PRoblem with differential form of Maxwell's third equation?

Matt1 · May11-06, 12:44 PM

Why does it seem as if the standard differential form of Maxwell's third equation (Faraday's Law) for time varying fields not take into account motional EMF. The differential form simply says that the curl of E is equal to minus the time rate of change of B field. However, there could be a curl of E even in cases where B is constant with a time varying loop in a static B field, so the differential form fails in motional emf, correct? Why isn't the differential motional emf term added to the right hand side of Maxwell's third equation (in differential form)? If it were added in, Maxwell's third equation would work in all cases.

The integral form of Maxwell's third equation works fine in all cases, as long as the derivative is kept OUTSIDE of the flux integral. When it is brought in then there is trouble, and clearly two different answers will be obtained for the emf when comparing the derivative being inside of outside the integral (when the derivative is only applied to the B-field, and not ds). I feel as if this is a flaw in the logic in the derivation of the differential form of the third equation.

This is not independent research. I am just looking for answers to a question.

Thanks.

Matt1 · May11-06, 02:44 PM

Any opinions?

He said that the differential motional emf term that I added was just an approximation at low velocities. Also, when Maxwell formulated the third equation he assumed the moving charge to be the reference frame. However, it seems that it wouldn't be too difficult to add in a term that describes the motional emf at a reference frame other than the bar.

The motional EMF term I added was integral(u x B)*dl. Stokes theorem can be applied to obtain integral(curl(u x B))*dS Therefore, the term curl(u x B) can be added to the right hand side of the differential form of the third equation.

**vanesch** · May11-06, 03:25 PM

Originally Posted by leright

However, there could be a curl of E even in cases where B is constant with a time varying loop in a static B field, so the differential form fails in motional emf, correct?

No, the differential form doesn't fail, because BY DEFINITION, you have to calculate the curl (or the circular integral in the integral formula) in a time-fixed loop, that is, a curve which is time-independent wrt the coordinate frame in which you are expressing the E and B fields.

Why this apparent paradox ? The reason is that the E and B fields do not transform as vectors when you switch from a coordinate frame to another, moving frame (in Lorentz speak, a boost).
E and B aren't really vectors in fact (they only behave as vectors under translations and rotations of coordinates, but not under boosts). They are really elements of a 2-tensor.
So if you want to consider a "moving curve" in frame A, then you should first go to the frame B that is static wrt the curve, and then transform your E and B field from frame A into frame B. And now comes the miracle:

A pure B field in frame A transforms into a B field PLUS an E field in frame B. The extra E field is exactly what corresponds to the EMF.
Indeed, the B-field will be "moving" but that is just "changing" in the B frame, and hence there will be a changing B field and thus a curl of the E-field in the frame B.

Matt1 · May11-06, 05:20 PM

I know that the curl is only calculated in a time fixed frame and I realize that if the derivative of B field in a time fixed frame is 0 then the B field does not induce an E field, and if the derivative of B field in a time fixed frame is NOT 0 then the B field DOES induce an E field. My issue is that an e-field CAN be induced by a constant B-field in the event of the motion of free charge (non-free space conditions) in the field.

I think that the commonly used differential form of Maxwell's equation assumes free space conditions where there is no moving free charge.

I also understand that the commonly used differential form of Maxwell's equation assumes that if there IS moving free charge the reference frame should be taken to be the free charge itself, and this is why Maxwell's equation in differential form "works", but it simply does not convey all possible imformation.

Matt1 · May11-06, 10:32 PM

message has been deleted.

**Hurkyl** · May11-06, 11:33 PM

My issue is that an e-field CAN be induced by a constant B-field in the event of the motion of free charge (non-free space conditions) in the field.

That's your mistake.

EMF is being induced -- not an E-field.

In the case of the time-varying magnetic field, the reason EMF gets induced is because the time-varying magnetic field induces an electric field, and that electric field starts pushing your (stationary) charges around.

In the case of motional emf, what happens is that the magnetic field itself starts pushing your (moving) charges around. It has absolutely nothing to do with the electric field.

In fact, my textbook specifically says that the integral form:

$LaTeX Code: <BR>\\oint \\vec{E} \\cdot d\\vec{l} = - \\frac{d \\Phi_B}{dt}<BR>$

is "valid only if the path around which we integrate is stationary."

Matt1 · May11-06, 11:46 PM

Originally Posted by Hurkyl

That's your mistake.

EMF is being induced -- not an E-field.

In the case of the time-varying magnetic field, the reason EMF gets induced is because the time-varying magnetic field induces an electric field, and that electric field starts pushing your (stationary) charges around.

In the case of motional emf, what happens is that the magnetic field itself starts pushing your (moving) charges around. It has absolutely nothing to do with the electric field.

