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Ampere's circuital law

  1. Sep 12, 2015 #1
    Imagine an E field coming out of your screen that is constant everywhere in space and time (∂E/∂t=0). And in your reference frame, let's say that this the only field there is -- there is no B.

    Say there is a loop in the plane of your screen, and so the plane of this loop is perpendicular to the E. If the loop starts shrinking, will there be any non-zero value of ∫H⋅dl induced around it as it shrinks?

    There is no current through the the loop, and E is constant inside of it.
     
    Last edited: Sep 12, 2015
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  3. Sep 12, 2015 #2

    vanhees71

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    Perhaps I misunderstand the question, but how should this line integral not simply be just 0, because ##\vec{H}=\vec{B}/\mu=0## everywhere?
     
  4. Sep 12, 2015 #3
    The amount of E passing through a surface enclosed by the shrinking loop is lessening even though E itself is time-constant inside of it. Would that be like a displacement current through it?

    After all, in a Faraday law analogy, if it were a loop perpendicular to a B field, there would be 2 ways EMF could be induced in the loop:
    1) The contours of the loop remaining fixed, but B varying with time - transformer EMF
    2) B being time constant, but the contours of the loop changing - motional EMF

    or some combination of both.

    Isn't there some magnetic field analogy to EMF that would make moving the contours of the loop have the same displacement-current effects on it as holding the contours fixed and varying E through the bounded surface of it?
     
    Last edited: Sep 12, 2015
  5. Sep 12, 2015 #4

    vanhees71

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    That's indeed an interesting question. It has to do with the correct form of the Maxwell equations in integral form when moving surfaces and boundary curves are involved.

    First we note that for an arbitrary vector field ##\vec{V}(t,\vec{x})## and an area ##F## with surface-normal vectors ##\mathrm{d} \vec{f}## one has
    $$\frac{\mathrm{d}}{\mathrm{d} t} \int_F \mathrm{d} \vec{f} \cdot \vec{V}(t,\vec{x})=\int_F \mathrm{d} \vec{f} \cdot [\partial_t \vec{V}(t,\vec{x}) + \vec{v}(t,\vec{x})(\vec{\nabla} \cdot \vec{V}(t,\vec{x})]-\int_{\partial F} \mathrm{d} \vec{x} \cdot [\vec{v}(t,\vec{x})\times \vec{V}(t,\vec{x})].$$
    For a proof, see
    https://en.wikipedia.org/wiki/Leibniz_integral_rule

    Now we apply this to the "electric flux"
    $$\Phi_{\vec{E}}(t)=\int_F \mathrm{d} \vec{f} \cdot \vec{E}(t,\vec{x}).$$
    Taking the time derivative gives according to the above mathematical theorem
    $$\frac{\mathrm{d}}{\mathrm{d} t}=\int_F \mathrm{d} \vec{f} \cdot [\partial_t \vec{E}(t,\vec{x}) + \vec{v}(t,\vec{x})(\vec{\nabla} \cdot \vec{E}(t,\vec{x})]-\int_{\partial F} \mathrm{d} \vec{x} \cdot [\vec{v}(t,\vec{x})\times \vec{E}(t,\vec{x})].$$

    Using the Ampere-Maxwell Law,
    $$\vec{\nabla} \times \vec{B}-\frac{1}{c} \partial_t \vec{E}=\frac{1}{c} \vec{j},$$
    and Gauss's Law,
    $$\vec{\nabla} \cdot \vec{E}=\rho,$$
    we find after using Stokes's theorem
    $$\frac{1}{c} \frac{\mathrm{d}}{\mathrm{d} t} \Phi_{\vec{E}} =-\frac{1}{c} \int_F \mathrm{d} \vec{f} \cdot [\vec{j}-\rho \vec{v}] + \int_{\partial F} \mathrm{d} \vec{x} \cdot \left (\vec{B}-\frac{\vec{v}}{c} \times \vec{E} \right ).$$
     
  6. Sep 12, 2015 #5
    is this a yes?
     
  7. Sep 13, 2015 #6
    Hi,

    It's obvious that the flux will decrease since the loop is shinking but I'm afraid this reduction will be compensated with the term - dx · (v/c x E) so that the magnetic field will not change.

