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If a slim conductor of some length $l$ and diameter $d\ll l$ is placed in a magnetic field $B$, and the field is changed by $\frac {dB}{dt}$, what (if any) is the voltage $V$ induced across the ends of the conductor?

In my case of interest, the slim conductor is a wire, fixed in space, that is victim of interference by an adjacent current, where the $\frac {dB}{dt}$ is caused by $I\ \sin(\omega t)$ in a source wire.

I am particularly interested in calculating a specific case (given $I_0$, $\omega$, and $r$ the distance between the two wires), as well as the fundamental connections to Maxwell's laws, probably the Maxwell- Faraday equation.

I am familiar with Lenz's law , but in my case of interest there is no return path or "ground plane", and so the victim wire has no current loop, or EMF loop. I can't form a curl integral, and no area is determined, and thus no time varying flux. Nevertheless, I would expect the above "wire rod" case to be the Maxwellian foundation of, or at least a step towards the Lenz "loop around flux" case. (Or perhaps I am terribly on the wrong foot here.)

The closest I come to this problem is by the Lorentz force , as it also involves a rod, and it involves a EMF on a charge in motion in a magnetic field. In contrast, my question centers around a time-varying magnetic field, without motion.

To be clear, the rod is fixed in space, and I am interested in the voltage calculation, not the motion or forces.

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(Image from https://www.aplusphysics.com/courses/regents/electricity/images/InductionProblem.png)

A related answer is provided in Induced electric field from homogeneous magnetic field but this leads to a inhomogeneous current distribution and E-field, and I am not sure what to make of it in the case of a rod.

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3 Answers 3

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When the magnetic field is changing, there is no notion of "voltage" only of EMF. If you were to couple the two ends of your rod to a voltmeter, its reading depends on the time rate of change of the magnetic flux in the loop composed of the rod and wires connecting it to the voltmeter. Different paths for the wires will give different "voltage" readings, so the "voltage" is not meaningful. The voltmeter records the EMF which is the force trying to drive a current through the the meter.

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  • $\begingroup$ I think this is what I have the most trouble with: if EMF creates a current, why does it not create a voltage (or an E field through a resistive rod)? (it always puzzled me during univ classes covering EM fields, and I should have asked...) If the coil had a break, would there be a voltage, and can that thinking be extended by gradually opening the coil and straightening it? Is it the same as the voltage across the terminals of an open induction coil? Thank you for joining and letting me ask these questions. $\endgroup$
    – P2000
    Commented Nov 25, 2023 at 3:31
  • $\begingroup$ The EMF certainly corresponds to an ${\bf E}$ field, and there is an ${\bf E}$ field between the ends of a coil but the ${\bf E}$ field is no longer the gradient of a potential. It is the difference in the potential at two points that is called voltage, so no potential means no notion of "voltage." $\endgroup$
    – mike stone
    Commented Nov 25, 2023 at 13:02
  • $\begingroup$ Yes, my concern with EMF is that I'd have to form a curl and it would go through the voltmeter. That's the classic way it's taught (at least to me - as I recall it) But without such loop or meter, is there a voltage (or internal E field - let there be non-zero resistance (or rho) and non-infinite mu if it helps the argument )? The E field across terminals of the cut loop would counter the field inside the coil. Enlarging the gap somehow does not change the theoretical (just the magnitude) but I wonder if straightening it introduces a pole or some other aberration. $\endgroup$
    – P2000
    Commented Nov 26, 2023 at 17:59
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If the (time-varying) field surrounds the wire equally to the right and left of the wire, symmetry surely dictates that the voltage is zero: neither end of the rod gains a charge.

Try formulating a 'hand rule' to predict the direction of the induced emf. Point the index finger (say) of either hand in the direction of increasing magnetic field. Second finger or thumb can still point in any direction at right angles to the index finger; no information about the direction of an induced emf can be obtained. The situation is too symmetrical – and too symmetrical for there to BE an emf!

If the field arises from a changing current in a parallel wire, the symmetry is broken and there is an induced voltage in the 'first' wire. For details, look up 'Mutual inductance of parallel wires'.

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  • $\begingroup$ Thanks Philip, you prompted me to look at this again and I updated my question with a related find. I still don't know what to make of the rod. Wouldn't your symmetry reason rule out induction in a coil altogether? $\endgroup$
    – P2000
    Commented Nov 24, 2023 at 19:14
  • $\begingroup$ "Wouldn't your symmetry reason rule out induction in a coil altogether?" You can relate the sense of the emf induced in the coil to the direction of increasing field by a left-handed screw rule. Is there any rule that could relate a direction of induced current in a straight wire to the direction of increasing field if the magnetic field totally surrounds the wire? $\endgroup$
    – Philip Wood
    Commented Nov 24, 2023 at 22:02
  • $\begingroup$ Philip, yes that's the thinking: EMF is talked about as a current, yet what happens if the coil is cut and then straightened? I have a hard time formulating the thought experiment in which the current is interrupted and an E-field or Voltage is produced. Thank you for the discussion. $\endgroup$
    – P2000
    Commented Nov 25, 2023 at 3:36
  • $\begingroup$ "EMF is talked about as a current" But wrongly. Although the name 'electromotive force' is misleading because emf is work per unit charge, it gives the right feel: if there is an emf in a wire the charges do experience forces. So the free electrons will flow – at least for a short time, until their movement is opposed by accumulations of charges on the ends of the wire if the circuit is incomplete. $\endgroup$
    – Philip Wood
    Commented Nov 26, 2023 at 12:27
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    $\begingroup$ P2000 I've tried to enhance my answer, but I'm afraid that the enhancement is a bit hand-wavy. Would you believe, the pun wasn't intended? I'll keep thinking. $\endgroup$
    – Philip Wood
    Commented Nov 26, 2023 at 18:00
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If a slim conductor of some length l and diameter d<<l is placed in a magnetic field B, and the field is changed by dB/dt, what (if any) is the voltage V induced across the ends of the conductor?

When $B$ is varying, a varying $E$-field (call it external) also appears along the wire. Therefore, I think, if your induced voltage produces an electric field inside the wire (call it internal) which is in the direction of the external varying $E$-field, the wire or rod accelerates along the $E$-fields, and in the meanwhile, it rotates about its center of mass because the internal $E$-field, due to your induced voltage, has accumulated the positive and negative charges, respectively, at each end of the rod, and thus the motion of these charges in the $B$-field produce a torque on the rod due to the Lorentz forces exerted on the rod's ends in the opposite directions.

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  • $\begingroup$ Thank you for your answer, Mohammad. Maybe I should clarify that the rod is fixed in space, and I am interested in the voltage, not the motion or forces. You write "has accumulated the positive and negative charges" which is the question. Has it? And if so, how is it calculated? $\endgroup$
    – P2000
    Commented Aug 3, 2020 at 15:05

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