In fact, my textbook specifically says that the integral form:

$LaTeX Code: <BR>\\oint \\vec{E} \\cdot d\\vec{l} = - \\frac{d \\Phi_B}{dt}<BR>$

is "valid only if the path around which we integrate is stationary."

In order for there to be an induced EMF (capacity to do work on charges) there MUST be an corresponding e-field. In order for there to be an induced EMF that can drive current around a closed path there MUST also be a "non-conservative" e-field produced, and the only way a non conservative e-field can be produced is from the energy stored in a b-field. The b-fields themselves do not do work. The energy in the b-fields INDUCE e-fields that can do work, and work over a closed path at that (conservation of energy is not violated since the energy allowing the e-field to do work over the closed path is coming from the b-field)

And in motional EMF, the magnetic field is not doing the work. The magnetic field is storing the ENERGY to produce an e-field that does the work. The force on a moving charge or current element is due to an induced e-field which comes from the energy in the b-field.

This is all getting more and more clear to me, and the more I think about it, the more things make sense. Before, the more I thought about it and the deeper I went, the more confused I got. I've reached the point where the deeper I get and the more I think about it, the more clearly things become.

Just because something has energy in some form does not mean that it itself can do work, even though energy is defined as the capacity to do work. The energy has to be converted to other forms of energy that can do the work. For instance, b-fields themselves cannot do work because the force is ALWAYS perpendicular to the direction of motion, and the b-field itself cannot increase the kinetic energy of a charge. However, the energy in the b-field induces an e-field which does the work for it. So, even if the energy itself cannot do work that doesn't imply that it isn't really energy. It can be converted to other forms of energy which can the work. So b-fields aren't DOING work, but they are facilitating work by inducing e-fields.

**Hurkyl** · May11-06, 11:58 PM

Work it out yourself:

(1) electric field that is zero everywhere.
(2) Constant magnetic field.
(3) Conducting wire.

When the wire starts to move, you have moving charges in a magnetic field. Thus, those charges start to move along the wire, and you get current, even though the electric field is zero, and the magnetic field is constant.

Your entire line of thought is based on the fact that magnetic forces cannot do work, correct? So all that means is that something else is doing the work. Why do you assume it's the electric field? There's a much simpler explanation that doesn't involve rejecting established physical laws!

I wonder how that wire started moving...

Matt1 · May12-06, 12:08 AM

Originally Posted by Hurkyl

Work it out yourself:

(1) electric field that is zero everywhere.
(2) Constant magnetic field.
(3) Conducting wire.

When the wire starts to move, you have moving charges in a magnetic field. Thus, those charges start to move along the wire, and you get current, even though the electric field is zero, and the magnetic field is constant.

Your entire line of thought is based on the fact that magnetic forces cannot do work, correct? So all that means is that something else is doing the work. Why do you assume it's the electric field? There's a much simpler explanation that doesn't involve rejecting established physical laws!

I wonder how that wire started moving...

ok, the b-field still is not doing WORK on the charges in the rod. When the rod moves in the constant magnetic field motional emf occurs. Motional emf is caused by an e-field that pushes the charges and that e-field is induced by the magnetic field. I am not rejecting any physical laws.

It is an electric field that is induced perpendicular to the magnetic field that causes the emf. The force is perpendicular to the magnetic field so it is not doing work...the induced e-field is doing the work, but the energy to produce the e-field is coming from the b-field.

Why is the force due to a magnetic field called Fsubm and when you divide Fsubm by Q it's called Esubm? Perhaps because it is an e-field doing the work. Also, take a look at the lorentz transformations (relationships between E and B fields with relativistic considerations) and you'll notice a UxB term when getting the e-field from b-field. The UxB is the force due to the B-field.

http://farside.ph.utexas.edu/teachin...s/node123.html - look at the determination of the perpendicular e-field due to the b-field.

**Galileo** · May12-06, 02:44 AM

Originally Posted by leright

In order for there to be an induced EMF (capacity to do work on charges) there MUST be an corresponding e-field. In order for there to be an induced EMF that can drive current around a closed path there MUST also be a "non-conservative" e-field produced, and the only way a non conservative e-field can be produced is from the energy stored in a b-field. The b-fields themselves do not do work. The energy in the b-fields INDUCE e-fields that can do work, and work over a closed path at that (conservation of energy is not violated since the energy allowing the e-field to do work over the closed path is coming from the b-field)

And in motional EMF, the magnetic field is not doing the work. The magnetic field is storing the ENERGY to produce an e-field that does the work. The force on a moving charge or current element is due to an induced e-field which comes from the energy in the b-field.

