Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Experiment with Faradays law

  1. Sep 13, 2012 #1
    Today we did an experiment with a solenoid through which we varied the current. Around this solenoid, was a different circuit (not coupled to the coil-circuit) where we measured the potential across a resistor. By varying the current my teacher showed how the potential over the resistor would vary.
    Now faradays law essentially says that a time varying magnetic field induces a rotating electric field.
    ∇xE = -dB/dt
    So I thought that the above experiment could be explained by the fact that the changing magnetic field induces an electric field which accounted for the electromotive force induced in the other circuit and thus any drop or increase in potential.
    BUT! As my friend correctly stated, the only place where the B-field is non zero is inside the solenoid itself. The circuit of the resistor was not a part of the inside of the solenoid. So -dB/dt would have to be zero for all points in space except inside the it. And that means no electric field can possibly have been induced from the varying magnetic field in the solenoid into the circuit of the resistor.
    So how is the above experiment explained?
     
  2. jcsd
  3. Sep 13, 2012 #2
    Curl of E is only non-zero where a B field exists also, but this does not mean the circulation of E requires magnetic field lines coincident with E field. The integral version of Faraday's law simply states the line integral of E around a closed path equals the negative of the time-changing rate of magnetic flux threading through such a closed path. Strictly speaking, for such transformer action situation, E is given at any point by minus the time-changing rate of A, the vector potential, and there can certainly be a non-zero A in the absence of a B at that point. Just check out the Wiki page http://en.wikipedia.org/wiki/Faraday's_law_of_induction
     
  4. Sep 13, 2012 #3
    Come again please..

    If the B-field is zero outside the solenoid, so is also the change of it in time and therefore the curl of E at those points. Where am I going wrong with this statement?
     
  5. Sep 13, 2012 #4
    The B-field isn't really zero, just very small. But I don't think thats the point.

    Take the integral form of Faradays law: i.e. that the surface integral of dB/dt is equal to the line integral of E around a closed contour which bounds that surface. Imagine then drawing a closed loop around your solenoid and any open surface which has this closed loop as its boundary. Is there is a non-zero flux of dB/dt going through that surface? If so, then the line integral of E around that loop (which is not inside of the solenoid) has a non-zero E.
     
  6. Sep 13, 2012 #5
    Why should the B-field be zero outside the solenoid? You don't have an infinitely-long solenoid, do you?
     
  7. Sep 13, 2012 #6
    Well suppose we had. Would the experiment still yield the same? Either way we can savely assume that the field is very small outside and thus the change in that should not induce a very strong electric field.
     
  8. Sep 13, 2012 #7
    As per my #2, and #4, one has to distinguish between the differential property curl E as 'vorticity' of the field, which is an intensive property of E, and induced emf around a closed path, which is an extensive or 'global' property. Stick with the definition E = -∇phi -∂A/∂t and you won't go wrong.
     
  9. Sep 13, 2012 #8

    Dale

    Staff: Mentor

    This is only correct if you have an infinitely long solenoid.
     
  10. Sep 13, 2012 #9
    Yes, see my post.
     
  11. Sep 13, 2012 #10
    If you had an infinitely long solenoid, then no, the experiment would not have the same result. Regardless of length, why do you assume the exterior B-field to be "very small?" If this were the case, then transformers wouldn't exist.
     
  12. Sep 13, 2012 #11
    I can see that the flux of b through a closed area bounding the solenoid is non-zero. Thus you conclude that the integral of E around a closed path bounding the solenoid is non-zero - is that correctly understand.
    Problem is just that doesn't get us very far. For where did this E-field generating a non-zero value for a closed line integral come from? It must come from the changing magnetic field. But that says ∇xE = -dB/dt
    problem is just that the above would give zero. It seems the integral and differential form conflict with each other.
     
  13. Sep 13, 2012 #12
    The magnetic field lines outside the solenoid tend to much "further" spaced apart than inside the solenoid (i.e. the flux outside << flux inside)
     
  14. Sep 13, 2012 #13
    I think you're getting a little bit confused. Think of Ampere's law. A magnetic field curls around some current. Imagine that current is due to a wire. Now imagine you are calculating the magnetic field far away from the wire. You still have a magnetic field even though there is no current close to where you are.
     
  15. Sep 13, 2012 #14
    Another thing, take the line integral of E around a closed loop. But now have the loop off to the side of the solenoid. That line integral will equal to zero because the - and + contributions will cancel each other out. BUT even though the line integral of E is zero, that doesn't mean that E itself is zero at any point. Does that help a little bit to give you a conceptual understanding?
     
  16. Sep 13, 2012 #15
    I think I found an article describing this exact problem. I would have to say it agrees with me, and that the reasons for the voltmeters to change measurements is far more subtle than the reasons above. Do you disagree? If so, I have misunderstood you.
    Emguy: I think you misunderstand the problem. The field is ZERO everywhere except inside the solenoid. Google the field of a solenoid.
     

    Attached Files:

  17. Sep 13, 2012 #16
    Yes. It is the manifestation of the integral form of Faraday's law - and that's the only one applicable to your case.
    No they don't. As stated in #7, the curl E version yields an intensive field property, valid 'at a point'. You want the net voltage around a circuit, then use the integral form, which only requires that some value of flux threads the circuit of interest. Note that this works perfectly in the case of a toroidal transformer configuration, which if properly constructed, has precisely zero magnetic field intersecting the secondary windings. Again, if in doubt always refer back to the definition for E, which here amounts to -∂A/∂t, as electrostatic fields are irrelevant to emf.
     
  18. Sep 13, 2012 #17
    I find it hard to distinguish between the cases. If you look at the line integral around the solenoid as a sum of E*dr and have at each point that E is zero then WHY would the integral not yield a big, whopping zero?
    Please take a look at the first page of the article attached above so you are sure what my problem is.
    You are probably right, but I don't see it with your explanation.
     
  19. Sep 13, 2012 #18
    Just because curl(E) is zero doesn't mean that E is zero.
     
  20. Sep 13, 2012 #19
    Think of curl(E) as a line integral over an infinitesimally small loop. Thus curl(E) will be zero outside the solenoid because the sum of each of those infinitesimally small sides will equal to zero. Similarly you could deduce that curl(E) would be equal to zero over that infinitesimally small loop because if you use faradays integral form / Stroke's theorem you will see that there is no dB/dt going through the loop.
     
  21. Sep 13, 2012 #20

    Dale

    Staff: Mentor

    Not true.

    Not true.

    http://en.wikipedia.org/wiki/Soleno...ctor_potential_for_finite_continuous_solenoid

    You are starting from some false assumptions, and reaching erroneous conclusions. The field outside the solenoid in your lab experiment was NOT zero.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Experiment with Faradays law
  1. Faraday's Law (Replies: 5)

  2. Faraday's law (Replies: 6)

  3. Faraday's Law (Replies: 8)

  4. Faraday's law (Replies: 12)

Loading...