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Ampere's law question

  1. Apr 8, 2015 #1
    First, some background. In Faraday's law the reluctance/resistance by "mother nature" to changing the magnetic flux is explicitly recognized by the "-" sign (commonly referred to as Lenz's law), i.e. if the time rate of change of the magnetic flux is positive, an emf is induced so as to oppose this change. Application of this idea allows one to determine the direction of integration in the line integral that that determines the emf.

    Here's my question: Is there an analogous effect, like a Lenz's law (for current) if the electric field is increasing or decreasing? For example, when a capacitor is charging there is a "growing" electric field dE/dt > 0, producing a displacement current in the same direction as the conduction current flowing into the capacitor and consequently producing an induced magnetic field having the same sense "axially" as that produced by the conduction current. So if now the capacitor is discharged (let's say slowly through a large resistor), the electric field is in the same direction as before, but now dE/dt < 0. Does this reverse the sense of B (axially) and hence the direction of the line integral on the left side of the Ampere-Maxwell equation? Or is the displacement current still in the same direction and just getting smaller, without changing the sense of B?

    I'm trying to understand the application of Faraday's law and Ampere's law to an electromagnetic wave. I know from Faraday's and Lenz's law the correct "sense" of the line integral, but I'm having trouble convincing myself of the correct sense when applying Ampere's law.

    I'm sorry for the rather "wordy" question, but I'm hoping someone with a firmer grip on this that I have can help me. Thanks.
     
  2. jcsd
  3. Apr 9, 2015 #2

    rude man

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    Easy way is to just use the current direction in the capacitor leads. The displacement current IS the current in the leads.The right-hand rule tells you the direction of the B field around a capacitor or lead: thumb in the direction of current, fingers curl a round the direction of B. You shouldn't have any trouble relating the direction of the E field buildup (dE/dt) inside the capacitor given the direction of conventional current into the capacitor: I = C dV/dt with V = -d E, d = distance between two large parallel plates in vacuum, for instance.
     
  4. Apr 9, 2015 #3
    Thanks for the reply, but I don't think you quite understood my question. In this figure E is directed into the page and is increasing. By the right-hand rule the "induced B" is in a clockwise direction since dE/dt is in the same direction as the conduction current that's causing the increasing E. I understand that. What I'm saying is this: Let's suppose the attached figure shows E without B, i.e. a static case, the capacitor is charged, but not charging. Now assume you connect the plates of the capacitor so that the charge separation will dissipate through a resistor. E is still in the same direction but the positive charge is draining off the plate so that dE/dt < 0. Is the "induced B" still clockwise? because E is still in the same direction? Or is it counterclockwise since the current is flowing in the opposite direction? It seems to me that the Ampere-Maxwell equation "doesn't care" whether E is increasing or decreasing (there is no minus sign like in Faraday's law.) Am I missing something here?
     

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  5. Apr 9, 2015 #4

    rude man

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    Yes, you are. Just because there is no minus sign in the relevant Maxwell equation doesn't mean the sign of B is not related to the sign of dE/dt. It most definitely is.
    OK, start with Maxwell: ∇ x H = ∂D/∂t in the absence of actual current flow. D = εE and H = B/μ.
    Using Stokes' law we can rewrite to
    ∫∫(∇ x H)⋅dA = ∫H⋅ds evaluated around a closed curve s defining the surface area A.
    So you wind up with ∫H⋅ds = ε ∫∫∂E/∂t⋅dA. This clearly states that the circulation of H is proportional to dE/dt including sign.
    And since Ampere's law equates ∫H⋅ds to I you now know that I = ε ∫∫∂E/∂t⋅dA = displacement current.

    So the B field is proportional to dE/dt and the sign of B follows the sign of dE/dt.
    Just as the E field is proportional to dB/dt, but with a minus sign: ∇ x E = -∂B/∂t.
    It's a practically perfect analogy.
     
  6. Apr 9, 2015 #5
    Once again, thanks for responding. So in answer to my questions
    1. you are saying that as the capacitor drains, the sense of B is opposite to its sense when the capacitor is charging. Right?
    2. Further, that there is no "resistance" to the decreasing electric flux (analogous to a back emf), no attempt by nature to prevent the reduction as she tries to prevent a change in magnetic flux.
    I agree that there is a "practically perfect analogy", but that's really what I'm getting at. Is it?
     
  7. Apr 9, 2015 #6

    rude man

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    o the flow of current. This E fgield repels further charges from increasing this E field, setting up an rquilibrium charge distribution and voltage.
    Right.
    The Hall effect produces an E field orthogonal to the flow of current and the externally applied B field which then counters further charge migration, so an equilibrium is set up relating the E field to the flow of current. That represents a resistance to increasing the E field.
    Also, imagine a parallel-plate capacitor being charged. As it charges up, more and more energy is required to add an extra unit of charge since d(energy) = V dQ but dQ = C dV. So the E field resists being increased. Not sure how 'analogous' these examples are, though.
    I have to leave it here. Perhaps others will add to my posts.
     
  8. Apr 9, 2015 #7
    Thanks for your insights.
     
  9. Apr 9, 2015 #8

    rude man

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    Welcome. If I think of anything further I'll let you know.
     
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