An electron in the ground state of hydrogen atom is revolving in anti-clockwise direction in a circu

Q: Why is the current direction opposite to the electron's velocity?

By convention, electric current is defined as the flow of positive charge. Since an electron carries a negative charge, its physical motion in one direction constitutes a conventional current in the exact opposite direction.

Q: Why can we use the gyromagnetic ratio here?

The gyromagnetic ratio $\frac{M}{L} = \frac{q}{2m}$ is a universal constant for any classical system where charge and mass are distributed proportionally. Since the electron is a single point particle, its charge-to-mass ratio perfectly dictates the ratio of its magnetic moment to angular momentum.

Q: Does the angle being $30^\circ$ or $150^\circ$ change the torque?

No, it doesn't change the magnitude. The torque magnitude depends on $\sin\theta$. Since $\sin(30^\circ) = \sin(150^\circ) = 0.5$, the resulting twisting force is identical regardless of whether you measure the angle from the positive or negative normal.

Q: Why do we use Bohr's theory for this problem?

The problem explicitly states the electron is in the 'ground state of a hydrogen atom'. Bohr's model provides the exact quantization condition ($L = \frac{nh}{2\pi}$) needed to find the angular momentum for the $n=1$ state.

Q: What is the physical meaning of this torque?

The torque attempts to align the electron's magnetic moment with the external magnetic field. This is the fundamental mechanism behind phenomena like Larmor precession and the Zeeman effect in quantum mechanics.

Question

An electron in the ground state of hydrogen atom is revolving in anti-clockwise direction in a circular orbit of radius $R$.
(a) Obtain an expression for the orbital magnetic moment of the electron.
(b) The atom is placed in a uniform magnetic induction $\mathbf{B}$ such that the normal to the plane of electron's orbit makes an angle of $30^{\circ}$ with the magnetic induction. Find the torque experienced by the orbiting electron.

Sigma Percentile · Accepted Answer

Imagine you are shrinking down to the atomic scale, staring at a hydrogen atom. You see an electron zipping around the nucleus in a perfect circular orbit. But this isn't just a moving particle; it's a tiny, continuous loop of electric current! And wherever there is a current loop, there is a magnetic field. The atom itself is a miniature magnet.

### The Setup: An Atom as a Magnet

To understand this atomic magnet, we first need to figure out how much 'current' this single electron is generating. Current is simply the rate of flow of charge. So, if we take the charge of the electron ($e$) and divide it by the time it takes to complete one full orbit (the time period $T$), we get our equivalent current:

$i = \frac{e}{T}$

But how long does one orbit take? The time period is the distance traveled divided by the speed: $T = \frac{2\pi R}{v}$.

To find the velocity $v$, we turn to Niels Bohr. Bohr's first postulate tells us that the angular momentum of an electron in the ground state ($n=1$) is quantized:

$mvR = \frac{h}{2\pi}$

From this, we can isolate the velocity:

$v = \frac{h}{2\pi mR}$

### The Master Equation: Magnetic Moment

Now, let's plug this velocity back into our time period equation:

$T = \frac{2\pi R}{\frac{h}{2\pi mR}} = \frac{4\pi^2 mR^2}{h}$

With the time period in hand, our equivalent current becomes:

$i = \frac{e}{\frac{4\pi^2 mR^2}{h}} = \frac{eh}{4\pi^2 mR^2}$

The strength of our atomic magnet is measured by its **magnetic moment** ($M$), which is the current multiplied by the area of the loop ($A = \pi R^2$):

$M = iA = \left(\frac{eh}{4\pi^2 mR^2}ight)(\pi R^2)$

Notice how beautifully the $R^2$ and $\pi$ terms cancel out! We are left with a pristine, fundamental constant:

$M = \frac{eh}{4\pi m}$

### The Elegant Shortcut: Gyromagnetic Ratio

I know that derivation took a few steps, but let's take a breath... there is a much faster, more elegant way to see this.

For any revolving charge, the ratio of its magnetic moment to its angular momentum is a universal constant known as the **gyromagnetic ratio**. It is simply the charge divided by twice the mass:

$\frac{M}{L} = \frac{e}{2m}$

Since we already know from Bohr that the angular momentum $L = \frac{h}{2\pi}$, we can instantly find the magnetic moment:

$M = \left(\frac{e}{2m}ight)\left(\frac{h}{2\pi}ight) = \frac{eh}{4\pi m}$

Physics is incredibly consistent. Two different paths, one identical, beautiful result.

### The Twist: Torque in a Magnetic Field

Now for the grand finale. We place our tiny atomic magnet into an external magnetic field $\mathbf{B}$. The problem states that the normal to the orbit makes a $30^\circ$ angle with the magnetic field.

Because the electron is negatively charged, its conventional current flows opposite to its motion. This means its magnetic moment vector $\mathbf{M}$ points exactly opposite to the geometric normal of the orbit.

When a magnetic dipole is placed in a magnetic field, it experiences a twisting force—a torque ($\boldsymbol{	au}$)—that tries to align it with the field. The formula for this torque is the cross product:

$\boldsymbol{	au} = \mathbf{M} 	imes \mathbf{B}$

The magnitude of this torque is:

$|\boldsymbol{	au}| = MB \sin	heta$

Whether you consider the angle to be $30^\circ$ (with the normal) or $150^\circ$ (with the magnetic moment), the sine value is exactly the same: $\sin(30^\circ) = \frac{1}{2}$.

Substituting our magnetic moment and the sine value, we get our final answer:

$	au = \left(\frac{eh}{4\pi m}ight) B \left(\frac{1}{2}ight) = \frac{ehB}{8\pi m}$

By the right-hand rule, this torque acts perpendicular to both the magnetic moment and the magnetic field. You've just calculated the fundamental twisting force that governs phenomena like Larmor precession in quantum mechanics. Incredible work!

Visualized Solution

$i = \frac{e}{T}$

$T = \frac{2 π R}{v}$

$M = i A$

$\frac{M}{L} = \frac{q}{2 m}$

$External Field$

$τ = M \times B$

$τ = \frac{e h B}{8 π m}$

The Sigma Insight: Magnetic Moment of Current Loop

The Setup

The Master Equation

The Elegant Shortcut

The Twist

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } i=Te​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T=v2πR​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M=iA

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } LM​=2mq​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } External Field

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } τ=M×B

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } τ=8πmehB​

The Sigma Insight: Magnetic Moment of Current Loop

The Setup

The Master Equation

The Elegant Shortcut

The Twist

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A loop carrying current I lies in the x-y plane as shown in the figure. The unit vector k^ is coming out of the plane of the paper. The magnetic moment of the current loop is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A wire of length L metre carrying a current i ampere is bent in the form of circle. The magnitude of its magnetic moment is ...... in MKS units.

$i = \frac{e}{T}$

$T = \frac{2 π R}{v}$

$M = i A$

$\frac{M}{L} = \frac{q}{2 m}$

$External Field$

$τ = M \times B$

$τ = \frac{e h B}{8 π m}$

A loop carrying current $I$ lies in the $x$ - $y$ plane as shown in the figure. The unit vector $\hat{k}$ is coming out of the plane of the paper. The magnetic moment of the current loop is

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