Error in Form of Maxwell's Equations?

[nonequilibrium]

Hello,

First of all, I have no objections against Faraday's Law in differential form, i.e.
$$\vec \nabla \times \vec E = - \frac{\partial \vec B}{\partial t}.$$

But in integral form, I usually encounter it in the form
$$\oint \vec E \cdot \mathrm d \vec l = - \frac{\partial \Phi_B}{\partial t}.$$

But surely this is wrong? (after all when B is constant, but the loop-surface is changing, there is no induced E-field; then it's the B-field delivering the emf)
The correct expression should be
$$\oint ( \vec E + \vec v \times \vec B ) \cdot \mathrm d \vec l = - \frac{\partial \Phi_B}{\partial t}. (*)$$

Does everyone agree?

(*) for clarification: with the "v" in the LHS I mean the netto speed of the particles in each point (of the loop) at the time of integration

ADDENDUM: Okay, another point of view is "but I'm looking at the E-field as viewed by the electrons moving along with the possibly changing loop, and then there is purely an induced E-field"; I suppose that could work (honestly, I'm not even sure if that is allowed, because what if the loop is accelerating, then you're using the laws of Newton in an accelerating frame?), but anyhow it sounds like a weird implicit thing to do, and it messes up the idea that the four laws of Maxwell describe the E and B-field in your own reference frame.

NB: I wasn't sure if I should make the partial derivative a total derivative when going to the integral form, but I suppose it's irrelevant

 

[vanhees71]

The correct integral form of Faraday's Law reads

$$\int_{\partial F} \mathrm{d} \vec{r} \cdot \vec{E}(t,\vec{r})=\frac{\mathrm{d}}{\mathrm{d}t} \int_F \mathrm{d}^2 \vec{F} \cdot \vec{B}(t,\vec{r}),$$

where the surface, F, and its boundary can be time dependent, and the time derivative gives precisely the additional term you mention in your posting.

 

[nonequilibrium]

I'm sorry, I'm probably being daft, but how does your equation correspond to my last equation? It looks like the erroneous equation in the middle of my post (but with the definition of phi written out)

 

[netheril96]

I think the E in
$$\oint \vec E \cdot \mathrm d \vec l = - \frac{\partial \Phi_B}{\partial t}.$$
indicates the EMF-associated non-electrostatic field.

 

[jfy4]

I think it is fine. You are right that the reason the current starts in magnetic.

Consider this case: let's have a circular wire with changing radius $$r(t)$$. Then our formula reads

$$\oint \vec E \cdot \mathrm d \vec l = - \frac{\partial \Phi_B}{\partial t}.$$

which equals

$$\oint \vec E \cdot \mathrm d \vec l=-\frac{d\Phi}{dt}=-B\pi\frac{d}{dt}(r(t))^2 =-2\pi Br(t)\frac{dr(t)}{dt}$$

This then shows that there is an electric field that results. And it should be detectable in the lab.

I also know that because the cause of the emf is magnetic, many physicists don't call it Faraday's law.

 

[netheril96]

I think it is fine. You are right that the reason the current starts in magnetic.

Consider this case: let's have a circular wire with changing radius $$r(t)$$. Then our formula reads

$$\oint \vec E \cdot \mathrm d \vec l = - \frac{\partial \Phi_B}{\partial t}.$$

which equals

$$\oint \vec E \cdot \mathrm d \vec l=-\frac{d\Phi}{dt}=-B\pi\frac{d}{dt}(r(t))^2 =-2\pi Br(t)\frac{dr(t)}{dt}$$

This then shows that there is an electric field that results. And it should be detectable in the lab.

I also know that because the cause of the emf is magnetic, many physicists don't call it Faraday's law.

Where is electric field in this example?

 

[jfy4]

I think the E in
$$\oint \vec E \cdot \mathrm d \vec l = - \frac{\partial \Phi_B}{\partial t}.$$
indicates the EMF-associated non-electrostatic field.

Where is electric field in this example?

Just as you said here in the first quote, on the left hand side of the equal sign.

 

[netheril96]

Just as you said here in the first quote, on the left hand side of the equal sign.

In my quote, I didn't say E is an electric field.

 

[jfy4]

In my quote, I didn't say E is an electric field.

My mistake. The E in the integral is the electric field.

 

[netheril96]

My mistake. The E in the integral is the electric field.

So you mean, by stretching a circular wire inside it, a constant B-field induces an electric field?

