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

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)?

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.

• "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." Alright, so we have a stationary straight rod lying horizontal and North-South. We steadily increase a vertically downward magnetic field in the region of the rod. In which direction is the emf that (I think you are claiming) is induced in the rod? Apr 27, 2020 at 10:27
• For your rod, there's no flux change so no emf. However if you connected a voltmeter to its two ends you would in fact get a voltage reading that depends on the voltmeter setup. Apr 27, 2020 at 18:51
• In other words, there is no emf defined for the rod in the ordinary sense. Similarly, if the rod was connected to a loop in the xy plane, the rod's emf would depend on the loop's configuration. Apr 27, 2020 at 18:52
• Good. We're agreed about the Physics. The sentence of yours that I quoted suggested that you thought there would be an emf "in the ordinary sense", which I took to mean one that would separate charges in the rod. Apr 27, 2020 at 18:58

"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).

• So as I understand it then, the Lorentz forces are not sufficient to explain induced EMF, but rather rather the fact that there is an electric field due to the changing magnetic field, and the relationship between them is what is described in the F-M law. Apr 25, 2020 at 23:32
• @Joshua Gardner The Lorentz force, $\vec F=q(\vec E + \vec v \times \vec B)$, is the vector sum of the force on a charged particle due to (1) an electric field (2) a magnetic field. For the rod moving through a magnetic field it's the second (magnetic) term that gives the force and hence the emf. When the magnetic flux through a stationary loop changes, it is the first (electric field) term that gives the emf. In both cases we may talk of an $induced\ emf$, and may use the equation $\mathscr E = -d \Phi/dt$. Apr 26, 2020 at 8:06
• @Joshua Gardner So your understanding (in your comment above) is basically sound, but I'd re-word it thus: "the Lorentz forces are not sufficient to explain induced EMF in a stationary loop, because as well as the 'electric' part of the Lorentz force we also need the fact that there is an electric field accompanying the changing magnetic field, as described in the F-M law." [A subtle point: I use 'accompanying' rather than 'due to' because I don't want to imply cause and effect.] Apr 26, 2020 at 17:17
• Awesome, that definitely sharpens my understanding of the interactions at play. Apr 26, 2020 at 23:02
• It occurs to me that if the magnetic field were of finite size (such as between the poles of a large electromagnet), then a changing flux would produce an electric field looping around the center of symmetry. This electric field would act on the electrons within a rod (placed away from the center of symmetry) and cause a separation of charge. Mar 31, 2021 at 17:54

Sorry for my poor english. My native language is french.

If you place a copper cylinder in the magnetic field $$\vec{B}=B_0cos(\omega t)\vec{e_z}$$ of a solenoid powered by a sinusoidal current, it is well known that currents flow in the conductive cylinder: these are the famous eddy currents, used in induction heating.

They are associated with the induced electric field which is orthoradial with respect to the axis of the solenoid: $$\vec{E_i}={\frac{1}{2}B}_0\omega r sin(\omega t)\vec{e_\theta}$$

If the cylinder of conductivity $$\gamma$$ is centered on the axis of the solenoid, by symmetry, these currents are circular, centered on the axis of the cylinder: $$\vec{j}=\gamma\vec{E_i}=\gamma{\frac{1}{2}B}_0\omega rsin(\omega t)\vec{e_\theta}$$

(To simplify, we neglect the magnetic field associated with the induced currents)

But if you offset the copper cylinder, these currents will still be centered on the axis of the conductive cylinder! How to explain it?

It is that the electric field to be taken into account is not only the induced electric field. It is necessary to add to it the electric field $$\vec{E_{ele}}$$ generated by the charges which accumulate on the surface of the conductor: $$\vec{j}=\gamma(\vec{E_i}+\vec{E_{ele}})$$: the currents which circulate in the piece of copper include a capacitive term, associated with the surface charges which vary with the field.

You might think that the above explanation has nothing to do with the question because the conductor is closed, while it is open in your question. But if you deform the copper cylinder, you end up with a piece of copper placed in a varying magnetic field. It's hard to tell if it's "open" or "closed".

For example, to fight against eddy currents, we laminate the material and the currents start to circulate differently.

So, in your question, the open portion of conductor is a piece of copper placed in a variable electric field associated with the variable magnetic field : currents will flow and charges will build up at the surface. The problem is complicated. Depending on the geometry, the currents will be rather closed or rather capacitive: in this last case going and coming from the charges as in a capacitor.