    In the link given by vanhees71, https://en.wikipedia.org/wiki/Leibniz_integral_rule, v is "the velocity of movement of the region", so when shinking a circular loop, v must go towards the center, so that v x E will be paralel to the direction of the line (dx).

    Think that this loop with out current (and I assume also with no charge) may be an imaginary artifact, so, how could it change any field?


    Best regards,
    Sergio
     
  8. Sep 13, 2015 #7
    Not an imaginary artifact. We have to be able to determine, yes or no, if compasses situated along the perimeter of the shrinking loop in the constant E field would deflect. Either they would or they wouldn't.
     
  9. Sep 14, 2015 #8

    vanhees71

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    I'm not so sure about this. Take the formula given above. In this case I guess we have ##\rho=0##, ##\vec{j}=0##, and ##\vec{B}=0##. Let's see whether this is consistent with the integral formula above (it's amazing that I couldn't find this straight-forward formula in any textbook, but unfortunately also the correct Faraday Law is rare in the literature; it's really strange, why nobody seems to address this in the more advanced books like Jackson, but of course the Faraday Law has important applications like the homopolar generator, while the analogous here discussed situation with the Ampere-Maxwell law seems to have not such applications).

    Let's take the loop to be circular with a radius ##a(t)## in the ##xy##-plane of a Cartesian coordinate system and ##\vec{E}=E \vec{e}_3##, ##E=\text{const}##. Then it's convenient to use cylinder coordinates ##(x,y,z)=(\rho \cos \varphi,\rho \sin \varphi,z)##. The electric flux through the circle bounded by the loop is
    $$\Phi_{\vec{E}}=\int_0^{a(t)} \mathrm{d} \rho \int_{0}^{2 \pi} \mathrm{d} \varphi \rho \vec{e}_3 \cdot \vec{E}=\pi a^2(t) E\; \Rightarrow \; \dot{\Phi}_{\vec{E}}=2 \pi a(t) \dot{a}(t) E.$$
    For the velocity field along the loop is
    $$\vec{v}(t,\vec{x})=\dot{a}(t) \vec{e}_{\rho},$$
    and the line integral thus is
    $$\int_{\partial F} \mathrm{d} \vec{x} \cdot \vec{v} \times \vec{E}=\int_0^{2 \pi} \mathrm{d} \varphi a(t) \vec{e}_{\varphi} \cdot \dot{a}(t) \vec{e}_{\rho} \times E \vec{e}_z = a(t) \dot{a}(t) E \int_0^{2 \pi} (-1)=-2 \pi a(t) \dot{a}(t).$$
    So above assumptions are consistent with the integral formula for our case:
    $$\dot{\Phi}_{\vec{E}}=-\int_{\partial F} \mathrm{d} \vec{x} \cdot (\vec{v} \times \vec{E}).$$
    So it seems as if simply no magnetic field will be induced.

    Let's also check qualitatively the force on the electrons in the wire. In the local rest frame of each wire element there is a magnetic field ##\vec{B}'=\propto \vec{v} \times \vec{E}=-\dot{a}(t) E \vec{e}_{\varphi}##. The Lorentz force on an electron thus is ##\propto \vec{v} \times \vec{B}' \propto \dot{a} E \vec{e}_{\rho} \times \vec{e}_{\varphi} \propto -\vec{e}_z##. So there's no current induced in the wire and thus also no B-field in the lab frame.

    The conclusion is again that here no magnetic field is induced.
     
  10. Sep 14, 2015 #9
    But if in the local rest frame of each wire element there is a magnetic field B-prime, then presumably a compass situated one of the elements and riding along with it would be accordingly deflected.

    With no B-field in the lab frame, what is there in the body of electromagnetic law to account for this effect in the lab frame? Assuming, that is, that the compasses would also be deflected in the lab frame. After all, isn't a needle's deflection versus non-deflection kind of hard to "transform away"?
     
  11. Sep 14, 2015 #10

    vanhees71

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    When you move a compas in an electric field, it will react to the magnetic field in its rest frame, but that's not the situation described un the OP.
     
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