(15.1) Diamagnetic Susceptibility and Levitation Field Strength

(a) Estimate the diamagnetic susceptibility of a typical solid

We use the expression from equation 12.15, which models the diamagnetic susceptibility as:

\[ \chi_m = -\mu_0 \frac{q^2 Z r^2}{4 m_e V} \]

Using typical values for a solid:

• \( \mu_0 = 1.26 \times 10^{-6} \, \text{N/A}^2 \)
• \( q = 1.6 \times 10^{-19} \, \text{C} \)
• \( Z = 1 \)
• \( r = 10^{-10} \, \text{m} \)
• \( m_e = 9.1 \times 10^{-31} \, \text{kg} \)
• \( V = 10^{-30} \, \text{m}^3 \)

Substituting into the formula:

\[ \chi_m = -1.26 \times 10^{-6} \cdot \frac{(1.6 \times 10^{-19})^2 \cdot 1 \cdot (10^{-10})^2}{4 \cdot 9.1 \times 10^{-31} \cdot 10^{-30}} \]

\[ \chi_m = -1.26 \times 10^{-6} \cdot \frac{2.56 \times 10^{-38} \cdot 10^{-20}}{3.64 \times 10^{-60}} \approx -1.26 \times 10^{-6} \cdot 70.3 \approx -8.86 \times 10^{-5} \]

Final answer: \( \boxed{\chi_m \approx -8.9 \times 10^{-5}} \)

(15.1b) Estimate the field strength needed to levitate a frog

We are given the force from diamagnetic levitation:

\[ F = -\frac{V}{\mu_0} \chi_m H \frac{dH}{dz} \]

Assume the following properties for the frog:

• Mass \( m = 0.1 \, \text{kg} \)
• Volume \( V = 10^{-4} \, \text{m}^3 \)
• Height \( z = 0.1 \, \text{m} \Rightarrow \frac{dH}{dz} \approx \frac{H}{z} \)
• Diamagnetic susceptibility \( \chi_m = -8.9 \times 10^{-5} \)
• Permeability of free space \( \mu_0 = 1.26 \times 10^{-6} \, \text{N/A}^2 \)

The gravitational force on the frog is:

\[ F = mg = 0.1 \cdot 9.8 = 0.98 \, \text{N} \]

Solving for the magnetic field intensity \( H \):

\[ H = \sqrt{\frac{F z}{V \mu_0 \chi_m}} = \sqrt{\frac{0.1 \cdot 9.8 \cdot 0.1}{10^{-4} \cdot 1.26 \times 10^{-6} \cdot (-8.9 \times 10^{-5})}} \]

\[ H = \sqrt{ \frac{0.098}{1.1214 \times 10^{-14}} } \approx \sqrt{8.74 \times 10^{12}} \approx 2.96 \times 10^6 \, \text{A/m} \]

Convert to magnetic field strength:

\[ B = \mu_0 H = 1.26 \times 10^{-6} \cdot 2.96 \times 10^6 \approx 3.73 \, \text{T} \]

Final answer: \( \boxed{B \approx 3.7 \, \text{T}} \)

(13.2) Estimate the size of the direct magnetic interaction energy between two adjacent free electrons in a solid, and compare this to the size of their electrostatic interaction energy.

We are given that the field of a magnetic dipole \( \vec{m} \) is:

\[ \vec{B} = \frac{\mu_0}{4\pi} \left[ \frac{3\hat{x}(\vec{m} \cdot \hat{x}) - \vec{m}}{|\vec{x}|^3} \right] \]

The interaction energy between two magnetic dipoles \( \vec{m}_1 \) and \( \vec{m}_2 \) is approximately \( \vec{m}_1 \cdot \vec{B}_2 \). This is maximized when the dipoles are aligned and separated by the minimum distance.

