An insulating thin rod of length $$l$$ has a linear charge density $$\rho(x) = \rho_0 \frac{x}{l}$$ on it. The rod is rotated about an axis passing through the origin ($$x = 0$$) and perpendicular to the rod. If the rod makes $$n$$ rotations per second, then the time averaged magnetic moment of the rod is:

We place the thin insulating rod along the +x-axis with its left end at the origin, so the rod occupies $$0 \le x \le l$$. The rotation axis passes through the origin and is perpendicular to the rod; hence every small charge element at position $$x$$ moves in a circle of radius $$x$$.

The given linear charge density is

$$\rho(x)=\rho_0\frac{x}{l}\;,$$

so an element of length $$dx$$ contains charge

$$dq = \rho(x)\,dx = \rho_0\frac{x}{l}\,dx.$$

If the rod completes $$n$$ revolutions per second, its angular speed is

$$\omega = 2\pi n,$$

and one full revolution takes the period

$$T = \frac{1}{n}\;\text{s}.$$

For any charge that goes around once every $$T$$ seconds, the current associated with that charge is (definition of steady current)

$$I = \frac{q}{T} = q\,n.$$

Therefore the infinitesimal current produced by the element $$dq$$ is

$$dI = n\,dq = n\,\rho_0\frac{x}{l}\,dx.$$

Next, we need the magnetic moment produced by this small current loop. The standard formula for the magnetic moment of a current loop is

$$\mu = I \times \text{(area of the loop)}.$$

Here each element travels in a circle of radius $$x$$, whose area is

$$A = \pi x^{2}.$$

Hence the infinitesimal magnetic moment is

$$d\mu = dI \; A = \left(n\,\rho_0\frac{x}{l}\,dx\right)\!\left(\pi x^{2}\right) = n\,\pi\,\rho_0\,\frac{x^{3}}{l}\,dx.$$

To obtain the net magnetic moment of the entire rod we integrate from $$x=0$$ to $$x=l$$:

$$\mu = \int_{0}^{l} d\mu = n\,\pi\,\rho_0\,\frac{1}{l}\int_{0}^{l} x^{3}\,dx.$$

Carrying out the integral,

$$\int_{0}^{l} x^{3}\,dx = \left.\frac{x^{4}}{4}\right|_{0}^{l} = \frac{l^{4}}{4}.$$

Substituting this result, we get

$$\mu = n\,\pi\,\rho_0\,\frac{1}{l}\,\frac{l^{4}}{4} = \frac{\pi}{4}\,n\,\rho_0\,l^{3}.$$

Because the rod rotates uniformly, this value is already the time-averaged magnetic moment.

Hence, the correct answer is Option A.