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Electric flux versus magnetic flux

  1. Sep 11, 2014 #1
    I'm taking E & M II... Would anyone agree that the following statements are true?


    An electric field changing in time gives rise to a displacement current.
    A magnetic field changing in time gives rise to an electromotive force.
     
  2. jcsd
  3. Sep 11, 2014 #2

    ShayanJ

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    By definition, we have [itex] \vec J_{displacement}=\frac{\partial \vec D}{\partial t}=\varepsilon \frac{\partial \vec E}{\partial t} [/itex]. So displacement current is a changing electric field.
    Eelectromotive force is something that you can't define in parallel with displacement current. It is caused by collectively considering the Lorentz force of the magnetic field on all of the charged particles of a material.
     
  4. Sep 11, 2014 #3
    Thanks Shyan. ε=-[itex]\frac{d\Phi}{dt}[/itex], so a changing magnetic flux creates an electromotive force, is that right?
     
  5. Sep 11, 2014 #4

    ShayanJ

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    Yes. But I think there should be some clarification.
    The formula you wrote is always considered in relation with a closed circuit because otherwise you don't have a persisting current and also the flux is a surface integral and only a closed circuit defines a surface.
    But that doesn't mean such an effect occurs only for closed circuits. when you have a wire, which doesn't form a loop, in a magnetic field, the Lorentz force makes its charged particles to move and so creates an electromotive force and so a current in it. The only point is that the current can, at most, oscillate in the the wire and that only happens for some changes of configurations. Even changing the magnitude of the magnetic field(only) won't cause that. And because you have no surface for deriving a flux, you need other mathematics for deriving such effects.
    The point I wanna emphasize is that the more fundamental law is Faraday's law of induction i.e.[itex] \vec \nabla \times \vec E=-\frac{\partial \vec B}{\partial t} [/itex].
     
  6. Sep 11, 2014 #5
    Shyan, OK, thanks for your post. I think I see what you mean... the emf defined here as a change in the magnetic flux only applies for motional emf? There are other ways to generate emf... Does Faraday's Law then mean that whenever we have a changing magnetic field, we have an emf?
     
  7. Sep 12, 2014 #6

    ShayanJ

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    No, that's not what I mean.
    At first, the [itex] \mathcal{E} [/itex] in [itex]\mathcal{E}=-\frac{d\phi}{dt} [/itex] is defined by [itex] \mathcal{E}=\oint \vec E\cdot \vec{dl} [/itex]. Remember the definition of a conservative field? [itex] \oint \vec E\cdot \vec{dl}=0 [/itex] for all closed path! and its only for such fields that you have [itex] \vec{E}=\vec \nabla \Psi [/itex]. You also should notice that [itex] \oint \vec E\cdot \vec{dl}=0 [/itex] is just KVL! And so its only for conservative electric fields that you have KVL and can define a scalar potential that alone by itself, can give the electric field. All electric fields generated by electric charge distributions are conservative.
    So when you have [itex]\mathcal{E}=-\frac{d\phi}{dt} [/itex], it means you're talking about another kind of electric field which is not conservative. Such fields are created by time-dependent magnetic fields according to [itex] \vec \nabla \times \vec E=-\frac{\partial \vec B}{\partial t} [/itex]. Let's return to [itex]\mathcal{E}=-\frac{d\phi}{dt} [/itex].
    You know that [itex] \phi=\int_S \vec B \cdot \vec{dS} [/itex] where S is the surface which you want to calculate the flux through. But only a closed circuit defines a surface and so you only can use this for closed circuits.(Of course you can always consider an imaginary surface but when you want to use the flux for calculating an emf, that can only be done for closed circuits.) But this formula considers all kinds of flux changes. Lets see this:
    [itex]
    \frac{d\phi}{dt}=\frac{d}{dt}\int_S \vec B \cdot \vec{dS}=\int_S \frac{\partial}{\partial t}[\vec B \cdot(dS\hat{dS})]=\int_S dS \frac{\partial}{\partial t}[\vec B \cdot\hat{dS}]=\int_S dS \frac{\partial}{\partial t}[B \cos{(\theta(t))}]
    [/itex]
    As you can see, the flux change can be because of the change in the magnitude of the magnetic field or because of the change in the relative orientation of the circuit and the field lines. But here I assumed that the circuit doesn't change shape, if it does, there will be other terms added to the surface integral when passing from [itex] \frac{d}{dt}\int_S ...[/itex] to [itex] \int_S \frac{\partial}{\partial t}... [/itex] and those terms will take care of the change in circuit's shape. So anything that can cause a change in flux is taken into account here. But don't get confused, you don't see such calculations in particular problems because there, you just integrate for that particular problem and things are simpler.
    But about [itex] \vec \nabla \times \vec E=-\frac{\partial \vec B}{\partial t} [/itex]. I said this is more fundamental because there are cases when you have a piece of wire which doesn't form a closed circuit. In such cases there may arise a potential difference between two ends of the wire but you can't use [itex]\mathcal{E}=-\frac{d\phi}{dt} [/itex] here because there is no closed circuit. Here, one should first find the electric field from [itex] \vec \nabla \times \vec E=-\frac{\partial \vec B}{\partial t} [/itex] and then use [itex] \Delta V=-\int_{{p_1}_\gamma}^{p_2} \vec E \cdot \vec{dl} [/itex] to find the potential difference between two ends p1 and p2 (which depends on the wire's shape [itex] \gamma [/itex] because the electric field is not conservative). Now this potential difference can give you the current. But here the current won't persist if things don't change in particular ways. And because in introductory physics (and this introductory stage lasts very long) we only consider stable states of the system, you don't encounter such a situation, because at the stable state of this system, there is no current or in case that those particular changes happen and there is a current, the problem is too complicated I think.
     
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