Atom as a Magnetic Dipole

An electron revolving around the nucleus in an atom is not just executing circular motion — it also constitutes a tiny current loop. Since a current loop produces a magnetic field, the orbiting electron behaves like a microscopic magnetic dipole.

Analogy: Think of the electron's orbit as a small circular wire loop carrying current — similar to a coil in an electromagnet. Just as current in a coil creates a magnetic field, the electron's motion creates a magnetic moment.

Step 1: Current due to an orbiting electron

An electron completing one revolution in a time period T constitutes current: I = \[\frac {e}{T}\]

Step 2: Magnetic moment of current loop

For any current loop, magnetic moment = current × area: μ = IA = I(πr²)

Step 3: Express in terms of velocity

Since T = \[\frac {2πr}{v}\], substituting gives: \[{\mu = \frac{evr}{2}}\]   ...(Equation 1)

Step 4: Relate to angular momentum

Angular momentum of electron: L = m_evr

Dividing Equation (1) by L:

Important Point: The term \[\frac {e}{2m_e}\]​ is called the gyromagnetic ratio — a constant for all electrons.

Step 5: Apply Bohr's Quantisation Condition

Bohr postulated: L = \[\frac {nh}{2π}\]​, where n = 1, 2, 3...

Substituting into Equation (2): \[{\mu = n\mu_B}\]   ...(Equation 3)

where: \[\mu_B = \frac{eh}{4\pi m_e} = 9.27 \times 10^{-24}\ \text{A·m}^2\]

Problem: Two electrons revolve in circular orbits of radii 5.3 × 10⁻¹¹ m and 2.1 × 10⁻¹⁰ m in opposite directions, each with speed 2.2 × 10⁶ m/s. Find the net magnetic moment.

Given:

• r₁ = 5.3 × 10⁻¹¹ m, r₂ = 2.1 × 10⁻¹⁰ m
• v = 2.2 × 10⁶ m/s (same for both)
• Directions: opposite

Formula Used: μ = \[\frac {evr}{2}\]

Step 1 - Moment of electron 1: \[\mu_1 = \frac{(1.6\times10^{-19})(2.2\times10^6)(5.3\times10^{-11})}{2} = 9.32 \times 10^{-24}\ \text{A·m}^2\]

Step 2 - Moment of electron 2: \[\mu_2 = \frac{(1.6\times10^{-19})(2.2\times10^6)(2.1\times10^{-10})}{2} = 3.696 \times 10^{-23}\ \text{A·m}^2\]

Step 3 - Net moment (opposite directions → subtract):

Answer: Net magnetic moment = 2.764 × 10⁻²³ A·m²