A moving (conducting) loop in a constant magnetic field (like a generator) will induce an EMF, but not because of Faraday's law. It's simply because in the loop you now have moving charges in a magnetic field which are pushed along by the Lorentz force and this causes the current. The magnetic force is not doing work ofcourse, but the person pulling the loop is.

**arunbg** · May12-06, 03:14 AM

I believe Galileo is right, the phenomenon is similar to Hall's effect which is attributed to Lorentz's forces.

However what I don't understand is that in many cases such as a rod rotating in a uniform magnetic field the induced emf across the ends of the rod can also be calculated by the eqn for motional "induced" emf
E = Blv (it is done so in my text) even though there is no change in net flux.I calculated the answer independently using lorentz forces and got the exact same answer which seemed the right approach to me.

Does this imply some kind of relation or is this simply a freak coincidence?

Matt1 · May12-06, 03:43 AM

Originally Posted by Galileo

A moving (conducting) loop in a constant magnetic field (like a generator) will induce an EMF, but not because of Faraday's law. It's simply because in the loop you now have moving charges in a magnetic field which are pushed along by the Lorentz force and this causes the current. The magnetic force is not doing work ofcourse, but the person pulling the loop is.

It is because of Faraday's law. Faraday's law says that the induced EMF is equal to the time rate of change in the flux through a conducting loop. This is what Faraday's law says. You can produce a change in the magnetic flux by changing the area of the loop, even in a constant B field.

The induced EMF can be because of a number of physical phenomena, but all of the physical phenomena is contained in Faraday's law.

**arunbg** · May12-06, 06:15 AM

It is because of Faraday's law. Faraday's law says that the induced EMF is equal to the time rate of change in the flux through a conducting loop.

There is no time rate of change of flux through a fixed conducting loop moving in a uniform magnetic field.

**Hurkyl** · May12-06, 09:24 AM

Why is the force due to a magnetic field called Fsubm and when you divide Fsubm by Q it's called Esubm?

My book doesn't use that notation, so I don't know what you mean by F_m and E_m.

Also, take a look at the lorentz transformations (relationships between E and B fields with relativistic considerations) and you'll notice a UxB term when getting the e-field from b-field. The UxB is the force due to the B-field.

But we're not doing a change of reference frame, so the Lorentz transformations are irrelevant.

The Lorentz transformations say that the EMF induced by motion through a pure magnetic field in one frame has to have the same effect as the EMF induced by the electric field in some other reference frame where there is no motion.

But that does not mean that an electric field spontaneously appears in the original reference frame!

The force is perpendicular to the magnetic field so it is not doing work...the induced e-field is doing the work, but the energy to produce the e-field is coming from the b-field.

No -- there is no work being done in the production motional EMF. The energy to create the current comes from the kinetic energy of the wire. Something caused the conducting wire to start moving -- the magnetic field acts to change that motion into a current. If there was no friction, then given enough time and space, the magnetic field will convert all of the wire's kinetic energy into current, and the wire will stop moving.

I am not rejecting any physical laws.

You're rejecting the third of Maxwell's laws. And, it seems, the mathematical theorem that the integral form is true if and only if the differential form is true.

Matt1 · May12-06, 01:23 PM

Originally Posted by Hurkyl

My book doesn't use that notation, so I don't know what you mean by F_m and E_m.

But we're not doing a change of reference frame, so the Lorentz transformations are irrelevant.

The Lorentz transformations say that the EMF induced by motion through a pure magnetic field in one frame has to have the same effect as the EMF induced by the electric field in some other reference frame where there is no motion.

But that does not mean that an electric field spontaneously appears in the original reference frame!

No -- there is no work being done in the production motional EMF. The energy to create the current comes from the kinetic energy of the wire. Something caused the conducting wire to start moving -- the magnetic field acts to change that motion into a current. If there was no friction, then given enough time and space, the magnetic field will convert all of the wire's kinetic energy into current, and the wire will stop moving.

You're rejecting the third of Maxwell's laws. And, it seems, the mathematical theorem that the integral form is true if and only if the differential form is true.

no no no....my whole point in this thread is that the differential form of the third equation FAILS to take into consideration motional EMF (and yes, motional EMF induces an E field with a curl), which is where there is a change in reference frame. That is the correction to the equation I am making, and for this you would need Lorentz transformations.

**jtbell** · May12-06, 03:14 PM

Originally Posted by arunbg

There is no time rate of change of flux through a fixed conducting loop moving in a uniform magnetic field.

If the loop rotates (as in a simple generator), the flux through the loop does change. For a fixed-area loop in a uniform magnetic field:

$LaTeX Code: \\Phi = AB \\sin \\theta$

$LaTeX Code: \\frac{d \\Phi}{dt} = AB \\frac {d}{dt} (\\sin \\theta) = AB \\cos \\theta \\frac {d\\theta}{dt}$