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Commit 6bc88d50 authored by Erik Strand's avatar Erik Strand
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Answer 13.1

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...@@ -9,12 +9,52 @@ title: Problem Set 10 ...@@ -9,12 +9,52 @@ title: Problem Set 10
{:.question} {:.question}
Estimate the diamagnetic susceptibility of a typical solid. 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) ### (b)
{:.question} {:.question}
Using this, estimate the field strength needed to levitate a frog, assuming a gradient that drops to 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. 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) ## (13.2)
......
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