Animated Solution for Physics - Magnetic Effects of Current: A particle of charge $q$ and mass $m$ moves in a circular orbit of radius $r$ with angular speed $\omega$. The ratio of the magnitude of its magnetic moment to that of its angular momentum depends on

Imagine a tiny particle, carrying a charge $q$ and possessing a mass $m$. It is not just sitting there; it is performing a continuous, elegant dance in a perfect circle of radius $r$.

This particle is moving with a constant angular speed $\omega$. As it moves, it is doing two very important things simultaneously: it is creating a magnetic field, and it is carrying momentum around the circle.

First, let's look at the magnetic aspect. A moving charge is the very definition of an electric current.

Even though it is just a single particle, its continuous circular motion creates an effective current loop. The equivalent current $i$ is the total charge passing a point per unit time.

Since the particle completes one revolution in a time period $T=\frac{2\pi }{\omega }$, the effective current is:

Now, any current loop possesses a magnetic moment $M$, which is the product of the current and the area $A$ of the loop. The area of our circular orbit is simply $\pi r^2$.

Multiplying these together, we find the magnetic moment:

Now, let's switch gears and look at the mechanics. Our particle has mass and is moving in a circle, which means it has angular momentum $L$.

The angular momentum of a particle in circular motion is the product of its moment of inertia $I$ and its angular velocity $\omega$.

For a point mass $m$ at a distance $r$ from the axis of rotation, the moment of inertia is $I=m{r}^2$.

Therefore, the angular momentum is:

We have our two key quantities: the magnetic moment $M$ and the angular momentum $L$. The question asks for the ratio of their magnitudes.

Let's divide $M$ by $L$:

Look closely at this expression. The radius squared ($r^2$) appears in both the numerator and the denominator. The angular speed ($\omega$) also appears in both.

They beautifully cancel each other out!

This final result is profound. The ratio $\frac{M}{L}$ depends only on the charge $q$ and the mass $m$ of the particle.

It is completely independent of how fast the particle is spinning ($\omega$) or how wide its orbit is ($r$).

This constant ratio is known in physics as the gyromagnetic ratio. It is a fundamental property that bridges the gap between the mechanical rotation of a particle and its magnetic properties, playing a crucial role in everything from classical electromagnetism to quantum mechanics!