Solving Electric Field & Displacement Current in Coaxial Cable

In summary, the problem deals with finding the displacement current and electric field in a long straight wire with an alternating current and a coaxial tube. By applying the Maxwell-Faraday law and considering the induced electric field to be zero at large distances, the electric field and displacement current can be calculated. However, there may be some complications in accounting for the time to propagate and the varying B-field outside the tube. The displacement current can be derived from the electric field once it is obtained.
  • #1
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Homework Statement


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].
  1. By constructing an appropriate Amperian loop or otherwise, and assuming that the induced electric field E goes to zero as the distance from the wire, [itex]s[/itex], becomes large, find [itex]{\bf{E}}(s,t)[/itex].
  2. Find the displacement current density, [itex]\bf{J_D}[/itex], and hence the total displacement current, [itex]I_D[/itex].


Homework 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]


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 Maxwell-Faraday, 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{s-a}{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 B-field 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 B-field 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 Maxwell-Faraday 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 z-dependence. 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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  • #2
The displacement current density should be easy enough to derive from E once I have it, but I can't really go any further until I work out E.
 

1. What is the difference between electric field and displacement current?

Electric field is a physical quantity that describes the force experienced by a charged particle in an electric field. Displacement current, on the other hand, is a phenomenon that describes the flow of electric charge in a changing electric field. While electric field is caused by the presence of stationary charges, displacement current is caused by the changing electric field itself.

2. How is electric field calculated in a coaxial cable?

In a coaxial cable, the electric field can be calculated using the formula E = V/d, where E is the electric field strength, V is the potential difference between the two conductors, and d is the distance between the two conductors. This assumes that the cable is in a vacuum or air, and there are no other materials present that could affect the electric field.

3. What is displacement current in a coaxial cable?

In a coaxial cable, displacement current is the flow of electric charge that occurs between the two conductors due to the changing electric field. This phenomenon was first described by James Clerk Maxwell in his famous set of equations, known as Maxwell's equations.

4. How does the presence of a dielectric material affect the electric field and displacement current in a coaxial cable?

The presence of a dielectric material, such as an insulating material, between the two conductors of a coaxial cable can affect both the electric field and displacement current. The dielectric material can change the distance between the conductors, which in turn can change the strength of the electric field. It can also affect the displacement current by altering the capacitance between the conductors, which is a measure of how much charge can be stored between them.

5. Can the electric field and displacement current in a coaxial cable be measured?

Yes, the electric field and displacement current in a coaxial cable can be measured using various instruments such as voltmeters and ammeters. These instruments can measure the potential difference and the flow of electric charge, respectively, between the two conductors. These measurements can be used to calculate the electric field and displacement current using the appropriate formulas and equations.

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