Answer 13.1

_psets/10.md

@@ -9,12 +9,52 @@ title: Problem Set 10
 {:.question}
 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)
 
 {:.question}
 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)