Reading: Griffiths, from Chapter 7: Page 300-307: you can skip examples, skip the proof on the bottom of page 307, and then pick back up the text on the bottom of page 308 “There is a sign ambiguity…” Read through 327 skipping the examples and the problems.

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, Problem 7.3
  2. Griffiths, Problem 7.11.
  3. Griffiths, Problem 7.23.
  4. Griffiths, Problem 7.62. This type of transmission line is referred to by electrical engineers as a “strip line.”

Some hints if you want them:

Hints for HW9

7.3 You usually find capacitance by assuming there’s a charge Q on each conductor. You can do that here, too. Ohm’s law will tell you how the current density is related to the electric field. And Gauss’s law will tell you how the electric field is related to the charge. I think you can take it from there.

7.11 You do have to look up the density and conductivity of aluminum. If you express things in terms of the cross-sectional area of the loop (the cross-section lying in a plane perpendicular to the direction its falling) then the cross-sectional area cancels out.

7.23 First compute the mutual inductance M. To do this, put I in the outer loop and compute the flux in the inner loop. Check out 7.22 and then notice (7.24) that the mutual inductance is the same whether you’re asking about the flux induced in the big loop by the current in the small loop or the the other way around. I believe Griffiths even calls that “cute.” To make your life EVEN EASIER, instead of calculating the flux through the small loop due to the current in the large loop, only calculate the flux through the small loop due to the LOWER SIDE of the large loop, and then argue that the total flux through the small loop should be twice that by symmetry.

7.62 I’d use the parallel plate formula for capacitance. (You could also derive it from Gauss’s law and the definition of capacitance.). Then I’d set 7.35 and 7.30 equal to each other to get the inductance. Griffiths hint in part (d) is okay, but I have a better one. Replace the \(\epsilon_0\) with \(\epsilon\) and replace \(\mu_0\) with \(\mu\) and Bob’s your uncle.