Below are three circuit diagrams for each of Faraday's experiments that allowed Faraday to come up with Faraday's Law. In Griffiths' Introduction to Electrodynamics Griffiths states (on page 302 of the third edition) the universal flux rule:

Whenever (and for whatever reason) the magnetic flux through a loop changes, an emf

$$\mathscr{E}=-\frac{d \Phi}{d t} \quad\quad (7.17)$$

will appear in the loop.

On page 303 of the third edition Griffiths further says:

Many people call this "Faraday's law."

Griffiths then says (on page 303 of the third edition)

In Faraday's first experiment it's the Lorentz force law at work: the emf is magnetic. But in the other two it's an electric field (induced by the changing magnetic field) that does the job.

My question is why are no electric fields produced if there is a changing magnetic flux in the first circuit diagram?

enter image description here

  • $\begingroup$ Why do you think there are no electric fields produced in the first one? $\endgroup$
    – hft
    Feb 5, 2015 at 3:14
  • $\begingroup$ Because in the text, Griffiths stated that the emf is produced only through the magnetic field. So if there were an electric field, it would play some role in producing the emf. $\endgroup$ Feb 5, 2015 at 3:23
  • $\begingroup$ In each case, the changing magnetic flux induces an electromotive force. The first and second diagrams are equivalent by relativity. In both the first and second diagram an electric field is induced in the wire (that causes the EMF) by the changing magnetic flux (changing area). In the third case it is the field strength that changes. $\endgroup$
    – hft
    Feb 5, 2015 at 3:33
  • $\begingroup$ @hft Griffiths describes the emf only using the magnetic field. So why isn't the eletric field used to describe the emf in the first case? $\endgroup$ Feb 5, 2015 at 3:45
  • $\begingroup$ In all the cases EMF = negative of time-derivative of Magnetic Flux. The EMF effectively is the electric field; you can't use it to describe itself... $\endgroup$
    – hft
    Feb 5, 2015 at 4:08

2 Answers 2


When you change the flux through a circuit, there are two reasons the flux changes:

1) First, the $\vec{B}$ field in a surface instantaneously spanned by the circuit (at that moment) is changing, in which case there is an electric field in that surface with a circulation $\oint \vec{E}\cdot d\vec{\ell}$ s around the loop that equals $\int -\frac{\partial \vec{B}}{\partial t}\cdot d\vec{a}$, so that:

$$ \oint_{\partial S} \vec{E}\cdot d\vec{\ell}=\int\int_S -\left(\frac{\partial \vec{B}}{\partial t}\right)\cdot d\vec{a}.$$

And this is rightly Faraday's Law (not the "universal" flux rule), because it is the mathematically equivalent integral version of: $$\vec{\nabla}\times\vec{E}=-\frac{\partial \vec{B}}{\partial t}.$$

So Faraday's law says that circulating electric fields cause the $\vec{B}$ field to change (popular sayings get the causality backwards). And get this straight, a circulating electric field is what causes the $\vec{B}$ field to change and the changing $\vec{B}$ field through an instantaneous surface between the circuit is one (of two) things that can make the flux change. The second and third experiments fall in this category. So what is the other reason the flux can change?

2) Second, the circuit itself can have velocity, $\vec{v}$, so the change in location of the circuit in the instantaneous $\vec{B}$ field could result in the $\vec{B}$ field being integrated through a surface whose boundary is changing. In this case (because there are no magnetic monopoles), the change in flux due to the moving circuit equals the circulation $\oint_{\partial S} -\left(\vec{v}\times\vec{B}\right)\cdot d\vec{\ell}.$ The nonmobile charges in the moving circuit are stressed by the magnetic force, but in the quasistatic limit the strain on the nonmobile charges is neglected (and already included in the motion of the circuit) and also in the quasistatic limit the actual motion of the mobile charges differs from the motion of the circuit $\vec{v}$ only by something parallel to the circuit direction $d\vec{\ell}$ so that $\oint_{\partial S} -\left(\vec{v}\times\vec{B}\right)\cdot d\vec{\ell}$ is actually numerically equal to (the negative of) the circulation of the magnetic force per unit charge around the circuit. The first experiment falls in this category. Well, technically the current produces its own $\vec{B}$ field, and it moves so there is a changing $\vec{B}$ field, so there is a little bit of circulating electric field even in the first experiment. This is called self inductance, so the first experiment includes both effects. But it is the only example amongst the three experiments listed that has this second effect where the magnetic force per unit charge is contributing to the emf $\mathscr E$ because the circuit element is moving through a $\vec{B}$ field.

