Question

An insulating thin rod of length $\ell$ has a x linear charge density $p(x) = \rho_0 \frac{x}{\ell}$ 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 :

• A. $\frac{\pi }{4} n \rho \ell^3$
• B. $ n \rho \ell^3$
• C. $ \pi n \rho \ell^3$
• D. $\frac{\pi }{3} n \rho \ell^3$

Answer 1

The Correct Option is A

The problem involves calculating the time-averaged magnetic moment of a rotating charged rod. We are given a linear charge density and need to find the average magnetic moment when the rod rotates about an axis perpendicular to its length.

1. First, let's consider the linear charge density given: p(x) = \rho_0 \frac{x}{\ell}. This implies that the charge density varies linearly along the rod.
2. The total charge dq on an infinitesimal segment dx of the rod is:
   dq = p(x) \cdot dx = \rho_0 \frac{x}{\ell} \cdot dx
3. We need to find the contribution to the magnetic moment from this segment. Using the formula for magnetic moment (with respect to an area swept by the rotation) for a small segment: d\mu = \frac{1}{2} \times \text{(Current)} \times \text{(Area)}
4. The current I associated with the rotating rod is given by its charge passing per unit time, and for n rotations per second, the effective current due to this rotation of charge is:
   dI = n \cdot dq = n \cdot \rho_0 \frac{x}{\ell} \cdot dx
5. The area traced out by the charge dq upon rotating is a circle of radius x with area \pi x^2. Thus:
   d\mu = \frac{1}{2} \cdot dI \cdot \pi x^2 = \frac{1}{2} \cdot n \rho_0 \frac{x}{\ell} \cdot \pi x^2 \cdot dx
   d\mu = \frac{1}{2} n \pi \rho_0 \frac{x^3}{\ell} \cdot dx
6. To find the total magnetic moment \mu of the entire rod, integrate over the entire length of the rod from 0 to \ell:
   \mu = \int_0^\ell \frac{1}{2} n \pi \rho_0 \frac{x^3}{\ell} \, dx
   \mu = \frac{n \pi \rho_0}{2 \ell} \int_0^\ell x^3 \, dx
7. The integral of x^3 is:
   \int_0^\ell x^3 \, dx = \left[\frac{x^4}{4}\right]_0^\ell = \frac{\ell^4}{4}
8. Substitute back to find \mu:
   \mu = \frac{n \pi \rho_0}{2 \ell} \cdot \frac{\ell^4}{4} = \frac{n \pi \rho_0 \ell^3}{8}
9. By comparing to the options given and simplifying, the correct answer matches:
   \mu = \frac{\pi }{4} n \rho \ell^3, which is the desired result.

Thus, the time-averaged magnetic moment of the rod is \frac{\pi }{4} n \rho \ell^3.