charge \( q \) uniformly distributed along its length. The charge per unit length \( \lambda \) can be expressed as:
\[ \lambda = \frac{q}{L} \]

### Step 2: Consider an Element of the Rod
Take a small element of the rod at a distance \( x \) from the pivot (midpoint). The length of this small element is \( dx \). The charge \( dQ \) on this small element can be expressed as:
\[ dQ = \lambda \cdot dx = \frac{q}{L} \cdot dx \]

### Step 3: Determine the Current Due to Rotation
As the rod rotates about the pivot with a frequency \( f \), the charge \( dQ \) moves in a circular path of radius \( x \). The time \( T \) taken to complete one full rotation is given by:
\[ T = \frac{1}{f} \]
The current \( dI \) due to the charge \( dQ \) can be calculated as:
\[ dI = \frac{dQ}{T} = dQ \cdot f = \left(\frac{q}{L} \cdot dx\right) \cdot f \]

### Step 4: Calculate the Magnetic Moment Contribution
The magnetic moment \( dM \) due to the small current element \( dI \) at a distance \( x \) from the pivot is given by:
\[ dM = \text{Area} \cdot dI = \pi x^2 \cdot dI \]
Substituting \( dI \) from the previous step:
\[ dM = \pi x^2 \cdot \left(\frac{q}{L} \cdot f \cdot dx\right) \]
Thus,
\[ dM = \frac{\pi q f}{L} x^2 \, dx \]

### Step 5: Integrate to Find Total Magnetic Moment
To find the total magnetic moment \( M \), we integrate \( dM \) from \( -\frac{L}{2} \) to \( \frac{L}{2} \):
\[ M = \int_{-\frac{L}{2}}^{\frac{L}{2}} \frac{\pi q f}{L} x^2 \, dx \]
The integral of \( x^2 \) is:
\[ \int x^2 \, dx = \frac{x^3}{3} \]
Evaluating the integral:
\[ M = \frac{\pi q f}{L} \left[ \frac{x^3}{3} \right]_{-\frac{L}{2}}^{\frac{L}{2}} = \frac{\pi q f}{L} \left( \frac{(\frac{L}{2})^3}{3} - \frac{(-\frac{L}{2})^3}{3} \right) \]
This simplifies to:
\[ M = \frac{\pi q f}{L} \cdot \frac{2 \cdot \left(\frac{L^3}{8}\right)}{3} = \frac{\pi q f L^2}{12} \]

### Final Answer
Thus, the magnetic moment of the rod system is:
\[ M = \frac{\pi q f L^2}{12} \]