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Relation between quasi-static and fully dynamic $vec E$ and $vec H$

Physics Asked by ZeroTheHero on April 17, 2021

Imagine an infinitely long coaxial cable with an inner wire of radius $a$ and outer radius $b$. The space in between the cable is filled with air ($epsilon=epsilon_0$).

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Suppose the inner cable carries a uniform current $I_0$. Then one easily finds, using Ampere’s law or Biot-Savart:
begin{align}
vec H(rho)=hat{phi}frac{I_0}{2pirho}, . tag{1}
end{align}

where $rho$ is the cylindrical distance in the plane perpendicular to $hat z$.

Now, suppose instead we turn off the current and set the potential difference between $a$ and $b$ to some constant uniform value $V$. Using Gauss’ law or simple superposition:
begin{align}
vec E(rho)=hat{rho}frac{V}{rho log(b/a)}, . tag{2}
end{align}

Solutions (1) and (2) depend critically on translational invariance along $hat z$, the cylinder axis of the system.

Now consider the time-dependent fields
begin{align}
vec H(rho,t)&=hat{phi}frac{I_0}{2pirho}cos(omega t-kz), , tag{3}
vec E(rho,t)&=hat rhofrac{V}{rho log(b/a)}cos(omega t-kz), . tag{4}
end{align}

One easily verifies that (3) and (4) satisfy Maxwell’s equation, v.g.
begin{align}
nabla times vec H&=- hatrho frac{I_0k}{2 pi rho } sin (omega t-k z), ,
frac{partial vec E}{partial t}&=-hatrho frac{V omega }{rho epsilon_0log left(frac{b}{a}right)} sin (omega t-k z )
end{align}

provided $I_0$ and $V$ are related through
begin{align}
frac{I_0k}{2 pi }=frac{V omega }{epsilon_0log left(frac{b}{a}right)} , , tag{5}
end{align}

without any condition on $omega$ or $k$. In other words, up to fine tuning of $V$ so that (5) holds, the quasistatic form of the fields given in (1) and (2) remain the $t$– and $z$-independent part of the time-dependent fields (3) and (4).

Why should should this quasi-static part remain given there is no restriction in $omega$
or $k$ in (3) or (4)?

I can see how one can recover some quasi-static limit if the wavelength $lambda=2pi/k >> b$ and $omega<<1$: with such assumptions the current looks locally uniform in $z$ since the changes in $z$ occur over a length scale of $lambda$ if $omega$ changes slowly, but this argument restricts $k$ and $omega$ whereas (3) and (4) hold for any $k$ and $omega$.

In other words, is there a way to go from (1) and (2) to (3) and (4) without the quasi-static argument which seems to restrict $omega$ and $k$?

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