Problem Set 10

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(13.1)

(a)

Estimate the diamagnetic susceptibility of a typical solid.

Starting from equation 12.15,

$\begin{align*} \chi_m &= -\mu_0 \frac{q^2 Z r^2}{4 m_e V} \\ &= -\num{1.26e-6} \si{N/A^2} \frac{(\num{1.6e-19} \si{C})^2 \cdot 1 \cdot (10^{-10} \si{m})^2} {4 \cdot \num{9.1e-31} \si{kg} \cdot (10^{-10} \si{m})^3} \\ &= \num{-8.9e-5} \end{align*}$

(b)

Using this, estimate the field strength needed to levitate a frog, assuming a gradient that drops to zero across the frog. Express your answer in teslas.

From 12.7,

$F = -V \mu_0 \chi_m H \frac{d H}{d z}$

I’ll assume the frog is 0.1 meters tall, has a mass of 0.1 kg, and a volume of $10^{-4} \si{m^3}$ (this is consistent with the frog being mostly water). I’ll also assume the magnetic field gradient is constant, so $dH/dz = H / 0.1$ . Solving for $H$ ,

$\begin{align*} H &= \sqrt{-\frac{F z}{V \mu_0 \chi_m}} \\ &= \sqrt{-\frac{0.1 \si{kg} \cdot 9.8 \si{m/s^2} \cdot 0.1 \si{m}} {10^{-4} \si{m^3} \cdot \num{1.26e-6} \si{N/A^2} \cdot \num{-8.9e-5}}} \\ &= \num{3e6} \si{A/m} \\ \end{align*}$

Thus the magnetic field is

$\begin{align*} B &= \mu_0 H \\ &= \num{1.26e-6} \si{N/A^2} \cdot \num{3e6} \si{A/m} \\ &= 3.7 \si{T} \end{align*}$

(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. Remember 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 force between two magnetic dipoles $m_1$ and $m_2$ (with associated fields $B_1$ and $B_2$ ) is $F = -\nabla (m_1 \cdot B_2)$ , and the characteristic interaction energy is $m_1 \cdot B_2$ . This will be maximized when $m_1$ and $B_2$ are parallel. From the equation above, we can see that the field strength will be maximized when $\hat{x}$ is antiparallel to $\vec{m}$ . Assuming a separation of 1 angstrom, the total energy is thus

$\begin{align*} E_m &= m \cdot \frac{\mu_0}{4 \pi} \left( \frac{4 m}{r^3} \right) \\ &= \frac{\mu_0 m^2}{\pi r^3} \\ &= \frac{\num{1.26e-6} \si{N/A^2} (\num{-9.28e-24} \si{J/T})^2}{\pi (10^{-10} \si{m})^3} \\ &= \num{3.5e-23} \si{J} \end{align*}$

Meanwhile their electrostatic potential is

$\begin{align*} E_e &= q E \\ &= q \frac{q}{4 \pi \epsilon_0 r} \\ &= \frac{q^2}{4 \pi \epsilon_0 r} \\ &= \frac{(\num{1.6e-19} \si{C})^2}{4 \pi \cdot \num{8.85e-12} \si{F/m} \cdot 10^{-10} \si{m}} \\ &= \num{2.3e-18} \si{J} \end{align*}$

(13.3)

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

(a)

A permanent magnet is attracted to an unmagnetized ferromagnet.

The equation of interest for energy density is

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

For this problem I assume the electric field is zero, so the total energy is

$U = \frac{1}{2 \mu} \int B^2 \mathrm{d} V$

An unmagnetized ferromagnet has a very high $\mu$ , so $U$ is reduced by packing more field lines into its extent. The permanent magnet’s field lines are densest closest to its body, so the gradient of the energy describes an attractive force.

(b)

The opposite poles of permanent magnets attract each other.

We know that $B = \mu (H + M)$ , so

$\begin{align*} U &= \frac{1}{2 \mu} \int \mu (H + M) \cdot H \mathrm{d} V \\ &= \frac{1}{2} \int (H^2 + M \cdot H) \mathrm{d} V \\ \end{align*}$

This is reduced when $H$ and $M$ are anti-aligned.

(13.4)

Estimate the saturation magnetization for iron at 0 K.

I looked up the density, molar mass, and number of valence electrons of iron. My answer also uses the proton mass and electron magnetic moment.

$\begin{align*} 7,874 \si{kg/m^3} &\cdot \frac{1 \text{ nucleon}}{\num{1.67e-27} \si{kg}} \cdot \frac{1 \text{ atom}}{55.845 \text{ nucleons}} \\ &\cdot \frac{2 \text{ electrons}}{1 \text{ atom}} \cdot \frac{\num{9.28e-24} \si{J/T}}{1 \text{ electron}} \\ &= \num{1.6e6} \si{A/m} \end{align*}$

(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.

(b)

Estimate the power dissipated if 1 kg of iron is cycled through a hysteresis loop at 60 Hz; the coercivity of iron is $\num{4e3} \si{A/m}$ .

(13.6)

Approximately what current would be required in a straight wire to be able to erase a $\gamma \text{-} Fe_2 O_3$ recording at a distance of 1 cm?

As found in problem 6.4 in problem set 4, the magnitude of the magnetic field a distance $r$ away from an infinitely long and thin conductor carrying a current $I$ is $I/(2 \pi r)$ . To erase information stored on $Fe_2 O_3$ we need this field to be about as strong as the coercivity $H_C = 300 \si{Oe}$ . Thus the current needed is

$\begin{align*} I &= 2 \pi r H_C \\ &= 2 \pi \cdot 10^{-2} \si{m} \cdot 300 \si{Oe} \cdot \frac{79.6 \si{A/m}}{1 \si{Oe}} \\ &= 1500 \si{A} \end{align*}$