B-cutting paradox about emf generation

[pengkuan]

B-cutting paradox about emf generation

 

An electromotive force can be generated either by a variation of magnetic flux or by a moving conductor cutting the magnetic force lines (B-cutting). For example, in the set-up shown in the Figure 1 the bar conductor with length l moves at the velocity v cutting the force lines of the magnetic field B. The generated electromotive force is:

U = B l v

 

A current source provides a current I which will do an electrical work against U. For constant I, when the bar moves from the position a to the position b, the quantity of work is:

dW(emf) = I U dt= I l B dx

 

On the other hand, the current-carrying bar will experience a Lorentz force:

F=I B l

 

Let us calculate the mechanical work the Lorentz force does to the exterior when the bar travels the same distance:

dW(emf) = F dx= I l B dx

 

We see that the mechanical work is equal to the electrical work:

dW(emf) =dW(mec)

 

 

 

 

So, no extra energy is left in the coil. However, the energy conservation law is not respected. Indeed, the magnetic energy stored in the coil depends on the positions of the bar. For positions a and b, the energies are respectively:

Ea =½ La I² and Eb=½ Lb I²

 

As the self-inductance Lb is greater than La, the magnetic energy stored in the coil is greater for position b than a:

Lb >La =>Eb> Ea

 

This implies that the coil has gained an extra quantity of energy:

Eb -Ea>0

 

But from where? If this were true, we could make the following machine. Imagine the above coil in a space free of magnetic field. The machine functions in the following way:

 

1) The bar is at the position a, I=0, Ea=0.

2) The current increases to I, the current source does the electrical work.

3) The magnetic energy stored in the coil is Ea =½ La I²

4) The bar slides in the magnetic field created by the other 3 sides. The bar goes to the position b, the current source does the electric work W(emf), the Lorentz force does the mechanical work W(mec). W(emf) and W(mec) cancel out and the coil has not gained any energy.

5) But, the stored magnetic energy is now. Eb=½ Lb I²

6) The current decreases to 0, the coil gives back the stored magnetic energy to the current source. In this cycle, the current source has gained the energy Eb-Ea.

7) The bar goes back to the position a with no current, thus no work.

8) Restart from the step 2).

 

After n cycles, the quantity of energy the current source would get is n( Eb-Ea ). This is impossible because energy cannot be created. Thus, the principle of this process is incorrect. I call this inconsistency the "B-cutting paradox".

[pengkuan]

Lorentz' EMF paradox

 

The "B-cutting paradox" concerns the energy balance. Let us see the creation of electromotive force by the moving bar. The conductor bar's free electrons feel a vertical Lorentz force in moving horizontally, which is denoted . But, is there really an electric tension?

 

The force is constant for steady velocity and constitutes an electrostatic field, . Obeying electrostatic laws the free electrons flow and distribute them-self to induce an electrostatic field so that the resultant electrostatic force on all free electrons inside the conductor bar is 0 and the surface becomes an equipotential. So the 2 points of contact with the 2 horizontal wires have the same potential and the tension is 0.

 

In order to understand this phenomenon, we imagine an identical conductor bar jointed on the one connected to the 2 horizontal wires. But this bar is insulated from the conducting bar and the horizontal wires. The 2 bars are called connected bar and non connected bar (See the Figure 4).

 

Evidently, the electric field in the non connected Bar is 0 and the surface is an equipotential. The connected bar is in exactly the same electromagnetic situation but caries the current of the coil. According to the superposition principle, the electric field is not influenced by the current. Thus, the surface of the connected bar is also an equipotential and the tension between the 2 contact points is 0.

 

Another point of view is that the potential is proportional to the energy that a free electron acquires in movement. Along any path inside the connected bar the line integral of the resultant electrostatic force is 0. Thus, the free electrons of the current get no energy at all. Again, the tension is 0.

 

This conclusion puts the Lorentz force law in opposition with the electrostatic law. I call this opposition the "Lorentz' EMF paradox".

