Add answers for 6.3

_psets/4.md

@@ -140,12 +140,29 @@ they'd be 81 kilometers high when stacked on top of each other.
 Use Stokes’ Law to find the magnetic field of an infinite solenoid carrying a current I with n
 turns/meter.
 
+Let the solenoid lie along x axis, with current flowing according to the right hand rule (about the
+x axis). Consider an axis aligned square in the xy (or xz) plane, where the edges aligned with the x
+axis are one meter long, with one fully inside the solenoid one fully outside. By Ampère's
+Law, the integral of the magnetic field along the boundary of this square is nI. This is independent
+of the length of the y (or z) aligned edges. So the field must be constant inside and out. Very far
+away from the solenoid the field should be zero, since the fields from the opposing sides cancel.
+Thus the field outside the solenoid is zero everywhere, and the field inside the solenoid is nI
+everywhere (pointing along the x axis). So assuming a vacuum inside the solenoid, we have $$H = nI$$
+and $$B = \mu_0 n I$$.
+
 ### (b)
 
 {:.question}
 Integrate the energy density to find the energy stored in a solenoid of radius r and length l, once
 again neglecting fringing fields.
 
+In this case we have no electric field, so $$U = B \cdot H / 2 = \mu_0 n^2 I^2 / 2$$. Integrated
+over a cylinder of radius $$r$$ and length $$l$$, we have a total energy of
+
+$$
+\frac{1}{2} \mu_0 n^2 I^2 \pi r^2 l
+$$
+
 ### (c)
 
 {:.question}
@@ -153,6 +170,18 @@ Consider a 10 T MRI magnet (Section 10.4) with a bore diameter of 1 m and a leng
 the outward force on the magnet? Remember – force is the gradient of potential for a conservative
 force.
 
+$$
+\begin{align*}
+\frac{\partial}{\partial r} U
+&= \frac{\partial}{\partial r} \frac{1}{2} \frac{B^2}{\mu_0} \pi r^2 l \\
+&= \frac{B^2}{\mu_0} \pi r l \\
+&= \frac{100 \si{T^2}}{4 \pi \times 10^{-7} \si{H/m}} \cdot \pi \cdot 1 \si{m} \cdot 2 \si{m} \\
+&= \num{50e7} \si{N}
+\end{align*}
+$$
+
+Yes, there is a sign error that I'm ignoring for now.
+
 
 ## (6.4)