Animated Solution for Physics - Magnetic Effects of Current: A thin circular disk of radius $R$ is uniformly charged with density $\sigma >0$ per unit area. The disk rotates about its axis with a uniform angular speed $\omega$. The magnetic moment of the disk is

Imagine a frisbee, but instead of plastic, it is made of pure electric charge, spinning rapidly around its central axis. Because the charge is in motion, it creates an electric current. And as we know from electromagnetism, any closed loop of current generates a magnetic moment. Our mission is to find the total magnetic moment of this entire spinning disk.

Since the velocity of the charge depends on its distance from the center ($v=\omega x$), charges near the edge move much faster than charges near the center. Because of this variation, we cannot just use a single, simple formula for the whole disk.

Instead, we must slice the disk into infinitely many thin, concentric rings. Let us focus on one such elemental ring of radius $x$ and infinitesimally small thickness $dx$.

First, we need to determine how much charge resides on this thin ring. The area of the ring is its circumference multiplied by its thickness, which is $dA=2\pi xdx$. Multiplying this area by the uniform surface charge density $\sigma$ gives us the charge $dq$ on the ring:

Now, this charge $dq$ is spinning around. A rotating charge is exactly what a current is! The equivalent current $dI$ is the charge $dq$ divided by the time period of one revolution, $T$. Since $T=\frac{2\pi }{\omega }$, we can write:

Substituting our expression for $dq$ into the current equation, the $2\pi$ terms beautifully cancel out, leaving us with a very neat expression for the elemental current:

We know the magnetic moment of any current loop is the current multiplied by the area it encloses. For our specific ring, the enclosed area is simply the area of a circle, $\pi x^2$. Multiplying our current $dI$ by this area gives the elemental magnetic moment $dM$:

To find the total magnetic moment of the entire disk, we just need to add up the magnetic moments of all these tiny rings. We do this by integrating $dM$ from the center, where $x=0$, all the way to the edge, where $x=R$:

The integral of $x^3$ is simply $\frac{x^4}{4}$. Applying the limits from $0$ to $R$, we get our final, elegant result:

The Gyromagnetic Ratio

Here is a brilliant trick for competitive exams: For any uniform charge distribution that shares the exact same geometry as its mass distribution, the ratio of its magnetic moment $M$ to its angular momentum $L$ is a constant known as the gyromagnetic ratio:

For a uniform solid disk, the angular momentum is $L=I\omega =\left(\frac{1}{2}m{R}^2\right)\omega$. The total charge is $q=\sigma \pi R^2$. Substituting these into the ratio:

Now, plug in $q=\sigma \pi R^2$:

In just two lines of algebra, we arrive at the exact same result without performing a single integration!