Reading from Griffiths, Chapter 10:

  • Read problem 10.12. Keep it in mind as something you need to figure out how to do.
  • Read the chapter up to 10.1.4 (skip 10.1.4), and skip the examples. I actually want you to get the derivations here.
  • Read 10.2.1 up until Equation 10.26 which is what you’ll use for Problem 10.12. You’ll need the concept of a “retarded” potential, which we will call a “delayed” potential. Spend some time on Example 10.2, because this is a useful blueprint for the problem you need to do.
  • Skip 10.2.2.
  • Read 10.3.1 Your goal is to understand where Equations 10.46 and 10.47 come from.
  • Don’t read 10.3.2, but I want you to notice where Griffith’s is going with it. He’s taking 10.46 and 10.47 and using them to get E and B. And after a lot of pain, he gets 10.72, which is what you need for Problem 10.20.
  • Look at Figure 10.10 and the paragraph below it. That is an incredibly important result. If you only remember one thing from this week, I want you to remember that the field of a fast-moving point charge is flattened out like a pancake in the direction perpendicular to the motion.

Note: two years ago students requested that I not use the term ‘retarded’ to refer to potentials that are delayed. You will definitely encounter that term elsewhere (and Griffiths uses it extensively). It’s historical and it employs the literal use of the word to mean `delayed’, but we decided to use ‘delayed’ in this class, and I will do that again.

You may work together and get help from other students. Your solutions must be written in your own words, without looking at someone else’s solutions while you write them.

Don’t forget the 9 points that we are looking for in your solutions (see Moodle).

In order to make sure you get your context and meaning/make sense points, next to your answers, please put a “c” with a circle around it for context, and an “m” with a circle around it for meaning.

  1. Griffiths 10.12 Calculate the deferred potential of a current loop with increasing current. (Hints: I’m going to show you this on the board…you probably are already having an inkling that you’ll end up doing 4 integrals, one for each of the little sections of wire. However, don’t do this right away. Once you put the delayed time into the current, i.e. \(I(t_r) = k(t - \mathscr{r}/c)\) you’ll have two terms in the integral (the \(kt\) part and the \(k\mathscr{r}/c\) part. One of those integrals is actually the integral of \(\vec{dl}\) around a closed loop, which is zero. Then the other part you can actually break into the 4 integrals. Also, you’re only calculating the potential at the center. So for the two parts of the loop that are semicircles \(\mathscr{r}\) can be replaced by the distance from the ring to the center.)

  2. Griffiths 10.20 A relatively straightforward application of equation 10.72 - the fields derived from the Leinard-Weichert potentials. (My hint is to replace \(\vec{mathscr{r}}\) (where that’s Griffiths’ fancy script r) and \(\vec{v}\) and \(\vec{a}\) with expressions that specify what direction they are in cartesian coordinates, for example \(\vec{a} = a \hat{x}\) (because you know the acceleration is in the x direction.} If you’re like me you’ll be tempted to replace that script \(\mathscr{r}\) with something, but don’t do it, because the script \(\mathscr{r}\) is in the answer.}

We’re only doing the first two (above). You can do the next two if you really want to.

(3) Griffiths 10.24 Glue charge onto a plastic ring and spin it. As in number 1, you’re only calculating the potential at the center, so \(\mathscr{r}\) can be replaced by the distance from the ring to the center.

(4) Griffiths 10.30 A charged rod with constant velocity. You’re asked to verify that your answer is consistent with Leinard-Weichert potential.