Answer 13.2

_psets/10.md

@@ -65,7 +65,34 @@ field of a magnetic dipole $$\vec{m}$$ is
 
 $$
 \vec{B} = \frac{\mu_0}{4 \pi}
-    \left[ \frac{3 \hat{x} (\hat{x} \cdot \vec{m}) - \vec{m}}{|\vec{x}|^3} \right]
+    \left[ \frac{3 \hat{x} (\vec{m} \cdot \hat{x}) - \vec{m}}{|\vec{x}|^3} \right]
+$$
+
+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
+
+$$
+\begin{align*}
+E_e &= q E \\
+&= q \frac{q}{4 \pi \epsilon_0 r} \\
+&= \frac{q^2}{4 \pi \epsilon_0 r} \\
+&= \frac{(\num{1.6e-19} \si{C})^2}{4 \pi \cdot \num{8.85e-12} \si{F/m} \cdot 10^{-10} \si{m}} \\
+&= \num{2.3e-18} \si{J}
+\end{align*}
 $$