 

[Matterwave]

The integral form of Faraday's law actually encompasses both changes in the B field, as well as changes in the area. The differential form only incorporates the changes in the B field. In fact, the emf induced by the change in the area is caused by the Lorentz force law which is external to the Maxwell's equations. Faraday's law in integral form encompasses both.

 

[vanhees71]

We can put the flux rule (integral form of Faraday's Law of induction) in the form

$$I R=-\frac{\mathrm{d}}{\mathrm{d} t} \Phi.$$

This rule holds for both cases, a moving wire or a time-varying magnetic field or both. In the latter case one has to take into account the change of the surface in the definition of the magnetic flux with time. A standard formula says that for any vector field, $$\vec{a}$$, one has

$$\frac{\mathrm{d}}{\mathrm{d} t} \int_{F} \mathrm{d}^2 \vec{F} \cdot \vec{a} = \int_F \mathrm{d}^2 \vec{F} \cdot [\partial_t \vec{a}+\vec{v} (\vec{\nabla} \cdot \vec{a})]-\int_{\partial F} \mathrm{d} \vec{r} \cdot (\vec{v} \times \vec{a}).$$

Setting $$\vec{a}=\vec{B}$$ and usding $$\vec \nabla \cdot \vec{B}=0$$ and Maxwell's equation $$\vec{\nabla} \times \vec{E} + \partial_t \vec{B}=0$$, we find

$$\int_{\partial F} \mathrm{d} \vec{r} \cdot (\vec{E}+\vec{v} \times \vec{B}) =-\frac{\mathrm{d} \Phi}{\partial t}=I R.$$

It's clear that for a wire loop, moving in a static magnetic field with no electric field present, the whole EMF comes from the $$\vec{v} \times \vec{B}$$ term on the left-hand side of this equation. It can be understood from the Lorentz force $$\vec{v} \times \vec{B}$$ per unit charge acting on the electrons in the wire, making up the current, $$I.$$

A very thorough discussion of this issue can be found in the Feynman Lectures on Physics Vol. II, Chpt. 17.

 

[jfy4]

Hello,

The correct expression should be
$$\oint ( \vec E + \vec v \times \vec B ) \cdot \mathrm d \vec l = - \frac{\partial \Phi_B}{\partial t}. (*)$$

$$\int_{\partial F} \mathrm{d} \vec{r} \cdot (\vec{E}+\vec{v} \times \vec{B}) =-\frac{\mathrm{d} \Phi}{\partial t}=I R.$$

These expressions make me nervous, and help me if I'm off here. But you have

$$d\vec{r}\cdot(\vec{v}\times\vec{B})$$

on the left hand side, and I believe this expression is always zero, since $$d\vec{r}$$ and $$\vec{v}$$ are always in the same direction.

If you finish my example (last step), you get the term I think you are looking for

$$\oint \vec E \cdot \mathrm d \vec l=-\frac{d\Phi}{dt}=-B\pi\frac{d}{dt}(r(t))^2 =-2\pi Br(t)\frac{dr(t)}{dt}$$

the left hand side is $$E(2\pi r(t))$$ which says that

$$\vec{E}=-B\frac{dr(t)}{dt}\hat{\mathbf{\phi}}.$$

where up is the z direction, and around it is $$\phi$$ and out is r.

Correct me if I made a mistake please.

EDIT: I should have made the $$d\vec{r}$$ at the beginning a different letter, it is not specific to the r direction in my coordinates. I mean the displacement vector in general.

 

[vanhees71]

I do not understand what is your problem with $$\mathrm{d} \vec{r}$$ and $$\vec{v}=\vec{v}(t,\vec{r})$$ since these quantities are not related somehow. The former quantity is the line element of the wire, along which you integrate at fixed time, $$t$$, and $$\vec{v}(t,\vec{r})$$ is the velocity of the material point $$\vec{x}$$ of this wire at the same time.

As the most simple example take a homogeneous magnetic field in $$z$$ direction and the loop consisting of an rectangle with three fixed sides and the fourth side (parallel to the $$y$$-axis) sliding along with constant velocity $$\vec{v}=v \vec{e}_x$$. Then $$\vec{v} \times \vec{B} =v B \vec{e}_x \times \vec{e}_z=-v B \vec{e}_y$$. Since $$\vec{E}=0$$ in this example, the line integral on the left-hand side of my equation thus gives $$-v B L$$, where $$L$$ is the length of the moving piece of wire.

