Solving Part b) of an Amperian Loop Problem

[jmcgraw]

If you know the net current enclosed in an amperian loop is zero, when can you conclude that the field at every point on the loop is zero?

Here's my actual problem.Part a), I'm pretty sure I did fine (B= u0*I/(2*pi*r)).

Part b) I'm guessing that B equals 0 at any radius r greater than c. The symmetry tells me intuitively that I'm correct, but I can't prove it to myself.What am I missing?

 

[berkeman]

If you know the net current enclosed in an amperian loop is zero, when can you conclude that the field at every point on the loop is zero?

What do you mean by "the field at every point on the loop..."? Which field?

 

[jmcgraw]

I attached the problem, but apparently it is still "pending."

What I mean is, take any point on the loop you created... When can you know that for each point on that loop the magnetic field is zero?

The line integral of B*ds is zero when the net enclosed current is zero. But that does not imply that the field, B, is zero on every point of the loop. When can you conclude that B is zero by knowing the enclosed current is zero?

 

[jmcgraw]

How long does it take to approve an attachment, anyway?

 

[jmcgraw]

O.k. Forget about the attachment.

Here's the problem:

http://img268.imageshack.us/img268/4214/cable4mm.jpg

Part b) is what I'm having trouble with

 

[berkeman]

Part b) I'm guessing that B equals 0 at any radius r greater than c. The symmetry tells me intuitively that I'm correct, but I can't prove it to myself.

One way to do it is to play with the paths that you would do the integration on outside the coax cable. Make part of the path be very far away, so that the field would be negligibly small there. And bring in the path only in one place near the coax. Since each of these paths must have a net of zero B field along them, you can conclude that B is zero everywhere outside.

For example, looking straight down the coax, draw an external path that is at infinity everywhere except on the right side. Draw the path coming in only from the right (at the 3 o'clock direction on a clock face), and bring it close to the outer surface of the coax, but not touching. Now draw the path going back out to infinity to the right, with the path out and the path in very close to each other. Now, the part of the path at infinity can't contribute anything, and the incoming and outgoing path segments both see the same B field because they are so close to each other. Therefore they must both see zero B field. Make other variations on the external path, and you should be able to convince yourself that B is identically zero outside the coax cable.

 

[jmcgraw]

Hmmmmmm... I don't think I'm getting what you're saying. This seems like it would prove that any enclosed set of currents would produce a 0 B field outside the outer tube, as long as the net current is 0. But that can't be.

For example, what if you moved the center cable off-center a bit. The field outside the outer tube would not be zero for all R > c. But it seems that what you are telling me to do would prove that it is 0.

 

[berkeman]

Hmm, good point. Perhaps I overspoke...

In the case of two parallel conductors, you get a net cancellation of the field as you go around a regular path integral that surrounds both conductors. The net of the integral is zero, but there is definitely B field outside the parallel conductors. I'll have to think some about why the path that I described has to also come out with a net zero...oh, I get it. The path direction is opposite coming in and going out, so that's why you get cancellation.

So that means that my argument to use that path shape for the coax case was wrong. I guess the only thing you can say about the coax case is that a circular path outside the coax (centered on the coax) has nothing special about any particular segment. So if the net of the integration is zero, it the integrand (the B field) has to be identically zero along the path. Not a very rigerous or strong argument, though.

 

[jmcgraw]

So if the net of the integration is zero, it the integrand (the B field) has to be identically zero along the path. Not a very rigerous or strong argument, though.

Hehehe. That's where I started... with the symmetrical intuition. But thanks for the help, anyway.

Also I think I remember Gauss' law and Ampere's law both being restricted to finite surfaces and loops. I think that any argument using anything at an infinite distance would be inherently flawed.