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Step-by-Step Solution
Step 1: Understand the Physical Situation
We have a thin insulating rod of length $l$. The linear charge density on the rod varies with the coordinate $x$ (measured from the origin) as
$$\rho(x) = \frac{\rho_0\, x}{l}.$$
The rod is rotated about an axis perpendicular to the rod and passing through the origin (one of its ends, at $x=0$). The rod makes $n$ full rotations per second.
We want to find the time-averaged magnetic moment $\langle M \rangle$ of this rotating charged rod.
Step 2: Express the Relevant Quantities
Angular velocity: Since the rod makes $n$ rotations per second, its angular velocity is
$$\omega = 2\pi n.$$
Charge element: A small element of the rod of length $dx$ at a distance $x$ from the origin carries charge
$$dq = \rho(x)\,dx = \frac{\rho_0 \,x}{l}\,dx.$$
Current element viewpoint: As the rod rotates, each charge element effectively describes a circular motion about the origin, thereby constituting a circulating current. This rotating element will contribute to the magnetic moment.
Step 3: Compute Contribution to the Magnetic Moment
The magnetic moment $dM$ due to a small charge element $dq$ rotating in a circle of radius $x$ can be written (in time average) as:
$$dM = \text{(charge) } \times \text{(angular frequency) } \times \frac{\pi x^2}{2\pi}.$$
However, a more standard approach is to view each charge element as constituting a current $dI$ flowing in a loop of radius $x$, with area $\pi x^2$, and the rotation frequency $n$:
Current contribution: Each point charge $dq$ completes $n$ loops per second. Hence, equivalent current for this small charge element is
$$dI = dq \times n = n \frac{\rho_0 x}{l}\,dx.$$
Area of the loop: The circular loop traced out by the element at distance $x$ has area
$$A = \pi x^2.$$
Differential magnetic moment: For a current $I$ in a loop of area $A$, the magnetic moment is $I \cdot A$. Thus,
$$dM = \bigl(dI\bigr)\cdot A = \left(n \,\frac{\rho_0 x}{l} \,dx\right)\,\bigl(\pi x^2\bigr).$$
So,
$$dM = n \,\frac{\rho_0 \,\pi\, x^3}{l}\,dx.$$
Step 4: Integrate to Find Total Magnetic Moment
Integrate $dM$ from $x=0$ to $x=l$ to cover the entire rod:
$$
M = \int_{0}^{l} dM
= \int_{0}^{l} n \,\frac{\rho_0 \,\pi\, x^3}{l}\,dx
= \frac{n \rho_0 \pi}{l} \int_{0}^{l} x^3 \,dx.
$$
Now compute the integral:
$$
\int_{0}^{l} x^3 \,dx = \left[\frac{x^4}{4}\right]_{0}^{l} = \frac{l^4}{4}.
$$
So,
$$
M = \frac{n \rho_0 \pi}{l}\, \frac{l^4}{4}
= \frac{n \rho_0 \pi l^3}{4}.
$$
Step 5: Identify the Final Expression
Let us rewrite $\rho_0$ as $\rho$ (as given in the answer), giving:
$$
M = \frac{\pi}{4}\,n\,\rho\,l^3.
$$
Step 6: Conclude the Correct Option
Comparing with the provided options, we see that the time-averaged magnetic moment is:
$$
\boxed{\frac{\pi}{4}\,n\,\rho\,l^3},
$$
which matches Option (2) in the list.
