Animated Solution for Physics - Magnetic Effects of Current: In a hydrogen atom, the electron moves in an orbit of radius $0.5\text{ A˚}$ making $10^{16}$ revolutions per second. The magnetic moment associated with the orbital motion of the electron is ......

Imagine you are looking deep into the heart of a hydrogen atom. What do you see? A single, solitary electron whizzing around a central proton at mind-boggling speeds. It’s not just moving; it’s completing a staggering $10^{16}$ revolutions every single second!

This isn't just a mechanical orbit. In the realm of physics, a moving charge is the fundamental definition of an electric current. Even though it's just one electron, its rapid, repetitive motion creates a steady loop of current.

To understand the magnetic properties of this atom, we first need to figure out how much "current" this single electron represents.

Current, denoted by $i$, is defined as the rate of flow of charge. If we stand at one point on the electron's circular track, how much charge passes by us in one second?

Since the electron passes by $f$ times per second (where $f$ is the frequency of revolution), the total charge passing per second is the charge of the electron $q$ multiplied by the frequency $f$.

Therefore, the equivalent current is given by:

This simple yet profound realization bridges the gap between kinematics and electromagnetism.

Now, any closed loop carrying an electric current acts like a tiny bar magnet. It generates a magnetic field and possesses a property called a magnetic moment ($M$).

The magnitude of the magnetic moment for a planar current loop is the product of the current $i$ and the area $A$ enclosed by the loop:

For our electron in a circular orbit of radius $r$, the area is simply the area of a circle:

Let's bring it all together. By substituting our expression for the equivalent current and the area into the magnetic moment formula, we get our master equation for the orbital magnetic moment of the electron:

This beautiful equation tells us exactly how the magnetic moment depends on the physical parameters of the orbit.

Now, it's time to plug in the numbers. We must be extremely careful with our units, ensuring everything is in standard SI units before we calculate.

We are given: - The radius of the orbit, $r=0.5\text{ A˚}=0.5\times 10^{-10}\text{ m}$ - The frequency of revolution, $f=10^{16}\text{ Hz}$ - The elementary charge, $q=1.6\times 10^{-19}\text{ C}$

Substituting these into our master equation:

First, let's square the radius:

Now, let's group the numbers and the powers of ten:

Using the approximation $\pi \approx 3.14159$:

Rounding to appropriate significant figures, we get our final answer:

$$M=1.26\times 10^{-23}{\text{ A-m}}^2$$

This calculation isn't just a textbook exercise; it's a gateway to quantum mechanics. The magnetic moment we just calculated is closely related to the Bohr magneton, a fundamental physical constant.

Furthermore, if you calculate the orbital angular momentum $L=mvr$ of this electron, you'll find that the ratio of the magnetic moment to the angular momentum $\frac{M}{L}$ is exactly $\frac{q}{2m}$. This constant ratio, known as the gyromagnetic ratio, is a cornerstone of atomic physics and plays a crucial role in phenomena like magnetic resonance imaging (MRI).