1) If a conducting loop was placed in a time varying magnetic field, the changes of $B$ over some time, would produce and electric field as Faraday's law indicates(Regardless if there is a conductor or not):

$$ \nabla \times E = -\frac{\delta B}{\delta t}$$

I cannot understand how the electric field would "drive" current in a closed loop, for the case of the electrostatic field produced by the seperation of charges(e.g ; charged capacitor or Lorentz force) how does this induced-electric field via the $\delta B$ cause a potential difference for current to flow?

2) If that same conducting loop was part of another circuit, with another emf source($emf_c$) in series, produced by a charged capacitor, how do they interact with one another? They are both emf's, but in series, are they all the same? Leading to the application of KVL?

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Of course, under the assumption that $emf_c$ = $emf_E$

The nature of the driving forces are different, how should they interact with one another?


1) The induced electric field is present near and inside the loop, due to changing magnetic field. No potential difference is necessary to have a current, it is the electric field, no matter if Coulombic or induced, which pushes the electrons. In a completely symmetrical cylindrical situation with closed loop, the changing magnetic field would not change electric potential at the points of the loop at all.

2) In this case the charges on the capacitor cause and guarantee non-zero potential difference on the terminals of the loop. The KVL law is about sum of voltage drops in a closed loop, not about emfs in that loop: thus the application of KVL would read

$$ U_{cap} + U_{loop} = 0. $$ In words, potential drop across the loop is the same magnitude as potential drop across the capacitor.

Sum of emfs in a circuit is a different and more tricky thing, because emf due to some circuit element is not necessarily determined solely by voltage on that element. The sum includes emfs due to all electromotive forces, including that of the induced electric field (this emf is not describable by the concept of voltage and is not present in the KVL equation above). If the conductor is ohmic (metals), then we can, based on the differential Ohm's law $j = \sigma E$ write this equation for the whole circuit: $$ emf_{due~to~cap} + emf_{due~to~induced~E} ~= RI $$ where $R$ is Ohmic resistance of the circuit.

One may wonder why on the left-hand side there are emfs due to capacitor and induced electric field in the loop, but no emf due to a possible resistor in series that can be in the way of the current (say, as way to model non-zero resistance of wire in the original setup). The reason is that the sum of emfs is really an integral of electric field along all the path segments the electric current really goes through, and $R$ is corresponding electric resistance of those path segments. Now, there is an integral of electric field through a resistor, but this is already taken into account by emf due to some other element, such as a capacitor causing the voltage in the first place, not to an emf of a resistor. Resistors do not contribute emfs.$^*$

Depending on the direction of change of magnetic field, the induced emf in the loop is either in the same sense as that of the capacitor (then it makes the current $I$ stronger), or the opposite (then it makes the current weaker).

Note that in case of the capacitor, magnitude of voltage $U_{cap}$ and magnitude of $emf_{due~to~cap}$ is the same, but for the loop, $U_{loop}$ does not have the same magnitude as $emf_{due~to~induced~E}~~~~~~~$ (due to induced electric field which contributes to emf but not to voltage drop on the loop).

$^*$ Some sources ignore this "identification of emf to its actual source" and assign any element in a circuit with two terminals $1,2$ ordered in the chosen sense of circulation, including resistor, some emf. This is done presumably to make resulting formulae apparently simpler as then we can write $\sum_k emf_k = 0$ similarly to KVL, where the sum includes both emf due to induced electric field and also all "element emfs", including those of resistors, but I think this is a bad idea. If there is magnetic flux changing through the whole circuit, the induced emf belongs to the whole circuit, not any lumped element in it. So there is no reason to assign emfs to path segments: if we did, their value would be not determined by those segment voltage and current, but also by external bodies (sources of magnetic flux). So it is better to have all emfs to be defined for the whole circuit and distinguish them based on the source.

