Since we don't have class on Friday, I thought I'd post some HW hints. Keep in mind that the HW is due Mon 31 Oct at midnight, so we still have Monday's class to discuss the problems, i.e., I have not necessarily ruined your long weekend :-)
For problem 1, you know that motion of the L and R resistors leads to induced potential differences. That means it looks as though two little batteries are to be inserted into the circuit. But where? Faraday's law tells us that the integral of E.dl around a closed loop is the time rate of change of flux. The "loops" are defined by, say, the right resistor and the central one, and the left resistor and the central one. If the potential difference is non-zero when walking around one of these loops, the situation would be the same as if we just inserted infinitesimal batteries in the horizontal rails, one for each loop. Thinking about the direction of the induced currents tells you the orientation of each battery. This leaves you with a simple two-loop circuit.
For problem 2, you know the flux carved out by the moving rod is the integral of B.dA. B is just that of a wire at some distance r, no problem. dA is how much area is swept out in a time dt, which would be lvdt, making the integral one over time. Set up that integral, and its time derivative is the induced voltage. If you actually integrated anything, you missed something ...
Problem 3 has a nice story to it. It is a problem out of another common intro book (not the one we're using), and I thought it was cute. However, the solution that book gives is rather obtuse and not easily generalizable. In looking for a better way, I stumbled across the simple trick for plane curves given as a hint. As I showed in class today, it is easy to derive - write dl in polar coordinates, cross it into r-hat, and that's that. So long as the curve lies within a plane, and you want the field at some point within that plane, it works. It turns out there is a recent paper in the American Journal of Physics (a physics education journal) that derives the same formula and gives some nice examples. The derivation there is more geometrical, and probably easier to grok at first, but I sort of liked the elegance of the method I showed you. Anyway, the paper is worth a read. [Download link will only work on campus.]
Problem 4 we just did in class. Think about this, though: the ring picks up angular momentum, though it started with none. That means the electromagnetic field must have contained some angular momentum that was imparted to the ring once we shut the field off!
For problem 5, think carefully about the direction of the induced currents in each resistor (presume the voltmeters draw no current, it as if they are not there). From Faraday's law you can find the magnitude of the current easily enough. Then since you know the voltage drops across a resistor by an amount IR, you can figure out the potential difference between b and a, which is what the voltmeter reads. Now think about this: the voltmeters are connected by perfect wires to the same points, but must read different values! How is that? It is because we have a time-varying B, and because the voltmeter measures the integral of E.dl over a given path. When B does not vary in time, E is conservative, and the path doesn't matter. When B *does* vary in time, E is not conservative, and the result of integrating E.dl, the voltage measured, depends on the path taken. It is a hard thing to get your head around.
Problem 6 we did in class. I will try to do the magnetic levitation of a superconductor (Meissner effect) demonstration before the end of the semester. Keep reminding me ...
Problem 7 we did in class. Someday, in a grad-level E&M class you may solve this problem directly using the cylindrical boundary conditions and Maxwell's equations, and it will be annoying and difficult (though exact). This way is limited in accuracy by your patience, but (I think) far more transparent. It was mainly included just to give you a physical picture for time-varying fields. There is a nice discussion in the Feynman lectures, vol. 2, ch. 23 [If you a PH major, buy these books. There is a common copy in the physics student lounge.]
For problem 1, you know that motion of the L and R resistors leads to induced potential differences. That means it looks as though two little batteries are to be inserted into the circuit. But where? Faraday's law tells us that the integral of E.dl around a closed loop is the time rate of change of flux. The "loops" are defined by, say, the right resistor and the central one, and the left resistor and the central one. If the potential difference is non-zero when walking around one of these loops, the situation would be the same as if we just inserted infinitesimal batteries in the horizontal rails, one for each loop. Thinking about the direction of the induced currents tells you the orientation of each battery. This leaves you with a simple two-loop circuit.
For problem 2, you know the flux carved out by the moving rod is the integral of B.dA. B is just that of a wire at some distance r, no problem. dA is how much area is swept out in a time dt, which would be lvdt, making the integral one over time. Set up that integral, and its time derivative is the induced voltage. If you actually integrated anything, you missed something ...
Problem 3 has a nice story to it. It is a problem out of another common intro book (not the one we're using), and I thought it was cute. However, the solution that book gives is rather obtuse and not easily generalizable. In looking for a better way, I stumbled across the simple trick for plane curves given as a hint. As I showed in class today, it is easy to derive - write dl in polar coordinates, cross it into r-hat, and that's that. So long as the curve lies within a plane, and you want the field at some point within that plane, it works. It turns out there is a recent paper in the American Journal of Physics (a physics education journal) that derives the same formula and gives some nice examples. The derivation there is more geometrical, and probably easier to grok at first, but I sort of liked the elegance of the method I showed you. Anyway, the paper is worth a read. [Download link will only work on campus.]
Problem 4 we just did in class. Think about this, though: the ring picks up angular momentum, though it started with none. That means the electromagnetic field must have contained some angular momentum that was imparted to the ring once we shut the field off!
For problem 5, think carefully about the direction of the induced currents in each resistor (presume the voltmeters draw no current, it as if they are not there). From Faraday's law you can find the magnitude of the current easily enough. Then since you know the voltage drops across a resistor by an amount IR, you can figure out the potential difference between b and a, which is what the voltmeter reads. Now think about this: the voltmeters are connected by perfect wires to the same points, but must read different values! How is that? It is because we have a time-varying B, and because the voltmeter measures the integral of E.dl over a given path. When B does not vary in time, E is conservative, and the path doesn't matter. When B *does* vary in time, E is not conservative, and the result of integrating E.dl, the voltage measured, depends on the path taken. It is a hard thing to get your head around.
Problem 6 we did in class. I will try to do the magnetic levitation of a superconductor (Meissner effect) demonstration before the end of the semester. Keep reminding me ...
Problem 7 we did in class. Someday, in a grad-level E&M class you may solve this problem directly using the cylindrical boundary conditions and Maxwell's equations, and it will be annoying and difficult (though exact). This way is limited in accuracy by your patience, but (I think) far more transparent. It was mainly included just to give you a physical picture for time-varying fields. There is a nice discussion in the Feynman lectures, vol. 2, ch. 23 [If you a PH major, buy these books. There is a common copy in the physics student lounge.]
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