Since these two effects completely determine the change in flux and the change in flux is the sum of these two changes (product rule), the (negative of the) total change in flux is equal to the sum of the circulation of the electric force per unit charge around the circuit and the circulation of the magnetic force per unit charge around the circuit. Their sum is the circulation of the Lorentz Force per unit charge around the circuit, which is the emf, $\mathscr E$, due to electromagnetic forces.

Thus, in quasistatics:

$$\mathscr E=-\frac{d \Phi}{dt}$$

Now, I have to say that I don't see any reason to think the "universal flux rule" actually holds outside quasistatics, since in general charges can move with a velocity other than the velocity of the wire plus a velocity term parallel to the wire. Thus the second effect due to the moving circuit will not always be equal to the circulation of the magnetic force per unit charge around the circuit. But it will be if charges aren't flying off of your circuit and instead the charges only going around it. So you still know when to expect it to hold. In the quasistatic limit, electrostatic forces have time to keep mobile charges flowing through the wire, and electrostatic fields don't contribute to the electromagnetic emf. But it does mean the name ``universal flux rule'' is a misnomer.

Finally as a caveat. I said the electrostatic forces didn't contribute to the emf, but since the circuit is moving, the electric fields responsible for keeping the mobile charges inside the wires (not flying out of the wires) can be non-electrostatic electric fields, which are then what is responsible for the self inductance.


In the first one notice that the loop is being moved and thus the charged particles in the loop. Hence there are electrons moving in a magnetic field, which is basically Lorentz force. Griffiths argues that the Emf created in this scenario is due to, or better yet can be explained by the Lorentz force.

In the second and the third one there is no charged particle moving in a magnetic field, which means that you cannot use Lorentz force law to explain the phenomena. Therefore you need another explanation, which is Faraday's Law. There is a changing magnetic field, therefore there is an Emf in the circuit.

Side Note: This particular phenomena led Einstein to think about relativity. The first and the second case is relativistically equivalent. You can say that I'm in the frame of reference, where the magnetic field moves or you can say that I'm in the reference frame, where the magnetic field is stationary. Disturbed by the fact that the same phenomena is explained by two different laws, he developed the special theory of relativity. In the very first paragraph of his famous paper “Zur Elektrodynamik bewegter Körper” he wrotes:

It is known that Maxwell's electrodynamics—as usually understood at the present time—when applied to moving bodies, leads to asymmetries which do not appear to be inherent in the phenomena. Take, for example, the reciprocal electrodynamic action of a magnet and a conductor. The observable phenomenon here depends only on the relative motion of the conductor and the magnet, whereas the customary view draws a sharp distinction between the two cases in which either the one or the other of these bodies is in motion. For if the magnet is in motion and the conductor at rest, there arises in the neighbourhood of the magnet an electric field with a certain definite energy, producing a current at the places where parts of the conductor are situated. But if the magnet is stationary and the conductor in motion, no electric field arises in the neighbourhood of the magnet. In the conductor, however, we find an electromotive force, to which in itself there is no corresponding energy, but which gives rise—assuming equality of relative motion in the two cases discussed—to electric currents of the same path and intensity as those produced by the electric forces in the former case.[Bolds and italics by me]

Reference: Zur Elektrodynamik bewegter Körper, A.Einstein (On the Electrodynamics of Moving Bodies)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.