First, regarding magnetic fields:
The magnetic field of a straight wire is: B=mu I/(2 pi r).
The magnetic field in the center of a wire loop is: B=mu I/(2 r) (larger by a factor of pi).
(r is the distance from the wire; B is the magnitude of the magnetic field associated with the current flowing in the wire.) These results come from the Biot-Savart equation and/or Ampere's relation. Mostly we will just use these results and not get bogged down in the derivations.
There is a problem on the homework in which I believe you are asked for the magnitude of the magnetic field due to the current in a wire with a single loop in it. I think the answer involves combining the two results above, thus the total field is the contribution from the straight wire, B=mu I/(2 pi r), plus the contribution from the loop, B=mu I/(2 r). (please post a comment if that doesn't work.)
A solenoid involves many loops of wire very close together with current flowing through all of them. The magnetic field everywhere inside a the solenoid is given by:
B=mu N/Z I = mu n I,
where I is the current, N is the number of loops in the solenoid, and Z is the length of the solenoid. ( Note that the units of B are the same as above because Z has units of length. Also, in truth, this i an approximation, in case you were wondering. It is not exactly correct, but it is a pretty good approximation in many cases of interest.)Chapter 26 might be summarized by saying: current flowing in wires produces magnetic fields, and we are particularly interested in the field produced inside a wire loop or a solenoid. Chapter 27 can be summarized as: changing magnetic fields generate electric fields.
In Chapter 27 the most important thing to learn and remember is that when a magnetic field changes, as a function of time, that creates an electric field. That is what Faraday's law is all about. Faraday's law is expressed in terms of a quantity called magnetic flux, or just flux, which is magnetic field strength multiplied by an area. For a solenoid, the flux is:
Phi = pi r^2 B
= pi r^2 mu (N/Z) I
The reason that flux is important to us is that when it changes it produces an emf (a voltage-like quantity) given by:
emf = -d(Phi)/dt
i.e., an emf equal to the time derivative of the flux.This relationship is the basis for the new circuit device known as an Inductor. It means that when the current in a solenoid changes, that change will produce a voltage. An inductor is a solenoid. The name inductor focuses on its function. The name solenoid focuses on its physical appearance. They both refer to the same thing.
The magnetic field inside the solenoid is B= mu N I/Z and the associated flux is pi r^2 times that. The emf produced by a changing current is then all those factors time dI(t)/dt, i.e.,
emf=-(pi r^2) mu N/Z dI/dt.
Actually, that is the emf per loop or turn. The total emf is N times that. From that, one obtains the inductance of a solenoid.With this knowledge you should be able to do many of the homework problems. If not, please feel free to ask any questions here.
**Note added on Sunday: Do you think that it might make sense to extend the due date for some of the homework problems? The reason I ask is that I believe that the homework problems can be divided into two categories: those that can be done using just the information provided above, and, those that can't. I would appreciate some feedback on this. Would some people go through the problems and analyze and explain which ones should be extended and why? Your efforts will be much appreciated. (PS. It is especially helpful if you include a summary of what the problem is asking as well as it's number.)
Hi professor schlesinger I was working on the first homework problem and I couldn't get it to work with the equation you gave us. Using B=mu(I)/(2pi*r) didn't work so I looked up this problem and I found a solution, which gave the right answer according to mastering physics but it omitted the pi in the denominator. so instead the equation looked like B=mu(I)/(2r). Is the equation above only right under certain conditions or did I make some other mistake? I was just wondering since I'm sure many if not most of the problems depend on this initial equation.
ReplyDeletenever mind I realize what I did
ReplyDeletehmmm. glad you brought this up. i think one is for a loop and the other is for a straight wire.
ReplyDeleteProblem 26.18 Part B:
ReplyDeleteShouldn't multiplying the acceleration of the particle by the time it is accelerating (t = 10^(-9)s) give the change in speed? This operation is not giving me the correct answer. Am I going about this wrong?
Nevermind, I realized that magnetic force always acts at right angles to a particle's velocity
ReplyDeleteyes could you please extend it! The homework is challenging and there's lots of lab reports and papers due this week for students.
ReplyDeleteproblem 27.56 is challenging. Given initial voltage, resistance, inductance, and current, they ask for the time the gate has been closed. Using the given values and the formula I= I(infinity) (1-e^-t/tao) is not working. I used initial voltage over resistance as an equivalent of the current as time approaches infinity, but I'm still wrong. I checked prefixes on the units, and everything is standard. whats the problem here?
ReplyDeletealso 27.63 says that current in an inductor rises to half its steady state value in 1.6 ms. it asks how long it takes for the magnetic energy in the inductor to rise to half its steady state value. how do i approach this?
27.55 is somewhat similar, stating that current rises to half its final value in 7.1 seconds. it asks for the time constant. what is the relationship here?
I will try to post something on those 3 problems (56, 63 and 55) in a little while.
ReplyDeleteAlso, I extended the HW to be due Sunday. Let me know if that does or doesn't show up on your MP accounts.
problem 27.56 is challenging....they ask for the time the gate has been closed. Using the given values and the formula I= I(infinity) (1-e^-t/tao) is not working. I used initial voltage over resistance as an equivalent of the current as time approaches infinity, but I'm still wrong. I checked prefixes on the units, and everything is standard. whats the problem here?"
ReplyDeleteI think you are approaching it perfectly. tau is L/R, right?? (not R/L) And I agree that the large t value of I is E_o/R.
"27.63 says that current in an inductor rises to half its steady state value in 1.6 ms. it asks how long it takes for the magnetic energy in the inductor to rise to half its steady state value. how do i approach this?"
ReplyDeleteI think that the magnetic energy is proportional to B^2, and also to I^2. So when the current is at 1/2 its steady state value, then the energy would be at ?/? of its steady state value, right? What is ?/? ? Once you get that, i think the rest will follow.
for 55, like 56, the formula I= I(infinity) (1-e^-t/tao) is the key. if (1-e^-t/tao)=1/2, the doesn't e^-t/tau = 1/2 also. so i think the answer involves a natural logarithm...
ReplyDeleteI'm a little lost on how to start 27.42, I don't know how to approach the problem.
ReplyDeleteThat is a difficult problem; it goes a little beyond what we have emphasized in class. First, I might suggest skipping it and doing other problems first.
ReplyDeleteTo do the problem you would want to understand and then use the emf-Flux relationship:
e=-dPhi/dt
I think that is the key.
Phi, the flux, is B times and area...
Then use: I = e/R.
an area
ReplyDelete