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Does a time varying magnetic field produce an EMF in an open conducting loop? (slidewire generator example)?

Physics Asked by Joshua Gardner on January 2, 2021

Consider a conducting rod placed in a magnetic field. I understand that if the rod moves normally to the field, it experiences an EMF. Furthermore, if the ends of rod are connected by a wire and the rod moves such that there is a flux change,the EMF would drive a current through the loop.

  1. If instead of the moving the rod to change the magnetic flux, if one were to increase the strength of the magnetic field itself would there still be an EMF generated and a current driven?

  2. Does this correspond to a moving rod in a constant field or does my classroom understandng(see below) of induction only applies to flux change caused by a conductor’s motion)?**

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My classroom understanding:

When my physics class was just learning about Faraday and Lenz’s law, we were shown the example of a slidewire generator to explain how a change in magnetic flux relates to an induced EMF and a corresponding current: An open-ended rod passed perpendicularly through a constant magnetic field will see positive and negative charges gather at opposite ends, as the relative motion between the charges and the field results in a magnetic force. There is no closed loop through which to analyze flux, but there is potential between the ends of the rod as long as it is moving through the field and the positive and negative charges are being forced apart. If we move the rod again, this time sliding it along the “rails” of a c-shaped piece of wire, we now have a closed loop, sitting perpendicular to the constant magnetic field. The potential in the rod can now push a current through the loop, generating a magnetic field in the opposite direction of the constant magnetic field. The motion of the rod now not only corresponds to a magnetic force on the charges in the rod, but also a change in the area enclosed by the loop, and thus a change in magnetic flux through the loop.

This all makes perfect sense.

But there are other ways to alter the magnetic flux than just changing the area of the loop. For example, we could increase the strength of the magnetic field while holding the loop area constant, and the loop would still see an EMF, according to Faraday.

So finally to the meat of my question: If we go back to the beginning, before the rod was sitting on the rails of our loop, and instead of pushing it along through the magnetic field, we instead started changing the strength of the field, would the rod (now open-ended) still experience a separation of charges like it did when we were pushing it through a constant field?
Does the changing magnetic field around a stationary rod count as relative motion between the field and the charges in the rod, corresponding to a moving rod and a constant field? Or does my class’s explanation of induction only apply to flux caused by a changing area (physical motion of the conductor)?

2 Answers

  1. As per Faraday's law, any time varying magnetic flux induces an EMF. So even when a rod doesn't move, an increasing magnetic field aka a time-varying magnetic field, would produce a time varying magnetic flux thereby producing an EMF.

  2. No this doesn't correspond to relative motion between the rod and the magnetic field*. Understanding the origin of charge separation in this scenario is a bit tricky. A time varying magnetic field actually generates an electric field** which causes charge separation and EMF production.


*Consider an everywhere homogeneous time varying field. The magnetic field strength experienced by the rod at any instant is independent of its configuration in the field and thereby its velocity.

**This electric field unlike that of electrostatics, is non-conservative and is therefore different.

Answered by lineage on January 2, 2021

"Does the changing "line density" of the magnetic field around a stationary rod count as relative motion between the field and the charges in the rod, corresponding to a moving rod and a constant field?"

The answer is: no, and there will be no separation of charges in a stationary straight wire due to a changing magnetic field.

When we move a metal rod through a magnetic field, the free electrons of the metal move with it and experience 'magnetic Lorentz forces' ($vec F = (-e) vec v times vec B$) along the wire, hence the emf. If the rod is open-ended one end acquires a surplus of electrons, the other, a deficit. These charges set up a pd across the wire and an electric field which soon exerts a force on the electrons that is equal and opposite to the Lorentz force, so stopping further electron migration.

When we change the magnetic field normal to a closed loop (real or imagined) we induce an emf in the loop. There are no moving electrons so no magnetic Lorentz forces. Instead there is an electric field that gives rise to an emf in the loop. This is in accordance with the Faraday-Maxwell (F-M) law that applies to all points and states that $$text{curl} vec E = -frac{d vec B}{dt}.$$ Whereas for a closed loop we can integrate the F-M law to give the magnitude and direction of an emf, there is no obvious way to do this for an open-ended conductor, and for your stationary straight rod, symmetry shows that there will be no emf (as the rod could be part of an 'anticlockwise' or a 'clockwise' loop).

Answered by Philip Wood on January 2, 2021

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