0
$\begingroup$

This question is related to Walter Lewin "Complete Breakdown of Intuition" question asked earlier. Imagine an area where the magnetics flux is changing and the electric field lines form circular loops with E-field uniform along the circular loop. Now I introduce a wire loop with a resistor in it. What are the Electric fields now around the same loop including the resistor and the ideal wires? The integral of the electric field is still the same = to rate of change of magnetics flux but at each location around the loop they are no longer uniform. How did the electric field across the resistor got generated? Is the E-field due to the magnetic flux causing the charge in the ideal wires to move and pile up against the resistor which generates the E fields necessary to generate the current across the resistor? To make the discussion simpler we can assume constant rate of change of magnetics flux so E-fields are stationary after the wire loop is introduced. Is there a way to simulate the electric fields before and after the current loop is introduced and also show where charges are accumulated if any?

As an answer to Walter Lewin demo question I wrote:

Instead of using voltage and voltmeters let's use just the fundamentals like Electric field, charges, and force. Get rid of the voltmeters and explain what the electric fields are inside resistor R1, R2 and the ideal wires. Before we inserted the current carrying wires, the electric field lines formed circular loops uniform across the loop. Once current carrying loop including R1 and R2 is introduced the electric fields change completely although the integral around the same circular loop is still equal to the rate of change of the magnetics flux through it. Explain why did the electric fields around the same circular loop changed? Is there a charge build up across the resistors that provide the field across the resistor? Is there any electric field across the ideal wire which will cause infinite current?

If you can explain the above, then it is simple to explain the voltage readings without reference to voltmeters. You can connect a capacitor across R1 and another across R2 and if they build up opposite and unequal charges I can see how the scope will show different voltages.

I know this is not really an answer and I should perhaps word it as a question. But it might help to answer this question as well.

$\endgroup$
  • $\begingroup$ Ask yourself what a wire does. Once you understand what wires do, you got your answer. $\endgroup$ – CuriousOne Jun 18 '16 at 1:10
  • $\begingroup$ Closely related: physics.stackexchange.com/questions/7936/… $\endgroup$ – dmckee Jun 18 '16 at 14:39
  • $\begingroup$ Could we have a link to "Walter Lewin "Complete Breakdown of Intuition" question" and a figure please. $\endgroup$ – jim Jun 18 '16 at 16:41
-1
$\begingroup$

You can use the Faraday's law (integral form) to find the integral of the electric field but not the field itself. For that you need to apply the some conditions. E.g., 1. There cannot be an electric field in an ideal wire, 2. The field is constant through the resistors, etc. Using this you can easily derive the electric field.

Secondly, the electric fields are not due to build up of charge; they are induced electric fields produced by fundamental laws of nature. There is another kind of emf due to build up of charge called motional emf, which also follows Faraday's law but that is when the wires move in a magnetic field to change the flux, instead of changing magnetic field producing changing flux, e.g. In a AC generator. Motional emf is due to Magnetic force (qv×B) and not electric field.

Thirdly, regarding the previous question, it is not possible to define a potential function (voltage) for a field whose curl is not 0. Therefore to explain the readings of the voltmeter you should use the currents, not voltages. The voltmeters are nothing but galvanometers calibrated in terms of voltage. (to calculate the current use microscopic form of ohm's law, $\vec j= \sigma \vec E$)

Finally, the situation described does not break intuition in any way. We were able to define electic potential by E = - Grad V in the begginning only beacuse for static situations, Curl E was 0. But here Curl E = -dB/dt so it is not possible to define V. (If you put the first equation into the second, you get Curl Grad V = 0 = -dB/dt. So V can be defined only if dB/dt=0) Electrical engineers usually forget this point because they usually assume that the fields are confined within the components and one component does not affect another. (They use it because this reduces Maxwell's equations to the simpler equations for KVL and KCL).

$\endgroup$
  • $\begingroup$ The reading of voltmeters represent "potential differences" at all times, there is never any real current flow trough a voltmeter (only a displacement current trough its capacitance while it initially charges up). The readings just don't represent a physical potential that exists independently of the path, instead they give us an amount of energy a charge would pick up or lose while moving along the path that is being physically implemented by the voltmeter leads. $\endgroup$ – CuriousOne Jun 18 '16 at 2:49
  • 1
    $\begingroup$ If there is a charged capacitor, by sending an electron beam through it, we can observe the deflection and hence determine the value of the E-field and the potential difference across the capacitor. This is basically what Lewin's scope was doing showing the "voltages". I agree with CuriousOne that you don't need currents to measure voltage. For constant change of magnetics flux there would be no currents after initial charging of the capacitor. My question to Kartik: How does the E-field change from a finite value inside the resistor to nothing inside the wire if there is no charge sheet? $\endgroup$ – InquiringMinds Jun 18 '16 at 4:52
  • $\begingroup$ @InquiringMinds: It's even worse than that: We can also decide to make the displacement current arbitrarily large by filling the empty space with a dielectric (like the gate capacitance of an electronic voltmeter, which is the practical version of the electron beam in the example). By changing the capacitance we can get as much charge to flow (i.e. as much current) as we want, without the induced voltage changing at all. Any and all attempts to correlate the flowing (displacement) current with the induced voltage are therefor hopeless. $\endgroup$ – CuriousOne Jun 18 '16 at 6:01
  • $\begingroup$ @InquiringMinds the thing you will measure is only $\int \vec E\cdot d\vec r$ but that is not the potential difference as it is path dependent. The potential as in E=-grad V is simply not defined. For more details you can read Feynman's lectures on physics Vol II. $\endgroup$ – Kartik Jun 20 '16 at 10:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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