An insulating rod of length L carries charge q distributed uniformly on it. The rod is pivoted at one end and is rotated at a frequency f about a fixed perpendicular axis. Then the magnetic dipole moment of the system is:

The correct option is:
$\frac{1}{3}\pi qf{L}^2$

Step-by-Step Derivation and Logical Reasoning:

1. Understanding the Physical Setup
We are given an insulating rod of length L carrying a total charge q distributed uniformly along its length. The rod rotates about a fixed axis passing through one of its ends (the pivot) and perpendicular to its length, with a rotational frequency f.

2. Finding the Linear Charge Density
Since the charge is distributed uniformly over the rod of length L, the linear charge density, represented by λ, is given by:
$\lambda =\frac{q}{L}$

3. Considering a Small Charge Element
Let us consider a small element of length dx at a distance x from the pivot. The charge dq contained in this small element is:
$dq=\lambda ·dx=\frac{q}{L}dx$

4. Determining the Effective Current
As the rod rotates with frequency f, this charge element dq moves in a circle of radius x. The time period of rotation is:
$T=\frac{1}{f}$
The equivalent current dI associated with this rotating charge element is the charge passing through a point per unit time:
$dI=\frac{dq}{T}=dq·f=\frac{qf}{L}dx$

5. Calculating the Magnetic Dipole Moment of the Element
The magnetic dipole moment dM of a current loop of area A carrying a current I is given by M = I · A.
For our circular path of radius x, the area enclosed is:
$A=\pi x^2$
Therefore, the magnetic dipole moment dM of the rotating charge element is:
$dM=dI·A=\left(\frac{qf}{L}dx\right)·\left(\pi x^2\right)$
Simplifying this expression yields:
$dM=\frac{\pi qf}{L}{x}^{2}dx$

6. Integrating to Find the Total Magnetic Dipole Moment
To find the total magnetic dipole moment M of the entire rod, we integrate dM from the pivot point (x = 0) to the outer end of the rod (x = L):
$M={\int }_0^L\frac{\pi qf}{L}{x}^{2}dx$
Since π, q, f, and L are constants, they can be pulled out of the integral:
$M=\frac{\pi qf}{L}{\left[}_{\frac{x^3}{3}}^{\right]}$
Evaluating this expression at the limits 0 and L:
$M=\frac{\pi qf}{L}·\frac{L^3}{3}$
Simplifying the powers of L gives us the final result:
$M=\frac{1}{3}\pi qf{L}^2$

Thus, we have successfully derived the magnetic dipole moment of the rotating rod.