From 47d1f4c5f1662832fd7fb9cbbf2195dc21a8f523 Mon Sep 17 00:00:00 2001
From: Erik Strand <erik.strand@cba.mit.edu>
Date: Wed, 1 May 2019 00:16:33 -0400
Subject: [PATCH] Answer 13.3
---
_psets/10.md | 27 +++++++++++++++++++++++++++
1 file changed, 27 insertions(+)
diff --git a/_psets/10.md b/_psets/10.md
index 5e2c544..7ebf43c 100644
--- a/_psets/10.md
+++ b/_psets/10.md
@@ -106,11 +106,38 @@ Using the equation for the energy in a magnetic field, describe why:
{:.question}
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)
{:.question}
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)
--
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