[swansont]

the bar conductor with length l moves at the velocity v

…

the current-carrying bar will experience a Lorentz force

 

 

Do you recognize the inherent contradiction of those two statements?

[pengkuan]

Do you recognize the inherent contradiction of those two statements?

 

I'm sorry, I do not see.

[swansont]

An object subject to a net force does not travel at a constant velocity. You get a contradiction when you overconstrain a problem in an unphysical way.

[pengkuan]

An object subject to a net force does not travel at a constant velocity. You get a contradiction when you overconstrain a problem in an unphysical way.

 

Thank you for the remark. I forgot to precise that there is an external force on the bar, which is equal to the Lorentz force.

[swansont]

Thank you for the remark. I forgot to precise that there is an external force on the bar, which is equal to the Lorentz force.

 

And this force does how much work, that you must also include in your analysis?

[pengkuan]

Internal Lorentz force paradox

 

A current in a magnetic field feels theLorentz force. For isolated coil, the Lorentz force on a current segment is dueto the magnetic field of the coil itself, that is, the Lorentz force isinternal to the coil. Let us study the coil shown in the Figure 1, which is rigid and is composed of a curved part and a straight part. The Lorentz force that the straight part acts on the curved part is F_curv, that the curved part acts on the straight partis F_str. F_curv and F_str are integrated to give the total Lorentz forces on the curved part, R_curv and on the straight part, R_strrespectively. R_curv and R_str are mutual Lorentz forces and, respecting the third Newton's law, their sum should be 0:

 

R_curv +R_str =0

But, does the Lorentz force law really predict 0 for this sum? The 2 total forces are calculated numerically. The calculated force are dimensionless. The length of the straight part is 2, the height of the curved part is 1 and the inclination 45°. The numeric values of the dimensionless forces are as follow:

 

The forceon the curved part R_curv : R_x=-6.4618, R_y =9.5457

The forceon the straight part R_str: R_x=0 R_y=-16.8930

 

 

 

The equations that give these values are:

 

 

 

The total forces R_curvand R_str are drawn in the Figure 1 and are not parallel. But why? In fact, the Lorentz force on the curved part is always perpendicular to the curved wire, and itsintegral has a x component. The Lorentz force on the straight part is vertical and has only y component, that is:

 

R_curv = Cx*i + Cy*j, R_str= 0*i + Sy*j

Now we add them up to find the resultant internal force:

 

R_res =R_curv +R_str= Cx*i + (Cy + Sy ) *j ≠0

R_res is the sum of all internal forces of the coil. As it has a non null x component, C_x, its y component is (C_y+S_y)and is non null for our numerical calculation, R_res violates the third Newton's law. If real internal force had non null value, one could make the following process:

 

Make a coil with inclined curve and straight wires, put the current on, and then let the coil move in the direction of the resultant force R_res. Since the magnetic flux passing through the coil is constant, the current will not do any work. But R_res would do a work in the movement, creating a quantity of energy.

 

 

 

This is impossible because energy cannot be created. R_res is predicted by the Lorentz force law, so, the Lorentz force law does not correctly predict internal magnetic force. I call this inconsistency the "Internal Lorentz force paradox".

 

And this force does how much work, that you must also include in your analysis?

 

 

yes, it was included.

Edited March 14, 2012 by pengkuan

[swansont]

Internal Lorentz force paradox

 

I thought we dealt with this elsewhere.

 

 

yes, it was included.

 

I see an electrical term and a Lorentz term. Where is the external force term?

[Phi for All]

!

Moderator Note

pengkuan, please stick to one speculative idea at a time. Especially don't bring in one unsupported idea to support another unsupported idea.

[pengkuan]

!

Moderator Note

pengkuan, please stick to one speculative idea at a time. Especially don't bring in one unsupported idea to support another unsupported idea.

 

ok

 

I thought we dealt with this elsewhere.

 

I gather the 3 paradoxes in one thread so that when I give this one address to people they could read all.

