enter image description here

I want to wind an air core solenoid around a copper tube. One end of the solenoid is soldered to the copper tube. (See picture above). Will a magnetic field be induced in the copper tube that cancels out a significant part of the field from the solenoid? (See picture below.) I regard anything smaller than 1% of primary field, or 20 dB down, to be insignificant.

enter image description here

Please note that, in the images, numbers in circles represent steps in the following logic.

Here is what I think will happen:

enter image description here

  1. Current flowing through solenoid sets up the primary magnetic field.
  2. The primary magnetic field induces an EMF.
  3. The induced-EMF causes a current to circulate around the tube.
  4. The induced current causes the induced magnetic field, which is in the opposite direction of the primary.

Here are my questions:

  1. Am I thinking through this correctly?
  2. How can I calculate the magnitude of the induced magnetic field?

Edit: Several people have asked about timespan. The solenoid will be energized for 3*tau (where tau = L/R) and then de-energized for >> 100*tau. Current is DC in the sense that it only ever moves in 1 direction. (Obviously current still has an AC component in Fourier domain.) But we're not talking about DC in the sense that the coil reaches steady state while energized. I hope this helps.

  • \$\begingroup\$ Why can't you use something non-conductive like wood, plastic, or composite? Copper is a bit of a strange choice to begin with due to its value and lack of suitability. \$\endgroup\$
    – DKNguyen
    Commented Oct 13, 2023 at 15:05
  • 5
    \$\begingroup\$ The copper tube will effectively be a shorted turn coupled to your solenoid. \$\endgroup\$
    – Hearth
    Commented Oct 13, 2023 at 15:10
  • \$\begingroup\$ When? "Net" implies total, but total over what time scale, or waveform? Where: do you want the flux applied to the coil itself, or the flux density, the field around it? What are you doing with the coil, is this like a coilgun thing or something? Does the rate of change matter at all? \$\endgroup\$ Commented Oct 15, 2023 at 12:21
  • \$\begingroup\$ @TimWilliams Thanks for pointing this out! I edited the question to hopefully clarify this. \$\endgroup\$ Commented Oct 15, 2023 at 13:49
  • \$\begingroup\$ Many tau, but which one? You show R_tube and R_DC but both have self-inductance as well. \$\endgroup\$ Commented Oct 15, 2023 at 15:45

2 Answers 2


The answer depends on whether the solenoid is being driven with AC or DC, and if so what frequency AC.

At DC, the copper tube with be irrelevant as the field is constant and there will be no long term induced currents.

As the frequency goes up, you will get increasingly large eddy currents (leading to heating, and reducing the net flux inside the tube). The effect of parasitic capacitance between the winding and the tube will also increase. This parasitic capacitance will significantly reduce the self-resonant frequency of the solenoid.

Whether any of this is an issue for you will depend on the details of your application.


Does Winding A Solenoid Around a Copper Pipe Reduce the \$\boxed{\text{NET}}\$ Magnetic Flux At All?

Not really and, let me be clear that I'm talking about the \$\boxed{\text{NET}}\$ magnetic flux. That's the sum of all the fluxes produced. Well, if you pushed me for a more complete answer that fully considered leakage fluxes then I'd say that the \$\boxed{\text{NET}}\$ flux reduces a bit.

It boils down to this: -

A circulating current in the pipe (a shorted turn) will produce a local flux that is wholly cancelled by a counter-flux from the solenoid. That counter-flux is due to the increase in the solenoid current.

You probably understand that if you removed the copper pipe, a small AC current will flow in the solenoid (due to its inductive reactance) and, if you inserted the copper pipe, extra current will be taken by the solenoid to drive a circulating current in the pipe (transformer action).

But, fluxes from the extra solenoid current and the pipe's circulating current are subtractive and cancel out. It's just like a transformer core; the \$\boxed{\text{NET}}\$ flux comes from the original current flow before the pipe was inserted.

Consider this counter argument: -

  • If the \$\boxed{\text{NET}}\$ flux increased then so would induction
  • If induction increases then so would the pipe circulating current
  • If circulating current increases then so does \$\boxed{\text{NET}}\$ flux
  • It's a vicious circle that ends in melt-down

Consider this alternative counter argument: -

  • If the \$\boxed{\text{NET}}\$ flux decreased then so would induction
  • If induction decreases then so would the pipe circulating current
  • If circulating current decreases then so does \$\boxed{\text{NET}}\$ flux
  • This ends with zero circulating current in the pipe and, we know that isn't true

Will a magnetic field be induced in the copper tube that cancels out any part of the field from the solenoid?

Sure it will but, let's get the terminology right...

  • The copper tube has an induced voltage and, that creates a circulating current due to the pipe acting like a short circuit
  • That circulating current produces a magnetic field that is entirely cancelled by the added current (and its magnetic field) in the solenoid primary
  • The \$\boxed{\text{NET}}\$ flux is the flux that would be seen should the pipe be entirely removed.

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.