Date |
Question |
Answer |

9/18 |
7.34 -- I'm not sure how to read
the expression for E: Is that theta(a
function of vt-r) or theta(a function of x)*(vt-r). Also, since E = E(r-vector, t), are we taking v to be |d(r-vector)/dt| = constant? |
Well, I don't see any x!
You are told it is the theta FUNCTION, so it is a function of
something. That something is vt-r. I don't understand that "since". v is a parameter in the given function, which you may assume to be constant. E is a function of r (position vector) and t . These are your variables. |

9/18 |
7.30 -- For part (b) where we're
comparing to equation 6.35, I'm not
sure I understand what Griffiths means by the 'interaction energy' of 2 magnetic dipoles. |
Did you try using the book's
index? |

9/18 |
31.79 --
1) On Thursday, you reminded me that the charged loop also has an E field to worry about, but since we're looking at the center of the loop, wouldn't that be zero? Did you just want us to mention that, or should we be looking at points _just above_ the center of the loop so that z << r but z != 0? 2) How far back do we need to go in deriving the expression for the electromagnetic energy density, u_EM? 3) The expression that I'm getting for u_EM doesn't seem that remarkable, so either I'm doing something wrong, or I'm not sure what kind of analysis you're looking for before we plug in numbers. Perhaps this goes back to my question (1)? |
(1) If the field is zero,
but te charge is not, you need to say why. z=0 is fine. (2) You don't need to derive it, just use it. (3) Well, I found something I thought was a bit remarkable. Look at how your answer depends on the physical parameters. Is that what you expected? |

Sept 13 |
I'm not sure what Griffiths is
getting at when he says "in the quasistatic case" in problem 7.24 part
(a). Is he implying that we shouldn't consider the back EMF yet, or are
we supposed to take the time average of dPhi/dt, or what? |
The current in the long wire
changes SLOWLY, so you can calculate the B it produces using static
methods (Ampere's law, for example, ignoring displacement current.) |

Sept 13 |
I'm
not sure if 1) the EMF in the toroid is the same as that through single loop [my gut feeling -- but that seems like the loops would be in parallel & that doesn't make sense since they're directly connected] or 2) the EMF in the toroid is N * (the EMF in one loop) [which I guess makes more sense if I think of the loops in series, but if the EMF is the potential difference around one loop, & they add all the way the toroid, how would you know where to start or end?]. And from there I'm not sure if the resistor is in series with the coils, or in parallel, or if that matters since it's the only resistor in the circuit. |
The emf is the integral around
the whole circuit, which includes all N coils. It doesn't matter
where you start and end, but in a real toroid there would be a place
where the wire connects out to somewhere else (like the resistor), so
you might want to start there. Remember, that EMF and potential difference are NOT the same thing! Only static fields contribute to potential difference. In series. |

Sept 13 |
Do we need to worry that the
resistor in the circuit is throwing off the rotational symmetry of the toroid about the z-axis? Or can we assume that the resistor can be placed arbitrarily at any point in the circuit and argue that it doesn't matter where? |
Imagine that the resistor is
connected to the toroid by very fine wire. It does not change the
geometry of the toroid at all. |

Sept 13 |
I'm thinking that if you had a
flat, symmetrical "figure 8" of wire, you'd just add the flux through each loop to get the total flux Phi = Phi1 + Phi2, and therefore you could find the total EMF = -dPhi/dt. If you took that figure 8 and folded it so that you had 2 parallel loops, would the EMF be cut in half, or would it stay the same? I keep waffling on whether you consider the flux through each loop (EMF stays the same), or the flux through area that the SYSTEM presents (EMF cut in half). |
This is a very good question, as
it really gets at the idea. By folding the figure 8, you don't
change the flux! That is really the whole issue, right there.
Notice that the integral of E dot dl doesn't change when you fold the
loop, and so phi, and hence d phi/dt can't change either. |

Sept 13 |
For Griffiths 7.20 must I derive the magnetic field a distance z above a current loop, and that for a magnetic dipole, or may I just quote the results from Ch 5 of the textbook? | You may quote the result.
But be sure that you COULD derive it if asked to. |