Is there a general rule that will allow one to find the parts where the emf is induced?

Consider for example the following two cases:

Case 1: A rod is made to move normally to the magnetic field at constant speed. Diagram

In this case, one can say that the emf is induced between the ends a and b. To be more precise the horizontal line segments are the parts where the emf is induced.

Case 2: Consider the following diagram: enter image description here Question is : enter image description here


enter image description here

If this door is moved about RS in the earth's horizontal magnetic field then in which parts is the emf induced? The answer states that as RS is fixed only the part QP cuts the magnetic field lines. Consequently, the emf is induced in either part QR or PS. I don't understand the last part. How does the first statement imply the second one ? It would be helpful if one could give a link the answer with Case 1.

  • $\begingroup$ Can I know the source of the question please? $\endgroup$ – Jimmy Kudo Apr 7 '16 at 18:55
  • 1
    $\begingroup$ It was asked as an examination question. $\endgroup$ – model_checker Apr 8 '16 at 5:34
  • $\begingroup$ Are you sure its a correct question, because when calculating, we have to integrate considering small vertical bars, and they are like batteries in parallel combination. The emf for each small bar will reduce as we move towards RS. But if we use KVL it will give us an inconsistency. Thus we need some information about resistance. as batteries with different emf cannot be added if they dont have resistance $\endgroup$ – Jimmy Kudo Apr 9 '16 at 6:30
  • $\begingroup$ I've made edits. This is all the information given. I hope that this clears some of the confusion... $\endgroup$ – model_checker Apr 9 '16 at 18:26
  • $\begingroup$ "The answer states that (...)": is this an exact quote of the answer? $\endgroup$ – L. Levrel Apr 9 '16 at 20:27

In a situation like this, it's always useful to go to the basic laws. Faraday's law in its usual formulation ("induced emf in a closed circuit equals minus the time variation of magnetic flux through the circuit") really covers two different phenomena: (i) induction of electric field by a time-varying magnetic field, and (ii) magnetic (Lorentz) force acting on moving charges. The first of these arises from one of Maxwell's equations, $\vec\nabla\times\vec E=-\partial\vec B/\partial t$. In both of your setups, however, the magnetic field is fixed, so we are dealing with the second case.

Think of the first example with the moving rod. As the rod moves, the magnetic field will push the electrons inside it sideways, since the magnetic force is proportional to $\vec v\times\vec B$. As a result, the ends of the rod will become polarized. The polarization charge will in turn induce an electric field inside the rod. This polarization process will continue until the forces of the magnetic field and the induced electric field on the charges inside the rod exactly cancel each other. Since the total force on a single charge $Q$ is $$ \vec F=Q(\vec E+\vec v\times\vec B), $$ the induced electric field at the end of the day will be $$ \vec E_\text{ind}=-\vec v\times\vec B. $$ From here, you can easily calculate the induced voltage between any two points in the bar.

The second problem can be answered in exactly the same way. There is no need to invent any closed "circuit" here and try to find the induced emf in it. In fact, there are some tricky examples where Faraday's law as formulated above in words does not hold: see Feynman's lectures, vol. II, sec. 17-2. Sticking to the basic laws, you can't make a mistake though.

  • $\begingroup$ After reasoning I've come to the conclusion that emf must be induced across PQ, then why does the answer state that the emf is induced across QR or PS? $\endgroup$ – model_checker Apr 15 '16 at 15:33
  • $\begingroup$ @User-3.14 Well, I'm not responsible for that answer so I don't know what exactly they meant. Perhaps the wording means an emf between QR and PS? In any case, as pointed out by L. Levrel in his/her answer, there will also be a tiny emf induced vertically inside QR/PS. $\endgroup$ – Tomáš Brauner Apr 16 '16 at 8:49

How does the first statement imply the second one ?

It does not. There's no reason why the emf in different parts would be "logically" linked to that in other parts.

The wording of the answer is also dubious to me (although I'm not 100% sure since not a native English speaker): "flux" is the surface integral (of B). Rods may cut magnetic lines, not magnetic flux.

RS and QP are easily understood in relation to Case 1. Now, for QR (resp. PS), view them as very large and short rods: you find emf is induced vertically, in the thickness of QR (resp. PS), not horizontally along QR (resp. PS). If you consider them to have no thickness, there is no emf altogether. Anyhow, the line integral of E along QR (resp. PS) is zero.

All in all, emf in a closed circuit is much easier to compute by Faraday's law.


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