Assuming the distance between them is 1 Å = \( 10^{-10} \, \text{m} \), and each magnetic dipole moment is approximately the Bohr magneton \( \mu_B = 9.28 \times 10^{-24} \, \text{J/T} \), the magnetic interaction energy is:

\[ E_m = \frac{\mu_0 m^2}{\pi r^3} = \frac{1.26 \times 10^{-6} \cdot (9.28 \times 10^{-24})^2}{\pi \cdot (10^{-10})^3} \approx 3.5 \times 10^{-23} \, \text{J} \]

Next, we compute the electrostatic interaction energy between two electrons:

\[ E_e = \frac{q^2}{4\pi \varepsilon_0 r} = \frac{(1.6 \times 10^{-19})^2}{4\pi \cdot 8.85 \times 10^{-12} \cdot 10^{-10}} \approx 2.3 \times 10^{-18} \, \text{J} \]

Conclusion: The electrostatic interaction between two electrons is about 5 orders of magnitude stronger than their magnetic interaction:

\[ \frac{E_e}{E_m} \approx \frac{2.3 \times 10^{-18}}{3.5 \times 10^{-23}} \approx 6.6 \times 10^4 \]

This shows that direct magnetic interactions between electrons are generally negligible compared to electrostatic forces.

(13.3) Using the equation for the energy in a magnetic field, describe why:

The attraction between a permanent magnet and an unmagnetized piece of ferromagnetic material (like iron) comes from the way magnetic fields store energy.

The energy density in a region with electric and magnetic fields is given by:

\[ U = \frac{1}{2} (\vec{E} \cdot \vec{D} + \vec{B} \cdot \vec{H}) \]

In our case, there’s no electric field involved, so \( \vec{E} = 0 \). That simplifies the equation to:

\[ U = \frac{1}{2} \vec{B} \cdot \vec{H} \]

Now, from the magnetic material relationship:

\[ \vec{B} = \mu \vec{H} \Rightarrow \vec{H} = \frac{\vec{B}}{\mu} \]

Substituting this into the energy density formula, we get:

\[ U = \frac{1}{2} \vec{B} \cdot \left( \frac{\vec{B}}{\mu} \right) = \frac{1}{2\mu} B^2 \]

To find the total magnetic energy in space, we integrate over the volume:

\[ U_{\text{total}} = \frac{1}{2\mu} \int B^2 \, dV \]

This tells us that magnetic energy is smaller when \( \mu \) (the magnetic permeability) is large. Ferromagnetic materials like iron have a very high \( \mu \), which means they can "soak up" magnetic field lines easily.

So when a magnet is near a ferromagnet, the magnetic field energy becomes smaller if more field lines go into the metal. Nature always tries to minimize energy, so the magnet is pulled toward the metal — that’s the attractive force!

(b) The opposite poles of permanent magnets attract each other.

Why do opposite ends of magnets pull toward each other?

We begin with the relationship between the magnetic field \( \vec{B} \), the magnetic field intensity \( \vec{H} \), and the magnetization \( \vec{M} \) in materials:

\[ \vec{B} = \mu (\vec{H} + \vec{M}) \]

This equation tells us how the total magnetic field \( \vec{B} \) is made up of the external field \( \vec{H} \) and the internal magnetization \( \vec{M} \), scaled by the permeability \( \mu \).

Now we consider the energy density stored in the magnetic field, given by:

\[ U = \frac{1}{2} \vec{B} \cdot \vec{H} \]

Substituting the expression for \( \vec{B} \):

\[ U = \frac{1}{2} \mu (\vec{H} + \vec{M}) \cdot \vec{H} \]

Distributing the dot product gives:

\[ U = \frac{1}{2} \mu \left( H^2 + \vec{M} \cdot \vec{H} \right) \]

From this, we see that the total energy depends on both the strength of the external field \( H \) and how aligned the magnetization \( \vec{M} \) is with that field.

When the magnetization \( \vec{M} \) and the field \( \vec{H} \) are anti-aligned (i.e., pointing in opposite directions), the dot product \( \vec{M} \cdot \vec{H} \) becomes negative. This reduces the total energy.

