- #1

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Using Lorentz force F = qv x B, we can conclude that emf produced.

Using Faraday's law, ∫E dl = 0 for any closed loops since there is no magnetic flux change.

So E = 0.

Contradiction ???

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- Thread starter HAMJOOP
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- #1

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Using Lorentz force F = qv x B, we can conclude that emf produced.

Using Faraday's law, ∫E dl = 0 for any closed loops since there is no magnetic flux change.

So E = 0.

Contradiction ???

- #2

- 7

- 0

If the magnetic field is parallel to the axis around which the ring rotates then you are right that there is no change in magnetic flux through the ring, however in this case the Lorentz force doens't predict an EMF either. Because the force is pointing inward or outward from the ring, and in opposite directions on both sides.

So there is no contradiction in this case.

Now, what if the ring is rotating so that its axis of rotation is perpendicular to the direction of the magnetic field. In this case the Lorentz force would predict that there is an EMF, but now you

So in all cases there is no contradiction.

- #3

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From the diagram

Lorentz force(radial component) drives the charge carrier (e.g. +q) to the centre of the disk, leaving negative charge on the rim.

so there is electric field, which is conservative inside the disk

Hence emf is built up from the centre to the rim of the disk.

Faraday's law in integral form can't predict that. @@

Lorentz force(radial component) drives the charge carrier (e.g. +q) to the centre of the disk, leaving negative charge on the rim.

so there is electric field, which is conservative inside the disk

Hence emf is built up from the centre to the rim of the disk.

Faraday's law in integral form can't predict that. @@

- #4

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I'm actually not sure how it works in this case, but this is called "Faraday's Paradox" so maybe you can find a suitable explanation online. There shouldn't be any contradition however.

I'm finding some explanations but none really seem to satisfy me. If I find one I'll link it.

- #5

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- 0

using the same condition as stated in the picture (see #3)

Faraday's law in integral form

∫E dl = 0 for

Faraday's law in differential form

curl E = 0, does not imply E = 0

self contradictory ??

or Faraday's law is incomplete, we need Lorentz force ??

- #6

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Use BLV

- #7

Jano L.

Gold Member

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Faraday's law in integral form

∫E dl = 0 for any closed loop, so E must be 0

This is the problem. Generally the equation ∫E dl = 0 for all loops does not imply E = 0, only that E is gradient of scalar function (potential).

If the disk is rotating freely in the magnetic field, the charges will redistribute on the rim and in the center in such a way that they produce electrostatic field which cancels the magnetic force:

$$

\mathbf E = -\mathbf v \times \mathbf B,

$$

so in equilibrium rotation there are no conduction currents.

- #8

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You never go wrong with the local form, which is anyway the natural way to describe fields. Using Heaviside-Lorentz it reads

[tex]\vec{\nabla} \times \vec{E}=-\frac{1}{c} \partial_t \vec{B}.[/tex]

The complete form of the integral equation is derived from that by using Stokes's integral theorem and then bringing the time derivative out of the surface integral. From this you get an additional contribution to the line integral, which you can combine with the left-hand side. In this way you get the only correct form of the integral form of Faraday's Law, with the complete EMF on the left-hand side:

[tex]\int_{\partial A} \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right ) = - \frac{1}{c}\frac{\mathrm{d}}{\mathrm{d} t} \int_A \mathrm{d}^2 \vec{F} \cdot \vec{B}.[/tex]

Now, there is no contradiction anymore, and everything is consistent with the derivation given in the postings before.

- #9

- 1,506

- 18

In the history of electrical power generation an arrangement of discs between the poles of magnets was an early form of generator.

There is an example in the science museum.

- #10

Jano L.

Gold Member

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In our case, the integral formulation with fixed loop and surface is still correct, only it is not very useful.

