4
\$\begingroup\$

Why is the magnetic field inside a coaxial cable the same as it would be near a current-carrying wire? Each side of the cable is conducting so shouldn't it be \$ \frac{I*μ0}{\pi*r} \$ because both sides contribute equally (at least in the middle between the radiuses)?

enter image description here

\$\endgroup\$

2 Answers 2

5
\$\begingroup\$

The outer shield of the conductor carrying the return current cannot generate a magnetic field inside the tube it forms hence, the only magnetic field between inner conductor and outer (tube) is generated by the current flowing through the inner central conductor. Try googling magnetic fields within a tube conductor carrying current; the field is zero. Laws of physics.

QuickField simulation of the outer shield (a tube) carrying 1000 amps with the inner wire open circuit: -

enter image description here

Image taken from here. As can be seen, there is no magnetic field between inner conductor and inside of shield tube.

\$\endgroup\$
1
  • \$\begingroup\$ Thank you for your answer! Apparently I got confused because I knew that there is magnetic field inside a solid current-carrying conductor and assumed that it is the same case for hollow conductors too. I guess I'll have to do some more reading :) \$\endgroup\$
    – CapBul
    Jun 23 at 23:20
2
\$\begingroup\$

There are two answers, the first essentially being of the "that's just the way it happens to be" kind: first, the magnetic field vector is the rotation (a differential operator often denoted by \$\nabla \times \dots\$) of real and "displacement" current densities. A feature of this vector rotation is that a closed line integral of it evaluates to exactly the same value as any surface integral of the original vector inside of that closed line would evaluate to. Now that surface integral of the current density indicates the total current \$\vec J + \frac d{dt} \vec D\$ going through this surface, and the line integral around tells the magnetic strength \$\vec H\$. Now if we pick a circle for the line integral and let current flow in its center (in any arrangement from line to filled circle to hollow circle that is in itself symmetric), the field \$\vec H\$ will have the same magnitude all around and will exclusively depend on the current density inside. The whole integral will actually remain completely unchanged regardless of how any current may be distributed outside (symmetric or not) or inside (symmetric or not), but if we want to stipulate that it's the same magnitude all around, we need to stipulate that symmetry.

Now that's sort of an apodictic statement of the "it is so, period" kind. Essentially, if you integrate small rotation pieces, they all cancel out everywhere except for the boundary where the integration stops.

The grindwork answer is to just integrate all the little pieces of magnetic field generated by each fragment of current according to the field equations. It's a whole bunch of ungrateful work where you can make a lot of mistakes. If you manage without mistakes, the summary solution predicted by the vector differential relations (Gaussian integral equivalences) will still magically fall out. It's a bit deflating, but those vector analysis equalities actually provide for immensively powerful shortcuts. Sort of like energy conversation laws in physics, except that we are talking about something purely mathematical, so it works in the abstract (though Noether's theorem does lend the abstract a stronger route into physics than just coincidence).

Maxwell's equations essentially tie the electric and magnetic field components into those differential vector frameworks.

\$\endgroup\$
1
  • \$\begingroup\$ The "rotation" operator is also commonly called the "curl", and the integral of it along a closed line is called the "circulation". And it might be worth pointing out explicitly that if you remove the center conductor and only consider the hollow tube, the current density inside of it is zero, so the magnetic field strength inside has to be zero as well. \$\endgroup\$
    – TooTea
    Jun 24 at 8:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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