Say we have a single, solid circular loop of wire rotating at a constant angular velocity in a uniform magnetic field. There is an induced EMF creating a current in the wire. At the absolute minimum, we should be able to say that the potential V is 2pi periodic (it should have a single value at a single position on the loop). In my understanding of the problem, the wire ought to be equipotential, even. While the wire is radiating energy due to its resistance, that energy is getting put back into the system by whatever is spinning it at constant angular velocity.

Now, break the loop and attach a voltmeter to the two ends of the wire. It will read a potential drop. I am struggling to rectify these two ideas. Help?

(I know there are some similar questions to this, but I have not seen an answer that specifically addresses this issue.)

  • $\begingroup$ Be aware that emf and potential difference are distinct and quite different concepts. I'll work on an answer if time permits. $\endgroup$ – Alfred Centauri May 12 '14 at 0:11
  • $\begingroup$ Yes, I guess that's what I am getting at in the first half of my question. My understanding has always been that this has (essentially) nothing to do with potential difference (or even the scalar potential at all). I am TA'ing an E&M course presently, and all the students seem to have been taught that they are interchangeable. I would not know how to respond, however, if I were to mention that there is a difference between the terms, and they were to ask why, then, a voltmeter measures a potential difference between the two ends of the probe... $\endgroup$ – nosirrahcd May 12 '14 at 1:34

emf is the line integral of the electric field along a closed path and is proportional to the rate of change of magnetic flux through a surface bounded by the closed path.

Consider a perfect conductor that almost forms a closed path; a loop of conductor with an arbitrarily small part removed.

Assume there is a changing magnetic field threading this loop.

Since, for a perfect conductor, there must be zero internal electric field, the line integral of the electric field within the conductor is zero. The only contribution to the integral must be within the gap in the conductor.

However, if we perform the same line integral without the conductor present, we must get the same result.

How is this possible?

Within the perfect conductor, charge has redistributed in such a way as to cancel any internal electric field. Since the electric field of this charge distribution is conservative, there is an associated potential difference, that can be measured with a voltmeter, and is, essentially, equal in magnitude and opposite in polarity to that of the emf. This is necessary in order to zero out the field within the conductor.


At the absolute minimum, we should be able to say that the potential V is 2pi periodic (it should have a single value at a single position on the loop)

This is a consequence (or a restatement) of Kirchoff's Voltage Law (KVL). KVL applies to lumped circuits.

In fact, the common definition of the scalar electrical potential is only valid when $\vec{\nabla}\times{}\vec{E}=0$, which is no longer the case when you introduce a changing magnetic field.

Therefore you should not expect to be able to evaluate your rotating loop using the scalar potential, and you must resort to Faraday's Law of Induction.


Refer the basic conduction mechanism in electric circuits. It is based on the surface charge feedback mechanism of electrons. In a dc circuit,when we say there is a voltage drop across the resistor, there is actually surface charge gradient along the resistor which is responsible for creating the necessary electric field inside the resistor. Due to this charge gradient,voltage is present.Voltmeter senses this charge gradient and current is established in the voltmeter.This current is proportional to the charge gradient and hence, to the voltage across the resistor.Voltmeter reads the voltage due to surface charge gradient only.Thus,voltmeter only reads static electric fields. For a conducting wire,the gradient of surface charge is so small that voltmeter draws almost negligible current,hence shows 0 voltage.That doesn't mean entire wire is at the same potential. Only for convenience,we assume so.

Now,in case of induced fields, there is no question of charges. They just APPEAR whenever there is a change in magnetic flux(It is governed by a beautiful part of physics,called Maxwell's equations). So voltmeter can't read any induced field.**However,if you break the loop and attach voltmeter in the gap,it sure will read the voltage. Because,**the induced field will cause the electrons to move opposite to the field so as to cancel it and they will accumulate at one end and start forming a gradient. This process will stop once the field due to gradient exactly equals(or cancels,precisely) the induced field. So,now we have voltage due to a static electric field which is equal to the induced emf. Thus,voltmeter reads it. This also explains why E-field inside a conductor placed in external field is 0,as stated by Gauss' theorem. (In a DC circuit,this Guass' statement isn't true as there is a field inside the wire. Is Gauss wrong?? Try this question..Its really interesting!!)

The process of formation of these electron gradients is transient in nature.It is called surface charge feedback mechanism.It occurs due to natural initial imbalance between fields and charges.It takes a few pico-seconds to complete.It is really the 'common sense' of circuit theory. Visit youtube for video lectures on this mechanism. Hope this helps...Good luck!


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