Explaining the Inconsistency and Solution in Faraday's Law of Induction

In summary, the Faraday Law of Induction does not suffer from the inconsistency encountered with the Ampere Law because Faraday's law is time dependent and any change in the magnetic field results in an instantaneous change in the electric field, while Ampere's law assumes steady currents and does not account for time dependence. Additionally, Faraday's law remains consistent even if magnetic monopoles exist, as an extra term involving the current of magnetic monopoles can be added to the right side of the equation. This difference between Faraday's and Ampere's law explains why the former does not require a displacement current term, unlike the latter.
  • #1
smantics
12
0

Homework Statement



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

Homework Equations



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

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?
 
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  • #2
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?
 
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  • #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.
 
  • #4
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?
 
  • #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.
 
  • #6
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)?
 
  • #7
Yes, thanks for helping me better understand this.
 

1. What is Faraday's Law of Induction?

Faraday's Law of Induction states that when a conductor is placed in a changing magnetic field, an electromotive force (EMF) is induced in the conductor. This means that a voltage is generated, which can cause an electric current to flow, in the conductor.

2. How does Faraday's Law of Induction work?

Faraday's Law of Induction is based on the principle of electromagnetic induction. When a conductor is placed in a changing magnetic field, the magnetic flux through the conductor also changes. This change in flux induces an EMF in the conductor, according to the equation EMF = -N(dΦ/dt), where N is the number of turns in the conductor and dΦ/dt is the rate of change of magnetic flux.

3. What are some real-world applications of Faraday's Law of Induction?

Faraday's Law of Induction has many practical applications, including generators, transformers, and electric motors. It is also used in devices such as induction cooktops and magnetic levitation trains. Additionally, it is the basis for various measurement devices, such as induction probes and magnetic flow meters.

4. What factors affect the magnitude of the induced EMF?

The magnitude of the induced EMF depends on several factors, including the strength and rate of change of the magnetic field, the number of turns in the conductor, and the area of the conductor. Additionally, the material and resistance of the conductor can also affect the magnitude of the induced EMF.

5. How is Faraday's Law of Induction related to Lenz's Law?

Lenz's Law is a consequence of Faraday's Law of Induction. It states that the direction of the induced current in a conductor will always be such that it produces a magnetic field that opposes the change in the original magnetic field. This means that the induced current will create a magnetic field that tries to cancel out the original changing magnetic field.

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