Since systems naturally evolve toward configurations that minimize energy, the magnets experience an attractive force when opposite poles (which correspond to anti-aligned \( \vec{M} \) and \( \vec{H} \)) face each other.

Conclusion: Opposite poles of permanent magnets attract because this configuration reduces the magnetic energy of the system. The interaction aligns internal magnetizations and external fields in a way that minimizes the total energy, resulting in an attractive force.

(13.4) Estimate the saturation magnetization for iron at 0 K.

"If we lined up all of iron's magnetic arrows, how strong would the total magnet be?"

(13.4) Estimate the saturation magnetization \( M_s \) of iron at absolute zero (0 K)

This is the maximum magnetization the material can have — when all the magnetic moments of the electrons are fully aligned.

What we need:

• Density of iron: \( \rho = 7874 \, \text{kg/m}^3 \)
• Molar mass of iron: \( M = 55.845 \, \text{g/mol} = 0.055845 \, \text{kg/mol} \)
• Avogadro's number: \( N_A = 6.022 \times 10^{23} \, \text{atoms/mol} \)
• Magnetic moment per iron atom (assume ~2 Bohr magnetons):
  \( \mu_{\text{atom}} = 2 \cdot \mu_B = 2 \cdot 9.28 \times 10^{-24} = 1.856 \times 10^{-23} \, \text{J/T} \)

Step-by-Step:

1. Atoms per cubic meter:

Use density and molar mass to calculate the number of atoms in 1 m³ of iron:

\[ n = \frac{\rho}{M} \cdot N_A = \frac{7874}{0.055845} \cdot 6.022 \times 10^{23} \approx 8.49 \times 10^{28} \, \text{atoms/m}^3 \]

2. Total magnetic moment:

Each atom contributes \( 2 \cdot \mu_B \), so the total magnetic moment per unit volume is:

\[ M_s = n \cdot \mu_{\text{atom}} = 8.49 \times 10^{28} \cdot 1.856 \times 10^{-23} = 1.575 \times 10^6 \, \text{A/m} \]

Final Answer:

\[ \boxed{M_s \approx 1.6 \times 10^6 \, \text{A/m}} \]

(13.5)

(a) Show that the area enclosed in a hysteresis loop in the (B,H) plane is equal to the energy dissipated in going around the loop.

The energy dissipated per cycle is given by:

\[ W = \oint H \, dB \]

This is equal to the area enclosed in the B-H hysteresis loop, representing the work done to flip the magnetic domains during each cycle.

(b) Estimate the power dissipated if 1 kg of iron is cycled through a hysteresis loop at 60 Hz, given the coercivity \( H_c = 4 \times 10^3 \, \text{A/m} \).

Assume loop area \( A \approx 4 H_c B_s \), and saturation \( B_s \approx 2 \, \text{T} \):

\[ W \approx 4 \cdot 4 \times 10^3 \cdot 2 = 3.2 \times 10^4 \, \text{J/m}^3 \text{ per cycle} \]

Density of iron: \( 7874 \, \text{kg/m}^3 \) → Volume = \( \frac{1}{7874} \approx 1.27 \times 10^{-4} \, \text{m}^3 \)

\[ P = W \cdot V \cdot f = 3.2 \times 10^4 \cdot 1.27 \times 10^{-4} \cdot 60 \approx 244 \, \text{W} \]

(13.6) Approximately what current would be required in a straight wire to erase a \( \gamma \)-Fe₂O₃ recording at a distance of 1 cm?

From Ampère’s law, the magnetic field around a long straight wire is:

\[ B = \frac{\mu_0 I}{2\pi r} \]

To erase magnetic data, we need a coercive field \( H_c = 300 \, \text{Oe} = 300 \cdot 79.6 = 2.39 \times 10^4 \, \text{A/m} \)

\[ I = 2\pi r H_c = 2\pi \cdot 0.01 \cdot 2.39 \times 10^4 \approx 1500 \, \text{A} \]

Final answer: \( I \approx 1500 \, \text{A} \)