Diamagnetism

Langevin Theory of Diamagnetism

Conditions: Spherical charge distribution of Z electrons per atom, weak external magnetic field B

• An electron of charge -e and mass m moving in a circular orbit of radius r experiences a change in angular frequency Δω equal to the Larmor frequency given by Δ$\omega = \frac{eB}{2m}$.
• This change in frequency induces an additional circular current $i = -\frac{e Deltaomega}{2pi} = -\frac{e^2 B}{4pi m}$.
• The induced magnetic moment μ_ind for a single electron orbit of area $A = \pi\langle r^{2} \rangle$ is given by μ_ind = i $A = -\frac{e^2 B}{4pi m} \pi \langle r^{2} \rangle = -\frac{e^2 B}{4m} \langle r^{2} \rangle$, where <r²> is the mean square projection of the orbit perpendicular to the field.
• Summing over Z electrons per atom and multiplying by the number density of atoms N gives the magnetization $M = N \sum \mu_{ind} = -\frac{Ne^2 Z \langle r^{2} \rangle B}{6m}$.
• Magnetic susceptibility $\chi = \frac{M}{H} = \mu_0 \frac{M}{B} (since B \approx \mu_0 H$ for weak fields) yields the final diamagnetic susceptibility expression.

Result: $\chi = -\frac{\mu_0 Ne^2 Z \langle r^{2} \rangle}{6m}$

Behavior of Diamagnetic Rod in a Non-Uniform Field

Conditions: Rod of volume V and susceptibility χ suspended in magnetic field gradient

• The potential energy U of a magnetic dipole of moment M in a magnetic field B is $U = -M \cdot B$.
• The induced magnetic moment per unit volume is $I = \chi H \approx \frac{\chi B}{\mu_0}$.
• The total induced magnetic moment is $M = V I = \frac{V \chi B}{\mu_0}$.
• Substituting M into the potential energy equation yields $U = -\frac{V \chi B^2}{\mu_0}$.
• The force F acting on the rod along the field gradient is $F = -\frac{dU}{ds} = \frac{V \chi}{\mu_0} \frac{d}{ds}(B^2)$. Since χ is negative, the force acts in the direction of decreasing magnetic field strength.

Result: $F = \frac{V \chi}{\mu_0} B \frac{dB}{ds} (pointing towards weaker field regions)$

Magnetization (I) vs Magnetic Field Intensity (H)

Draw: A straight line passing through the origin in the fourth quadrant, with a negative slope equal to χ.

Interpret: Shows that magnetization is directly proportional to applied field intensity but directed oppositely, resulting in negative susceptibility.

Exam use: Tested in MCQs to identify material type from I-H or M-H slope characteristics.

Susceptibility (&chi;) vs Temperature (T)

Draw: A horizontal straight line parallel to the temperature axis, indicating zero variation with T.

Interpret: Demonstrates the complete temperature independence of diamagnetic susceptibility.

Exam use: Differentiating diamagnetism from paramagnetism and ferromagnetism in conceptual questions.

Quincke's Method / Guoy's Balance Experiment Principle

Draw: A U-tube setup showing depression of a liquid level or apparent weight loss of a diamagnetic sample placed between pole pieces of an electromagnet.

Interpret: Measures the force exerted on a diamagnetic material in a magnetic field gradient to determine susceptibility.

Exam use: Experimental basis questions regarding measurement of magnetic susceptibility in PGT exams.

A diamagnetic substance has a magnetic susceptibility of -1.5 × 10^-5. Calculate its relative permeability.

• Recall the relation between relative permeability and magnetic susceptibility: μ_r = 1 + χ.
• Substitute the given value of susceptibility: μ_r = 1 + (-1.5 × 10^-5).
• Calculate the final numerical value: μ_r = 1 - 0.000015 = 0.999985.

Answer: 0.999985

Estimate the atomic diamagnetic susceptibility of a substance given that the atomic number density $N = 5 × 10^{28} m^{-3}$, Z = 10, and mean square radius <r²$ \rangle = 1 × 10^{-20} m^{2}$. Take μ₀ = 4π × 10^-7 T m A^-1, $e = 1.6 × 10^{-19} C$, $m = 9.1 × 10^{-31} kg$.

• Use Langevin formula: $\chi = -\frac{\mu_0 Ne^2 Z \langle r^{2} \rangle}{6m}$.
• Substitute the numerical values: $\chi = -\frac{(4pi \times 10^{-7}) \times (5 \times 10^{28}) \times (1.6 \times 10^{-19})^2 \times 10 \times (1 \times 10^{-20})}{6 \times (9.1 \times 10^{-31})}$.
• Calculate numerator: 4π × 10^-7 × 5 × 10²⁸ × 2.56 × 10^-38 × 10 × 10^-20 = 4π × 1.28 × 10^-8 ≈ 1.609 × 10^-7.
• Calculate denominator: 6 × 9.1 × 10^-31 = 5.46 × 10^-30.
• Divide and evaluate: $\chi = -\frac{1.609 \times 10^{-7}}{5.46 \times 10^{-30}} \times 10^{-20} \text{ (re-evaluating powers carefully)} -> Let us compute directly: \chi = - \frac{4pi \times 10^{-7} \times 5 \times 10^{28} \times 2.56 \times 10^{-38} \times 10 \times 10^{-20}}{5.46 \times 10^{-30}} = -1.96 \times 10^{-6}$.

Answer: -1.96 × 10^-6

A small bismuth sphere of volume 10^-6 m^-3 and magnetic susceptibility -1.6 × 10^-5 is placed in a non-uniform magnetic field where the product B(dB/dx) is 10² T² m^-1. Calculate the magnetic force acting on the sphere. (μ₀ = 4π × 10^-7 T m A^-1)

• Use the force formula for a magnetic sample in a field gradient: $F = \frac{V \chi}{\mu_0} B \frac{dB}{dx}$.
• Substitute $V = 10^{-6} m^{-3}$, $\chi = -1.6 × 10^{-5}$, B(dB/dx) = 10² T² m^-1, and μ₀ = 4π × 10^-7.
• $F = \frac{10^{-6} \times (-1.6 \times 10^{-5}) \times 10^{2}}{4pi \times 10^{-7}}$.
• Simplify the numerator: -1.6 × 10^-9.
• Simplify the denominator: 4π × 10^-7 ≈ 1.2566 × 10^-6.
• $F = \frac{-1.6 \times 10^{-9}}{1.2566 \times 10^{-6}} = -1.27 \times 10^{-3} \text{ N}$.

Answer: -1.27 × 10^-3 N (repulsive force away from higher field region)