 

 

I see an electrical term and a Lorentz term. Where is the external force term?

 

It is in the mechanical work.

[swansont]

It is in the mechanical work.

 

What you call mechanical work in the OP is the work done by the Lorentz force. There is an equal amount of work done against that to keep the bar moving at v. So there is still the electrical term left over.

[pengkuan]

What you call mechanical work in the OP is the work done by the Lorentz force. There is an equal amount of work done against that to keep the bar moving at v. So there is still the electrical term left over.

 

The force that keeps the bar moving at v does work W1, the emf of the bar does electric work W2 they cancel. W1-W2=0

[swansont]

The force that keeps the bar moving at v does work W1, the emf of the bar does electric work W2 they cancel. W1-W2=0

 

What about the Lorentz force? A problem I see here is that while a wire moving at v will induce a current via the EMF, it will not arbitrarily induce the current I that is coming from your current source; the induced current from it depends on v. You have assumed they are equal under all conditions.

 

Also, what about Joule heating of the circuit?

[pengkuan]

What about the Lorentz force? A problem I see here is that while a wire moving at v will induce a current via the EMF, it will not arbitrarily induce the current I that is coming from your current source; the induced current from it depends on v. You have assumed they are equal under all conditions.

 

Also, what about Joule heating of the circuit?

 

It is an ideal case, current is constant, independent of v. Joule heating is not cnsidered.

[swansont]

It is an ideal case, current is constant, independent of v. Joule heating is not cnsidered.

 

But the induced EMF, and thus induced current, IS dependent on v. You cannot arbitrarily set them equal.

[pengkuan]

But the induced EMF, and thus induced current, IS dependent on v. You cannot arbitrarily set them equal.

 

The property of a current source is to keep the current constant.

 

Wikipedia

In circuit theory, an ideal current source is a circuit element where the current through it is independent of the voltage across it.

[swansont]

The property of a current source is to keep the current constant.

 

Wikipedia

In circuit theory, an ideal current source is a circuit element where the current through it is independent of the voltage across it.

 

I'm not talking about the current source, I'm talking about the EMF. It is speed dependent. Thus, the current from it is speed dependent.

[pengkuan]

I'm not talking about the current source, I'm talking about the EMF. It is speed dependent. Thus, the current from it is speed dependent.

 

It is a loop. The current is the same all over the loop.

[swansont]

It is a loop. The current is the same all over the loop.

 

For v= 0, what is the induced EMF? What current flows as a result of the induced EMF?

[pengkuan]

For v= 0, what is the induced EMF? What current flows as a result of the induced EMF?

 

u=0

[swansont]

OK. So now the bar is moving at some arbitrary speed v. There will be an induced EMF causing a current i ≠ I in the loop. There is a Lorentz force, but in order to keep v constant, you must also have an external force on the bar, which does work on the system.

 

At some speed v1, I = i. There is now no current in the loop. Thus, no Lorentz force. But no current also means there is no stored magnetic energy. So, no conundrum.

[pengkuan]

OK. So now the bar is moving at some arbitrary speed v. There will be an induced EMF causing a current i ≠ I in the loop. There is a Lorentz force, but in order to keep v constant, you must also have an external force on the bar, which does work on the system.

 

At some speed v1, I = i. There is now no current in the loop. Thus, no Lorentz force. But no current also means there is no stored magnetic energy. So, no conundrum.

 

what is i? and I?

[swansont]

OK. So now the bar is moving at some arbitrary speed v. There will be an induced EMF causing a current i ≠ I in the loop. There is a Lorentz force, but in order to keep v constant, you must also have an external force on the bar, which does work on the system.

 

At some speed v1, I = i. There is now no current in the loop. Thus, no Lorentz force. But no current also means there is no stored magnetic energy. So, no conundrum.

 

I is the current from the current source in your setup. i.e. using your notation.

[pengkuan]

I is the current from the current source in your setup. i.e. using your notation.

 

I have only I, not i. If you consider another current i, which is different, you are not talking about my circuit.