Hints and review to problem set 4

1. A charged particle in an electric and magnetic field

Newton's equations can be written as $m\dot{\bm{v}} = q(\bm{E} + \bm{v} \wedge \bm{B})$. Use this expression in the differentiated kinetic energy. The other two parts of the question are straightforward. Note you can verify the proposed solution directly.

2. Forces between current-carrying wires

a) $\bm{B} = B\bm{k}$ where $B = (\mu_0 I/2) a^2/(a^2 + b^2)^{3/2}$.

b) You can use the fact that the force experienced by a charged particle in a current-carying wire is given by $I\bm{t} \wedge \bm{B}$ where $bm{t}$ is a tangential vector of the wire.

c) Use the formula \[ \bm{F}_{21} = -\frac{\mu_0}{4\pi} I_1 I_2 \int_{\Gamma_1}\int_{\Gamma_2} (\de{\bm{r_1}} \cdot \de{\bm{r_2}}) \frac{\bm{r}_2 - \bm{r}_1}{|\bm{r}_2 - \bm{r}_1|^3}. \]

3. Magnetic potential for a cylindrical conductor

We have $\bm{A} = A(r)\bm{k}$ where \[ A(r) = \begin{cases} -\mu_0 J \left[ \frac{r^2 - a^2}{4} + \frac{a^2}{2}\log a\right] & r < a \\ -\frac{\mu_0 J a^2}{2} \log r & r > a. \end{cases} \]

then you can compute the magnetic field using $\bm{B} = -A'(r)\bm{e}_\theta$ (can you prove this?).

4. Ampere's law

You can substitute $\bm{B} = -A'(r)\bm{e}_\theta$ into the line integral and evaluate directly. This gives $\mu_0 J \pi r^2$ for $r < a$ and $\mu_0 J \pi a^2$ for $r > a$.