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Voltage difference of a circle under changing magnetic field

  1. Dec 27, 2014 #1
    Assume that I am having a 3D Cartesian Coordinate and a linearly time-dependent magnetic field ##\textbf{B}(t) = B_0 \cdot k \cdot t \cdot \textbf{z}## where ##B_0## and ##k## are just constants.

    If circular conductor(a thin ring whose thickness is negligible) is put parallel to the ##XY## plane and one measures the voltage difference on 2 different points ##a, b##, how shall I predict which one will have a higher voltage?

    What confuses me is that according to Faraday's Law, the electric field around the circular conductor is given by

    ##\nabla \times \textbf{E} = - \frac{\partial \textbf{B}}{\partial t}##

    Thus for any point on the conductor the induced electric field is the same. Then going from any point along the induced electric field till reaching the same point one gets a voltage drop. For any 2 different points ##a, b## on the circular conductor I can always go from ##b## to ##a## or from ##a## to ##b## "along the induced electric field" and get voltage drops for both cases.

    Could any one help to resolve the part that confuses me?
     
  2. jcsd
  3. Dec 27, 2014 #2

    mfb

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    How do you plan to measure that voltage? How do you even define the voltage?

    With time-dependent magnetic fields, there is no clear, unique way to define an electric potential. You can still consider the vector potential A.
     
  4. Dec 27, 2014 #3
    @mfb, this is pretty new for me. Would you please show why the electric potential cannot be well defined (or a reference talking about this)?

    By voltage I mean ##\int_{path} \textbf{E} \cdot d \textbf{l}## and when talking about time-dependent fields I think both potentials are still defined as

    ## \textbf{E} = -\nabla \phi - \frac{\partial \textbf{A}}{\partial t}##

    ## \textbf{B} = \nabla \times \textbf{A}##

    While the uniqueness of the solutions of potentials can be restricted by choosing the Lorentz gauge.
     
  5. Dec 28, 2014 #4

    mfb

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    If you want the electric field to be the (spatial) derivative of a scalar potential, then this field always has zero curl, in clear contradiction to your setup.

    Your path integral depends on the path, that is not practical.

    Yes you can fix that with the vector potential A, but then it is unclear what "voltage" means.
     
  6. Dec 28, 2014 #5
    My definition could be path-dependent. Sorry for that I haven't checked it and I will check it today. However in AC circuit we do talk about "voltage" and it's all about time-dependent variables. I don't see why this term cannot be well defined here (but you can argue that I'm not defining it well).

    I don't presume that "voltage" has to be "scalar potential". Any definition is welcome as long as it is consistent with the Maxwell equations.

    I do want to buy equipments and make the measurement, however I'm afraid of that the unwanted features of equipments(resistance of conductor and voltage meter, imperfect circular shape etc) would give a misleading result. Thus I expect this problem to be first solved theoratically.
     
  7. Dec 28, 2014 #6

    mfb

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    You can define whatever you want. You just should invent a new name for it because it is not a standard quantity any more. The Maxwell equations do not have potentials at all (for exactly that reason!).

    Time-dependent voltages are no problem. Time-dependent magnetic fields (in regions where you are interested in a potential) are.
     
  8. Dec 28, 2014 #7
    @mfb, this is still confusing. What about I put an N-turns(##N \ge 1##) solenoid inductor whose axis is parallel to the Z-axis in the same configuration and use a galvanometer to measure the its 2 ends? Is this different from the case of a circular conductor? By which rule shall I predict the value I measure?
     
  9. Dec 29, 2014 #8

    mfb

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    Why the 2 ends? The length of the solenoid is not important here. The voltage reading can depend on the path of the cables.
     
  10. Dec 29, 2014 #9
    Understood. Thank you so much for your patience :)
     
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