If we connect the center and the rim by static wire, we create electric circuit and have current. Let's imagine static loop ##\partial A## going through the disk and the wire. Then the Faraday law implies

$$

\oint_{\partial A} \mathbf E\cdot d\mathbf S = - \frac{d}{dt}\int_A \mathbf B \cdot d\mathbf S.

$$

This is fine, because in fact both magnetic and electric field are static fields, and thus both sides of the equation vanish.

Clearly the law in this form is not very useful.

How to explain the current then?

The key is to realize that the electromotive force in a circuit is not given by the electric field alone, but there may be other contributions, sometimes denoted by ##\mathbf E^*## (magnetic force, effective driving forces due to chemical reaction in a cell, concentration gradients etc.)

The emf is then given by

$$

\epsilon = \oint_{\partial A} \mathbf E + \mathbf E^* \cdot d \mathbf l.

$$

In our case, this additional intensity is due to the magnetic field, so

$$

\epsilon = \oint_{\partial A} \mathbf E + \mathbf v\times \mathbf B\cdot d \mathbf l.

$$

Since the electric field is electrostatic, only the magnetic contribution to emf is non-zero, and turns out to be ##\omega BR/2##.

(As as sidenote, in a sense the integral formulation of the Faraday law is more general than the differential formulation, because the integrals can handle field discontinuities neatly, while with the differential form, we have to write down the equations for the jumps separately.)

- #11

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[tex]\vec{\nabla} \cdot \vec{E}=\rho.[/tex]

In integral form this translates into

[tex]\int_{\partial V} \mathrm{d}^2 \vec{F} \cdot \vec{E}=\int_V \mathrm{d}^3 \vec{x} \rho.[/tex]

- #12

Jano L.

Gold Member

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I never tried to write down the Gauss law. I tried to explain that the explanation of the current in the experiment under discussion is not in the Faraday law, but rather in the fact that the emf contains contributions due to other agents than the electric field.

- #13

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It is, however, my point that of course Faraday's Law is completely suitable for the situation at hand. The only important point is to use the full version for moving bodies. You can solve the problem either with this complete form of Faraday's Law or using the microscopic argument with the complete Lorentz force on the electrons in the wire. As shown in the thread both methods lead to the same result.

This is no surprise since the coupled set of equations of electromagnetism (i.e., Maxwell's equations and the equation of motion of charges with the Lorentz force) are derived from the underlying principle of least action and thus by construction must be consistent!

- #14

Jano L.

Gold Member

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- #15

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In our case, the integral formulation with fixed loop and surface is still correct, only it is not very useful.

If we connect the center and the rim by static wire, we create electric circuit and have current. Let's imagine static loop ##\partial A## going through the disk and the wire. Then the Faraday law implies

$$

\oint_{\partial A} \mathbf E\cdot d\mathbf S = - \frac{d}{dt}\int_A \mathbf B \cdot d\mathbf S.

$$

This is fine, because in fact both magnetic and electric field are static fields, and thus both sides of the equation vanish.

Clearly the law in this form is not very useful.

How to explain the current then?

The key is to realize that the electromotive force in a circuit is not given by the electric field alone, but there may be other contributions, sometimes denoted by ##\mathbf E^*## (magnetic force, effective driving forces due to chemical reaction in a cell, concentration gradients etc.)

The emf is then given by

$$

\epsilon = \oint_{\partial A} \mathbf E + \mathbf E^* \cdot d \mathbf l.

$$

In our case, this additional intensity is due to the magnetic field, so

$$

\epsilon = \oint_{\partial A} \mathbf E + \mathbf v\times \mathbf B\cdot d \mathbf l.

$$

Since the electric field is electrostatic, only the magnetic contribution to emf is non-zero, and turns out to be ##\omega BR/2##.

(As as sidenote, in a sense the integral formulation of the Faraday law is more general than the differential formulation, because the integrals can handle field discontinuities neatly, while with the differential form, we have to write down the equations for the jumps separately.)

I think it is e = ωBR

It is known as Faraday's disc

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