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Faraday Law of Induction

  1. Nov 16, 2013 #1
    1. The problem statement, all variables and given/known data

    Why does the Faraday Law of Induction not suffer from the inconsistency encountered with the Ampere Law?

    2. Relevant equations

    [itex]\oint[/itex] E * dl = - (N) d/dt [itex]\int[/itex]B * dA

    3. The attempt at a solution

    The Ampere Law is inconsistent with the time varying equation of continuity, so Maxwell put in the "displacement current" term to the current-density term. Faraday's law states that the EMF induced by a change in magnetic flux depends on the change in flux, change in time and the number of turns of coils(N). The minus sign in Faraday's law of induction means that the EMF creates a current I and magnetic field B that oppose the change in flux.

    So is the reason that the Faraday's Law of Induction doesn't have the inconsistency because of the presence of time derivative of the flux or because of the negative sign?
     
  2. jcsd
  3. Nov 16, 2013 #2

    WannabeNewton

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    Well it's two-fold. Let's take a look at Ampere's law without the displacement current term: ##\vec{\nabla}\times \vec{B} = \mu_0 \vec{j}##. Recall that for any smooth vector field, the divergence of the curl of the vector field always vanishes so in our case ##\vec{\nabla} \cdot (\vec{\nabla}\times \vec{B}) = 0## but this implies that ##\vec{\nabla} \cdot \vec{j} = 0## for all systems which violates conservation of charge: ##\vec{\nabla} \cdot \vec{j} + \frac{\partial \rho}{\partial t} = 0##. In order to make sure Ampere's law is consistent with conservation of charge, we add the displacement current ##\mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}##. At this point I'm basically just elaborating on what you already stated in the first paragraph of your post.

    Now let's take a look at Faraday's law: ##\vec{\nabla}\times \vec{E} = -\frac{\partial \vec{B}}{\partial t}##. Take the divergence of both sides. Is there an inconsistency? If not, why physically is there no inconsistency i.e. what is the crucial difference between Ampere's law without the displacement current and Faraday's law that frees the latter from the inconsistency suffered by the former?
     
    Last edited: Nov 16, 2013
  4. Nov 16, 2013 #3
    Taking the divergence of both sides would give you zero on both sides correct? So the difference between Faraday’s and Ampere’s law is that Ampere’s law assumes that the currents are steady and so it is not dependent on time. Then because an electric field is generated by a "changing" magnetic field then Faraday’s law is time dependent. Then because Faraday's law is time dependent and any change in B would then have an instantaneous change in E, there is no inconsistency.
     
  5. Nov 17, 2013 #4

    WannabeNewton

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    Right so taking the divergence of both sides shows that the same inconsistency doesn't occur with Faraday's law but it isn't exactly for the reason you mentioned. In seeing that Faraday's law carries no inconsistency, you used the fact that ##\vec{\nabla}\cdot \vec{B} = 0## always holds correct? What if there existed a magnetic charge such that ##\vec{\nabla}\cdot \vec{B} = \rho_{m}##? Would Faraday's law remain consistent in the above form?
     
  6. Nov 17, 2013 #5
    In the case of a magnetic monopole which I think is what you are asking, an extra term involving the current of magnetic monopoles would need to be added to the right side of Faraday's Law in differential form.
     
  7. Nov 17, 2013 #6

    WannabeNewton

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    Right. So do you see the difference now between ##\vec{\nabla}\times \vec{E} = -\frac{\partial \vec{B}}{\partial t}## and ##\vec{\nabla}\times \vec{B} = \mu_0 \vec{j}## with regards to why the latter needs to have a ##\mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}## term added but no change is needed for the former (under the framework that magnetic charges don't exist)?
     
  8. Nov 17, 2013 #7
    Yes, thanks for helping me better understand this.
     
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