An insulating thin rod of length l has a linear charge density rho(x) = rho_0 (x/l) on it. The rod i

Q: Why do we take current as $I = qn$?

Current is the rate of flow of charge ($I = q/t$). If a charge $q$ completes one rotation in time $T$, the current is $q/T$. Since frequency $n = 1/T$, the current becomes $I = qn$.

Q: Why can't we just use the total charge and the total area?

Because different parts of the rod are at different distances from the axis of rotation. They trace out circles of different areas. Therefore, you must integrate the magnetic moment of each infinitesimally small element.

Q: What does $\rho_0$ represent in the equation $\rho(x) = \rho_0 \frac{x}{l}$?

$\rho_0$ is a constant that represents the maximum linear charge density, which occurs at the very end of the rod where $x = l$.

Q: Why did the final answer use $\rho$ instead of $\rho_0$?

It's a notational choice in the options. By substituting $x=l$ into the density function, we see that the density at the end of the rod is exactly $\rho_0$. The options simply use $\rho$ to denote this value.

Q: What would happen if the rod was rotated about its center instead of one end?

You would set up the same integral, but your limits of integration would change from $-l/2$ to $l/2$. You would also need to be careful with the area, as radius is $|x|$, making the area $\pi x^2$, which is symmetric.

Question

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

Options:

(A) $n\rho l^3$

(B) $\pi n\rho l^3$

(C) $\frac{\pi}{3} n\rho l^3$

(D) $\frac{\pi}{4} n\rho l^3$

Sigma Percentile · Accepted Answer

The problem asks us to find the time-averaged magnetic moment of a rotating rod with a non-uniform linear charge density. This is a classic application of the concept that a moving charge constitutes an electric current.

### Analyzing the Setup

Imagine a thin rod of length $l$ lying along the x-axis, with one end at the origin. The rod is rotating about the y-axis (which passes through the origin and is perpendicular to the rod) with a frequency of $n$ rotations per second.

The charge density on the rod is not uniform; it's given by $\rho(x) = \rho_0 \frac{x}{l}$. This means the charge is more concentrated towards the outer end of the rod.

### The Master Equation

When a charge $q$ moves in a circle with frequency $n$, it crosses any given point on the circle $n$ times per second. Therefore, the equivalent current $I$ is simply the total charge passing per second:
$$I = qn$$

The magnetic moment $M$ of a current loop is the product of the current and the area $A$ enclosed by the loop:
$$M = IA$$

Since the charge is distributed continuously along the rod, we cannot use a single value for $q$ or $A$. Instead, we must consider an infinitesimally small element of the rod and integrate.

### Setting up the Integration

Let's take a small element of length $dx$ at a distance $x$ from the origin. 
The charge on this small element is:
$$dq = \rho(x) dx = \rho_0 \frac{x}{l} dx$$

As the rod rotates, this small element $dx$ traces out a circle of radius $x$. The equivalent current $dI$ due to this rotating element is:
$$dI = dq \cdot n = n \rho_0 \frac{x}{l} dx$$

The area $A$ of the circular path traced by this element is:
$$A = \pi x^2$$

Now, the elemental magnetic moment $dm$ produced by this small portion of the rod is:
$$dm = dI \cdot A = \left( n \rho_0 \frac{x}{l} dx \right) (\pi x^2) = \frac{\pi n \rho_0}{l} x^3 dx$$

### Final Calculation

To find the total magnetic moment $M$ of the entire rod, we integrate $dm$ from $x = 0$ to $x = l$:
$$M = \int_0^l dm = \int_0^l \frac{\pi n \rho_0}{l} x^3 dx$$

Since $\frac{\pi n \rho_0}{l}$ is a constant, we can pull it out of the integral:
$$M = \frac{\pi n \rho_0}{l} \int_0^l x^3 dx$$

The integral of $x^3$ is $\frac{x^4}{4}$. Applying the limits from $0$ to $l$:
$$M = \frac{\pi n \rho_0}{l} \left[ \frac{x^4}{4} \right]_0^l = \frac{\pi n \rho_0}{l} \left( \frac{l^4}{4} - 0 \right) = \frac{\pi n \rho_0 l^3}{4}$$

The problem's options use $\rho$ instead of $\rho_0$. Notice that at the end of the rod ($x = l$), the charge density is $\rho(l) = \rho_0 \frac{l}{l} = \rho_0$. If we denote this maximum charge density simply as $\rho$, our expression becomes:
$$M = \frac{\pi}{4} n \rho l^3$$

**This perfectly matches option (d).**

Select Answer:

Visualized Solution

Visualizing the Setup

Magnetic Moment of Rotating Charge

Elemental Charge and Current

Elemental Magnetic Moment

Integration Setup

Evaluating the Integral

Final Substitution

The Way Forward

The Sigma Insight: Magnetic Moment of Current Loop

Analyzing the Setup

The Master Equation

Setting up the Integration

Final Calculation

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