I'm having some trouble solving an electrodynamics problem.
There is a wire inside a coaxial cylindrical conductor of radius $a$.
An alternating current $I=I_0 cos(\omega t)$ flows in the wire and returns through the outer cylindrical conductor.
We suppose both the wire and the cylyndrical conductor have infinite lenght.
We want to find the electrical field generated by the variable magnetic field.
In first approximation, I computed $\overrightarrow{B}$ using the Ampére law and I found:
$$\overrightarrow{B}=\frac{\mu_0}{2\pi} \frac{I_0 \cos(\omega t)}{r}\hat{e}_{\theta}$$
only inside the cylinder, while $\overrightarrow{B}=0$ outside the cylinder.
Then I used the Faraday law to compute $\overrightarrow{E}$: I know $\overrightarrow{E}=E(r, t)\hat{e}_z$ because of the symmetry of the problem, so I chose a rectangular circuit with 2 sides parallel to the z axis (which is along the wire). One of them is at a generic $r=r<a$, the other one is at $r=a$. This is a good choice to compute $\overrightarrow{E}$ inside the wire, but I don't understand if there is an electrical field outside the wire too.
If I choose a rectangular circuit with a side at $r=r>a$ and the other one at $r=a$ of course I find $\overrightarrow{E}$ to be $0$ because there is no magnetic field outside the cylinder.
However, if I choose the other one at $r=r_0 <a$ there is a variation of the magnetic field's flux inside the part of the rectangle that happens to be inside the cylinder.
Also, if $\overrightarrow{E}$ is actually 0 outside the cylinder, can there be a displacement current outside the cylinder? My professor wrote in his solution that there is in fact a displacement current outside the cylinder, but I don't understand why if there is no induced electric field.
1 Answer
If outside the cylinder you have empty space, then the displacement current there is $ \varepsilon_0\partial{\bf E}/\partial t$, so if someone says that there "is in fact a displacement current" in that region, then that is the same a saying ${\bf E}\neq 0$ there. But is it? Let us calculate...
What you describe is an abnormal situation. Ordinary signal propagation in coaxial conductors would have $I=I_0 \cos(\omega (t - k z))$ where $k=\omega/c$, here the $z$ dependence is absent. But we can just accept that this is the current (driven by a non-EM force along the entire wire, presumably) and then solve for the fields exactly, but not by using Ampere's equation because that was changed in the Maxwell equations to include a term $\partial{\bf E}/\partial t$ term. If we use the full equations the field will consist of cylindrical Bessel functions: $$ E_z \sim Y_0(k r) + K J_0(k r) $$ $$ B_\phi \sim Y_1(k r) +K J_1(k r) $$ where the coefficient $K$ is then fixed by making $E_z =0$ on the cylinder wall and then the overall scale is fixed by requiring that close to the central wire $B_\phi \rightarrow \mu_0 I_0/(2\pi \rho)$. The current density on the cylinder wall must than be equal to the tangential $H$ at its inner surface, which one can find as: $$ J_z = \frac{-I_0}{2\pi\ a \ J_0(k a)} \Rightarrow I_\text{cyl}=\frac{-I_0}{J_0(k a)} $$ For $ka>0$, $J_0(ka)<1$, so we have a return current that is larger than the central wire current in this solution! The solution does have all fields zero outside the cylinder, because of the matching coefficients of the Bessel functions we chose.
You can make another choice, insisting that the return current should be exactly $-I_0$. For that we can keep the above solution and just add some positive current on the cylindrical surface. That additional current will give outgoing cylindrical waves, so in that case both the $E$ and the $B$ field will not be zero outside the cylinder. But also not on the cylinder so it doesn't satisfy the ordinary boundary conditions for a perfect conductor.
The first solution is interesting, it has a nonzero total current in the two conductors (a common-mode current) but still produces no outside fields. Of course if you add the displacement current from $E_z$ inside the cylinder (outside it is zero) then the total common-mode current is zero. Note that the normal TEM mode of a lossless coaxial system does not have an $E_z$ component, so there we only get a zero outside field if the common-mode current is also zero.