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  • $\begingroup$ Thank you for clarifying the distinction between PDs and the sum of emfs, quick add-on question here, if the sum of emf's cancel, it's possible to have a PD generated by the capacitor but no current flow due to the net-electromotive force being zero? $\endgroup$ – Geodesic Jan 28 '19 at 6:58
  • $\begingroup$ Indeed ! In any AC circuit with inductive load driven by sinusoidal source of voltage, this happens twice a period. $\endgroup$ – Ján Lalinský Jan 28 '19 at 14:40
  • $\begingroup$ You're welcome. If a post answers your question to your satisfaction, you can mark the post as accepted. $\endgroup$ – Ján Lalinský Jan 28 '19 at 18:49
  • $\begingroup$ @JánLakinsk´y Suppose a capacitor is discharging through a resistor. Could the pd across the resistor be classed as an emf? I want to say 'no', because I want to restrict emfs to energy changes arising from non-conservative fields, but then I note your term $emf_{cap}...$ $\endgroup$ – Philip Wood Feb 2 '19 at 16:26
  • $\begingroup$ @PhilipWood Good question! While the resistor under current has voltage across its terminals, it does not count as emf for the circuit, because such element has no energy stored, it does not act on the current in a way which increases current; on the contrary, resistor always opposes current, dissipates energy into heat. If we included such "resistor emf" just because there is voltage, we would get incorrect equation $Q/C - RI = RI$. The capacitor and inductor are different, because they can store energy inside and can release it by acting on the charge carriers with current-increasing force. $\endgroup$ – Ján Lalinský Feb 2 '19 at 21:32

Question (1) The electric field associated with the changing magnetic force drives the current directly. After all, an electric field is the force per unit charge on a charge, such as an electron in a wire! Suppose the wire is a circular loop through which the magnetic flux is changing. Electrons are driven round the loop by the electric field. I don't think we can even talk about potential difference in a case like this; there is, as you say, no separation of charges and no conservative electric field. But, as I said, the non-conservative Faraday's law electric field is all we need to explain the current that flows in the loop.

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I tend to think of it like this;

An electron moving at a velocity v perpendicular to a magnetic field B will experience a force acting on it which is proportional to its velocity v and proportional to the strength of the magnetic field B, and which is perpendicular to both the direction that it is moving in and to the direction of the magnetic field.

Now if instead of thinking of an electron moving relative to a magnetic field we think of a magnetic field moving relative to a stationary electron in a wire (there should be no difference right?) then the moving magnetic field will cause a force to act on the electron which again is perpendicular to the direction of motion of the magnetic field and to the direction of the magnetic field.

If you use the right hand rule you will find that the force acting on the stationary electron will tend to try to move it in a circle around the direction of motion of the magnetic field.

Voltage is energy per unit charge ie. E/Q in units J/C. Energy is related to force by E=F*v where v is the velocity of the object (in this case the electron), indeed it takes work to move an electron through something that has a resistance greater than 0 ohm, or to move an electron to a higher potential energy (ie. higher voltage).

In order to take it to the next step and relate all this to current in amps we need to know about drift velocity; we tend to think of current moving through a wire at close to the speed of light but that is not true, in fact only voltage propagates through a wire at close to the speed of light. because there are so many free electrons present in a wire the velocity each of them needs to move at on average to cause a measurable current is incredibly small. Now there equations become a bit harry at this point, so I'm not going to explain this in detail, but using the equations for drift velocity you can relate the force acting on the charge and the total charge to the current in amperes, and then we have come full circle, explaining how the force acting on the electron equates to voltage causing motion of electrons ie. current.

To answer your question of how the circuit you show would work is very simple, when there is no changing magnetic field the loop of wire would simply be a short, or very small resistance to be more precise, and once you have a changing magnetic field through the loop of wire it would act as if someone had cut the wire and put an electromotive force (voltage) in series with the circuit. If you know a little bit about electric circuits it is easy to figure out what the currents and voltages would be once the electromotive force is present due to the changing magnetic field.

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  • $\begingroup$ That perspective is simple, and good to imagine thanks! $\endgroup$ – Geodesic Jan 28 '19 at 7:00

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