A long coaxial cable, of length l, consists of an inner conductor (radius a) and an outer conductor (radius b). It is connected to a battery at one end and a resistor at the other (Fig. 8.5). The inner conductor carries a uniform charge per unit length λ, and a steady current I to the right; the o

Question

A long coaxial cable, of length l, consists of an inner conductor (radius a) and an outer conductor (radius b). It is connected to a battery at one end and a resistor at the other (Fig. 8.5). The inner conductor carries a uniform charge per unit length λ, and a steady current I to the right; the outer conductor has the opposite charge and current. What is the electromagnetic momentum stored in the fields?

Accepted Answer

The fields are

[latex]E=\frac{1}{2\pi\epsilon _0}\frac{\lambda }{s}\hat{s}, B=\frac{\mu_0}{2\pi}\frac{I}{s}\hat\phi. [/latex]

The Poynting vector is therefore

[latex]S=\frac{\lambda I}{4\pi^2\epsilon _0s^2}\hat{z}. [/latex]

So energy is flowing down the line, from the battery to the resistor. In fact, the power transported is

[latex]P=\int{S.da}=\frac{\lambda I}{4\pi^2\epsilon _0}\int_{a}^{b}{\frac{1}{s^2}2\pi sds }=\frac{ \lambda I}{2\pi\epsilon _0}\ln (b/a)=IV, [/latex]

as it should be.
The momentum in the fields is

[latex]P=\mu_0\epsilon _0\int{Sd au }=\frac{\mu_0\lambda I}{4\pi^2}\hat{z}\int_{a}^{b}{\frac{1}{s^2}l2\pi sds}=\frac{\mu_0\lambda Il}{2\pi}\ln (b/a)\hat{z}=\frac{IVl}{c^2}\hat{z}. [/latex]

This is an astonishing result. The cable is not moving, E and B are static, and yet we are asked to believe that there is momentum in the fields. If something tells you this cannot be the whole story, you have sound intuitions. But the resolution of this paradox will have to await Chapter 12 (Ex. 12.12).
Suppose now that we turn up the resistance, so the current decreases. The changing magnetic field will induce an electric field (Eq. 7.20):

[latex]E(s)=\left[\frac{\mu_0}{2\pi}\frac{dI}{dt}\ln s+K ight]\hat{z,} [/latex] (7.20)

[latex]E=\left[\frac{\mu_0}{2\pi}\frac{dI}{dt}\ln s+K ight]\hat{z}. [/latex]

This field exerts a force on ±λ:

[latex]F=\lambda l\left[\frac{\mu_0}{2\pi}\frac{dI}{dt}\ln a+K ight]\hat{z}-\lambda l\left[\frac{\mu_0}{2\pi}\frac{dI}{dt}\ln b+K ight]\hat{z}=\frac{\mu_0\lambda l}{2\pi}\frac{dI}{dt}\ln (b/a)\hat{z} . [/latex]The total momentum imparted to the cable, as the current drops from I to 0, is therefore

[latex]P_{mech} =\int{Fdt}=\frac{\mu_0\lambda Il}{2\pi}\ln (b/a)\hat{z}, [/latex]

which is precisely the momentum originally stored in the fields.

Question 8.3: A long coaxial cable, of length l, consists of an inner cond...

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