I am reading the Griffiths's Introduction to Electrodynamics, Example 8.3 and stuck at understanding some statements:

Example 8.3. 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 $\lambda$, 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 ?

enter image description here

Solution : The fields are

$$ \mathbf{E}=\frac{1}{2\pi \epsilon_0}\frac{\lambda}{s}\hat{\mathbf{s}}. $$ $$ \mathbf{B}=\frac{\mu_{0}}{2\pi}\frac{I}{s}\hat{\mathbf{\phi}}.$$

Q. My question is, Why are $\mathbf{E}$ and $\mathbf{B}$ determined that way? I'm still struggling to solve this on my own. And I've searched everywhere, but haven't been able to find anything related to it. I'm not necessarily asking for an answer ( altough it would be nice if anybody could give me the detailed calculation ). Can anyone give me some resources that I can refer as a hint?

And, he continus to argue as follows :

The poyinting vector is therefore $$ \mathbf{S}=\frac{\lambda I}{4\pi^2\epsilon_0s^2}\hat{\mathbf{z}}. $$

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

$$ P = \int \mathbf{S}\cdot d\mathbf{a}=\frac{\lambda I}{4 \pi^2 \epsilon_0}\int_{a}^{b}\frac{1}{s^2}2\pi s ds = \frac{\lambda I}{2\pi \epsilon_0}\ln(b/a)=IV, $$

as it should be.

Can I ask interlude question, although this is not main question? What the sentence "Evidently energy is flowing down the line, from the battery to the resistor." exactly means? And why the calculation $P = IV$ explains this sentence? And I don't know why the power $P$ is defined as $\int \mathbf{S} \cdot d\mathbf{a}$ and why $V= \frac{\lambda}{2\pi \epsilon_0} \ln(b/a)$. Is there any related resource?

I think that I am beginner of electrodynamics. Digesting these stuffs will deepen my understanding of electromagnetism. Can anyone helps?

  • 1
    $\begingroup$ Are you skipping ahead in the book? Have you thoroughly studied the earlier material on Gauss's law and Ampere's law? $\endgroup$
    – d_b
    May 8, 2023 at 0:35
  • $\begingroup$ @d_b : In fact, I studied it before, but I forgot about it, so I'm looking for it again now. I commented to naturallyInconsistent 's answer. I did my own calculations, but I'm not sure if I got it the right way. Can you see? Anyway thank you. $\endgroup$
    – Plantation
    May 8, 2023 at 0:48

1 Answer 1


The magnetic field part should be obvious. You have a current carrying wire, the expression for its magnetic field is something covered in the book much earlier. Now, because this is a coaxial cable with cylindrical symmetry, outermost and innermost you know that there must be exact cancellation, and so the magnetic field must only be between the gap.

The electric field part is quite similar. The outside of the wire should be considered as grounded, because no field exists between the outer shell and infinity. The inner wire, however, is at higher electrostatic potential, due to the battery setting it up at higher potential. So there must be an E field between the coaxial cable. How big this must be, is actually not so obvious. It depends upon battery $V$ and cable $b$ and $a$, and that determines $\lambda$, where $\lambda$ is the solution for a uniformly charged cable that you have solved before. Back then, $\lambda$ is the fixed value. Now, it is fixed by the system.

The battery is the true source of the electromagnetic energy that is being consumed by the resistor. The battery converts from chemical potential energy to electromagnetic energy, and the resistor is converting electromagnetic energy into heat. We know that $P=IV$ must hold. Maxwell's equations tell us that EM field carries energy via the Poynting vector, and that is where the $\int\vec S\cdot\vec{\mathrm dA}$ comes from.

The whole point is to show that all these definitions and mathematical work gives agreement with earlier, ad hoc results. That everything is correct and proper, not randomly thrown at you.

  • $\begingroup$ Thanks for unexpected detailed answer. But, in fact, through read it several times, I still don't understand it well-on a mathematical level- :) ( I feel like a fool ). It's a little ambiguous, but I'll try to implement calculation. $\endgroup$
    – Plantation
    May 7, 2023 at 13:17
  • 1
    $\begingroup$ It is given in order of difficulty. i.e. you should start by seeing if you get, step by step, the B field, and then the E field, assuming arbitrary $\lambda$ you do not yet know how to relate. Then find $\lambda=\lambda(V,a,b)$. Finally, just show that the integral gives you $P=IV$ $\endgroup$ May 7, 2023 at 14:50
  • $\begingroup$ Thank you. I got strength from your answer. For the $\mathbf{B}$, it seems that it is true by the Ampere's law ( c.f. Griffiths's book Example 5.7. ) And for $\mathbf{E}$, I guess that, by the cylindrical symmetry, by the Gauss's law, $|\mathbf{E}| 2 \pi s l = \oint _{S} \mathbf{E} \cdot d \mathbf{a} = Q_{enc}/\epsilon_0 = \lambda l / \epsilon_0$ so that $\mathbf{E}= \frac{\lambda}{2 \pi s \epsilon_0}\hat{\mathbf{s}}$. Is this argument correct? Does this argument reflect your intent well? $\endgroup$
    – Plantation
    May 8, 2023 at 0:42
  • 1
    $\begingroup$ correct. The battery sets up V, so do radial integral of E to relate $\lambda$ to those $\endgroup$ May 8, 2023 at 1:04
  • 1
    $\begingroup$ battery voltage or EMF. The long name of EMF is a mistake (it is not a force) and so typically it is just stated as the abbreviation and not expanded. It can also be the voltage set up by a power supply $\endgroup$ May 8, 2023 at 8:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.