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Finding electric field/displacement current for timevarying current in coaxial cable 
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#1
Jun1111, 10:12 PM

P: 12

1. The problem statement, all variables and given/known data
An alternating current [itex]I = I_0 \cos(\omega t)[/itex] flows down a long straight wire and returns along a coaxial tube of radius [itex]a[/itex].
2. Relevant equations [tex]\nabla\times{\bf{E}} = \frac{\partial{\bf{B}}}{\partial t}[/tex] [tex]\nabla\times{\bf{B}} = \mu_0{\bf{J}} + \mu_0\epsilon_0\frac{\partial{\bf{E}}}{\partial t}[/tex] [tex]\oint_C{\bf{E}}\cdot\,d{\bf{\ell}} = \frac{\partial}{\partial t} \int_S {\bf{B}}\cdot\,d{\bf{a}}[/tex] [tex]\oint_C{\bf{B}}\cdot\,d{\bf{\ell}} = \mu_0 I_{enc} + \mu_0\epsilon_0\frac{\partial}{\partial t}\int_S {\bf{E}}\cdot\,d{\bf{a}}[/tex] 3. The attempt at a solution Just to clarify, I believe the current in the coaxial tube is the negative of the current through the central wire. The fact that you're asked to compute displacement current in the second part of the question should mean that you can calculate E solely using MaxwellFaraday, and once you have E it ought to be fairly straightforward to calculate B using Ampere's law. The time derivative of B ought to look something like [itex]\sin(\omega t)[/itex] (diminishing with increasing s, so something like a [itex]\frac{1}{s}[/itex] or [itex]\frac{1}{s^2}[/itex] factor thrown in as well), but I'm not sure if I should replace [itex]t[/itex] with [itex]t\frac{s}{c}[/itex] (and possibly [itex]t\left(\frac{sa}{c}\right)[/itex]) to account for the time to propagate. In that case, because of the difference in distance from the wire and the tube, there should always be some Bfield outside the tube (if the currents were constant and opposite forever, then B would of course be 0 for s>a). I think the electric field runs parallel to the current, and it is constant at some time t for a fixed distance from the axis. If there is a Bfield outside the tube, varying with time, then the electric field outside the tube is not constant through time. The appropriate Amperian loop to apply the MaxwellFaraday law is a rectangle with two sides parallel to the axis. This would give: [tex]\oint_C{\bf{E}}\cdot\,d{\bf{\ell}} = {\bf{E}}(s_2,t)h  {\bf{E}}(s_1,t)h = \iint_S \frac{\partial{\bf{B}}}{\partial t} \cdot\,d{\bf{a}} = \int^{s_2}_{s_1}\int^h_0\frac{\partial{\bf{B}}}{\partial \,t} \,ds\,dz[/tex] Where h cancels because there is no zdependence. The problem is that I can't figure out how to get something that goes to 0 as [itex]s\rightarrow\infty[/itex]. At one point I ended up trying to integrate something like [itex]\frac{1}{s}\sin\left(\omega\left(t\frac{s}{c}\right)\right)[/itex], and that seems to give the incomplete gamma function in its result, so that can't possibly be right. I'm completely lost, but I'm certain that the answer is simpler than I'm trying to make it. 


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