This is a followup question to this one.

I already found some similar questions in PhysicsSE (as this one and this one) but I didn't found the answer I was looking for.

In a conducting loop wire moving inside a constant magnetic field the charges move, thus a potential difference is induced. Anyway the charges just accumulate to one side of the loop and, after the initial movement of electrons, no current is flowing in the loop. For instance, see the first example at page 128 of these notes.

(Of course things are different if the loop in entering or exiting the region with magnetic field).

The definition of the electromotive force is $$ \mathcal{E}=\oint\textbf{f}_s\cdot\text{d}\textbf{l}, $$ where $\textbf{f}_s$ is a non-conservative force per unit charge.

Using the integral form of Faraday's law, is it correct to state that in the loop moving inside a constant magnetic field $$ \mathcal{E}=-\frac{\text{d}}{\text{d}t}\int_\Sigma \textbf{B}\cdot \text{d}\textbf{A}=0? $$

Moreover, online I found some examples of derivations of the emf when there is no loop involved: 1-dimensional example, 3-dimensional example. But I don't get along which path they are integrating. In the examples involving loops, in order to have $\mathcal{E}\neq 0$, we should have that whether $\textbf{B}$, or the dimension of the loop, or the angle between $\textbf{B}$ and the surface (delimited by the loop) are varying (right?). But with no loop?

Moreover$^2$, the second point of the answer to this question states that emf inside an open ended (non-looped) conductor is $$ \mathcal{E}=\int_A^B \textbf{E}_s\cdot \text{d}\textbf{l},$$ where $\textbf{E}_s$ is the electrostatic field, and the integral doesn't seem to be around a circuit (otherwise it would be zero).

How is this true? Are we integrating over a loop which extends outside the conductor so that $\textbf{E}_s$ contributes only in one direction? What does this result prove? Am I mistaking the definition of emf?


2 Answers 2


How is this true? Are we integrating over a loop which extends outside the conductor so that $\mathbf{E}_{s}$ contributes only in one direction? What does this result prove?

The reason the loop goes to zero is that the integral: $$ \oint_{C} d\mathbf{l} \cdot \mathbf{E} \tag{0} $$ is a path integral of the electric field, $\mathbf{E}$, along segments $d\mathbf{l}$ which result in the calculation of an electric potential difference. If the conductivity of the conductor is infinite, there can be no internal electric fields in the static limit, which is another way of saying the entire conductor must be at the same electric potential. This is why adding up all the elements in a circuit loop using Kirchhoff's circuit laws have a net result of zero (specifically Kirchoff's second law).

If the in Equation 0 is not closed and only covers a small portion of a total loop, it is possible to find a finite result. However, this isn't the net effect of the system and can be very misleading (e.g., could give the impression of a net potential when there is none).

Am I mistaking the definition of emf?

I think so. An emf results from something like a non-electrical source (e.g., changing external electromagnetic fields). If a wire loop is moving at a constant velocity in a uniform, static external magnetic field, there is no net emf in the static limit (i.e., long after the loop started moving). During the initial motion or a case where a loop enters a region of magnetic field from a region without, then there would be a temporary emf until all the charges redistributed accordingly.

You can have a situation where there is charge separation (e.g., open circuit) that results in a net emf, but the resultant electric field cancels that which generated the emf in the first place.

In general, I tend to focus on the induced fields for this topic, as they are much easier to understand. The reason being that the time-varying fields can do work on particles and cause them to move, which is more intuitive. The open-ended circuit examples or non-closed integral examples are misleading in many ways because were the integrals done over closed loops, one would still get zero in the absence of time-varying fields or some source of external work.

  • $\begingroup$ I agree that the source of emf is a non-electrical one. But isn't the definition the one I stated above ($\varepsilon=\oint \textbf{f}_s\cdot \text{d}\textbf{l}$)? Is it correct to say that without a closed loop it does not make much sense trying to derive an emf? $\endgroup$
    – Charlie
    Commented Mar 5, 2021 at 15:13
  • $\begingroup$ But in that expression, $\mathbf{f}_{s}$ is a non-conservative force (i.e., not curl-free). In the limit of everything being static, the only non-conservative forces would be external to the system or something like a chemical energy (e.g., inside a battery, which incidentally, would not be static on long time scales). $\endgroup$ Commented Mar 5, 2021 at 15:19
  • $\begingroup$ Exactly. In the examples I have in mind this $\textbf{f}_s$ is due to magnetic field acting on the moving charges. Therefore the emf is a motional one. $\endgroup$
    – Charlie
    Commented Mar 5, 2021 at 15:26
  • $\begingroup$ Conservative magnetic field? I am not following. $\endgroup$
    – Charlie
    Commented Mar 5, 2021 at 15:33
  • $\begingroup$ Isn't work done by any magnetic field zero (even if its time-varying)? Even if work is not done by $\textbf{B}$ this doesn't mean it is conservative. $\endgroup$
    – Charlie
    Commented Mar 5, 2021 at 15:43

In the case of a closed loop of a conductor without resistance (say a superconductor), the initial transient effect of the applied B-field is to induce an emf, what results in an increasing current. Once the loop is not moving in a way to change to magnetic flow, the current keeps constant, and the emf is zero.

The case of a no-closed loop can be modeled as an (almost) closed loop, except for a capacitor that prevents the closure. The effect of the initial transient applied B-field is to charge the capacitor. After the static condition is achieved as before, there is an oscillating current in the circuit. Now, the emf of the loop (as a whole) is zero as before, but there is an oscillating emf between the capacitor plates.

The mechanical analogy for the first case is an object in the outer space, where a torque is applied for a while. After the torque comes to zero, the spin continues, with constant angular momentum around the COM.

For the second case, it is like a spring that receives an initial force and keeps oscillating after the force is taken out. The total emf corresponds to the external applied force, and the emf in the capacitor plates to the internal stresses of the spring.

  • $\begingroup$ I do not understand why an oscillating emf should arise in the open loop case. $\endgroup$
    – Charlie
    Commented Mar 5, 2021 at 20:01
  • $\begingroup$ @ClaudioSaspinski - Are you thinking about a non-ideal capacitor here? In the limit of an ideal capacitor, there will be no oscillating current. $\endgroup$ Commented Mar 5, 2021 at 20:11
  • $\begingroup$ If the magnetic flow doesn't change, there is no emf. So the charges in the plates of the capacitor cannot stay there. It will discharge. As the conductor has inductance, it is a LC circuit in its natural frequency. $\endgroup$ Commented Mar 5, 2021 at 21:03
  • $\begingroup$ I am not convinced. Moreover the first sentence of your answer seems to contradict what you usually read in books: no current flow in a loop wire moving at constant speed in a constant magnetic field. Emf is zero, thus no current. (For instance see web.mit.edu/sahughes/www/8.022/lec14.pdf at page 128). $\endgroup$
    – Charlie
    Commented Mar 5, 2021 at 22:31
  • $\begingroup$ @Charlie That is true in practice, because real conductors have resistance. And without emf, the current is zero. But with superconductors, without emf, any current stays constant. $\endgroup$ Commented Mar 5, 2021 at 23:14

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