The magnetic flux on the right-hand side is obviously given by

$$\Phi(t)=B A(t)=B L (L_0'+v t),$$

leading indeed to $$-\frac{\mathrm{d} \Phi(t)}{\mathrm{d} t}=-v B L.$$

 

[vanhees71]

The correct integral form of Faraday's Law reads

$$\int_{\partial F} \mathrm{d} \vec{r} \cdot \vec{E}(t,\vec{r})=\frac{\mathrm{d}}{\mathrm{d}t} \int_F \mathrm{d}^2 \vec{F} \cdot \vec{B}(t,\vec{r}),$$

where the surface, F, and its boundary can be time dependent, and the time derivative gives precisely the additional term you mention in your posting.

This is obviously wrong, the OP has been right. Sorry for my mistake. What's always right are Maxwell's equations in differential form, especially Faraday's Law

$$\vec{\nabla} \times \vec{E}+\partial_t \vec{B}=0$$.

For the derivation of the correct formula see my other postings.

 

[vanhees71]

Now, let's look on your example:

Let's have a homogeneous static magnetic field in $$z$$-direction and the circular wire in the $$xy$$ plane. Then the magnetic flux (in standard orientation of the boundary, given by the wire, as counter-clock wise) is indeed

$$\Phi_B(t)=B \pi r^2(t).$$

For the line integral, we use polar coordinates

$$\vec{r}=r(t) (\cos \phi,\sin \phi,0)$$.

The line element at fixed time is

$$\mathrm{d} \vec{r}=\mathrm{d} \phi r(t) (-\sin \phi,\cos \phi,0)$$.

The velocity obviously is

$$\vec{v}(t,\vec{r})=\dot{r}(t) (\cos \phi,\sin \phi,0)$$,

leading to

$$\vec{v} \times \vec{B}=\dot{r}(t) B(\sin \phi,-\cos \phi,0),$$

and the line integral is

$$\int_0^{2 \pi} \mathrm{d} \phi [-r(t) \dot{r}(t) B]=-2 \pi r(t) \dot{r}(t) B=-\dot{\Phi}_B(t)$$.

Again, the integral form of Faraday's Law is verified.

 

[jfy4]

I do not understand what is your problem with $$\mathrm{d} \vec{r}$$ and $$\vec{v}=\vec{v}(t,\vec{r})$$ since these quantities are not related somehow. The former quantity is the line element of the wire, along which you integrate at fixed time, $$t$$, and $$\vec{v}(t,\vec{r})$$ is the velocity of the material point $$\vec{x}$$ of this wire at the same time.

As the most simple example take a homogeneous magnetic field in $$z$$ direction and the loop consisting of an rectangle with three fixed sides and the fourth side (parallel to the $$y$$-axis) sliding along with constant velocity $$\vec{v}=v \vec{e}_x$$. Then $$\vec{v} \times \vec{B} =v B \vec{e}_x \times \vec{e}_z=-v B \vec{e}_y$$. Since $$\vec{E}=0$$ in this example, the line integral on the left-hand side of my equation thus gives $$-v B L$$, where $$L$$ is the length of the moving piece of wire.

The magnetic flux on the right-hand side is obviously given by

$$\Phi(t)=B A(t)=B L (L_0'+v t),$$

leading indeed to $$-\frac{\mathrm{d} \Phi(t)}{\mathrm{d} t}=-v B L.$$

This is obviously wrong, the OP has been right. Sorry for my mistake. What's always right are Maxwell's equations in differential form, especially Faraday's Law

$$\vec{\nabla} \times \vec{E}+\partial_t \vec{B}=0$$.

For the derivation of the correct formula see my other postings.

Yes yes, after some skimming and talking aloud I am convinced that the OP is right, as are you. Allow me to correct my self and restate these items on my own terms to ensure understanding.

The equation

$$\mathcal{E}=\oint\vec{E}\cdot d\vec{l}=-\int \frac{\partial\vec{B}}{\partial t}\cdot d\vec{A}$$

Holds for situations were the surface (line circle, contour) is stationary in the observers frame. This is the Maxwell-Faraday Law.

the equation

$$\mathcal{E}=\oint (\vec{E}+\vec{v}\times\vec{B})\cdot d\vec{l}=-\frac{d\Phi}{dt}$$

is a general case for when the contour is moving with respect to the observer (for both an electric field and a magnetic one).

 

[nonequilibrium]

It's been a few confusing posts, but it seems we all agree after all :)