Ampère's Law in its integral form (ignoring the term Maxwell introduced for capacitors) essentially says that the total magnetic field directed around a closed loop is proportional to the current flowing in it. This carries the implication that any loop of creates a corresponding magnetic field (the basis of solenoids etc). Please do correct anything incorrect in the above! My question is this: does the converse of the above work (i.e.: does a magnetic field create a current in a wire). My intuition says this is wrong since in most of EM you need some notion of a changing field for any effect to be produced. Thanks for any help!

  • 1
    $\begingroup$ You're right throughout – including your intuition about requiring a $changing$ magnetic field (or motion of a conductor through a field). [Faraday didn't know this when he first started looking for an electrical effect of magnetic fields to partner the recently discovered magnetic effect of an electric current.] $\endgroup$ – Philip Wood Jun 28 '17 at 12:29
  • $\begingroup$ "total magnetic field directed around a closed loop is proportional to the current flowing in it." - it's proportional to the current through the surface bounded by the loop. Is this what you mean? $\endgroup$ – Hal Hollis Jun 28 '17 at 14:08

You are right in your description and intuition of Ampère's law - the presence of a magnetic field does not imply a current will be created in a nearby wire. However, there are two things to note:

  1. The presence of the magnetic field implies there is a current somewhere (no current = no magnetic field).
  2. If you start in a situation with a superconducting loop of wire and no magnetic field, and they you create a magnetic field (by sending current through a loop somewhere), then you will indeed get a corresponding current in the superconducting loop (so as to keep the flux through the loop unchanged). But as you say, at some point during the experiment this required a changing field (but after that, it can remain static and the current will continue).

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.