Peng Kuan 彭宽titang78@gmail.com

Tuesday, May 22, 2012

In several articles I have shown one
inconsistency of the Lorentz force law, that Lorentz force internal to a coil
violates the third Newton ’s law. Below are some links:

Synthesis of the inconsistency of the Lorentz force law

http://pengkuanem.blogspot.com/2012/04/synthesis.html

Mathematical cause of the existence of the remaining resultant internal Lorentz force

http://pengkuanem.blogspot.com/2012/04/mathematical-cause-of-existence-of.html

Proof of the remaining resultant Lorentz force internal to a triangular coil

http://pengkuanem.blogspot.com/2012/04/synthesis.html

Mathematical cause of the existence of the remaining resultant internal Lorentz force

http://pengkuanem.blogspot.com/2012/04/mathematical-cause-of-existence-of.html

Proof of the remaining resultant Lorentz force internal to a triangular coil

The Lorentz force law is also used to interpret
the generation of an electromotive force in a coil with a wire moving in a
magnetic field. Let us look at the Figure
1 in which a rectangular coil with a movable wire is shown.
The movable wire is the bar conductor of length

*l*that is constrained to move at the velocity*v*. An electrical generator provides a current*I*in the coil that creates a magnetic field*B*. As the bar conductor moves, it cuts the force lines of the magnetic field*B*. This is the “B-cutting” action referred in the title.
Haha, that was a very tough problem to solve. Assuming no R and some mass of bar (so not external forces), the function for velocity looks like

ReplyDeleteInline image 1

Interestingly, the function for current looks like this:

Inline image 2

I think you didn't take the current into account. Yes I let the speed up while you just let it move at a constant rate against a resistance. The thing is, the bar in your case has to have started at no velocity..which implies a infinite current (assuming no R), and then had a velocity that is some value - which implies some back emf which reduces total voltage and total current. When you have a massless bar, and no resistance, then you have a situation where you are getting an infinite amount of force on a massless object (or your external resistive force). Point it..stuff starts to drop out.

Instead, I assumed all constants equal to 1. Permitivity, permeability, geometry, mass, voltage. I set R = 0 because that leaving it there makes the differential equation much more difficult.

Long story short,

The I before is so much greater than the I after, that even though you have increased the size of the loop, you have actually decreased the energy stored in the magnetic field.

The weird thing about this problem that made it challenging to think about is the following:

At first, no movement, so you have a high current and high magnetic field - high current means high acceleration.

As the bar starts moving, it produces emf, which reduces the current, and also reduces the acceleration.

I automatically said..well, eventually the emf will reach that of the battery, at which point there will be no acceleration. The problem with that is that if the emf is equal to the battery, there is no current - but if there is no current there is no magnetic field to produce the emf.

The result is that the bar gets faster and faster indefinitely (assuming no friction), while the current drops lower and lower. As the speed approaches infinity, the current approaches 0.

So I thought one step forward and asked what would happen if you used a constant external force like you said.

In that case, current is constant, speed is constant, acceleration is 0.

The work done by the battery is equal to the backEMF*I *time.

this work is equal to the F*d + change of .5LI^2

This may seem odd, but the back EMF will be much bigger if you start with the horizontal rectangle to be very narrow...as in (the bar slides to the right, but starts near the battery). The backEMF is related to the B field near the bar, and the B field near the bar would be much bigger when it's closer to the battery than when it's further away.

As the size of the circuit grows to infinity, the backEmf is purely proportional to I*dx/dt, but at the same time, the rate of change of inductance per location dL/dx becomes 0.

It is impossible to make a clear reasoning for a transient problem as that of no-constant speed. So, I use constant speed for the bar so that force and current are constant and EMF equals the battery voltage.

DeleteAt the time, I thought that there was no force between perpendicular currents and thus, no B-cutting EMF. This is proven to be false by my "Perpendicular action experiment with a long rectangular coil"

http://pengkuanem.blogspot.fr/2014/03/perpendicular-action-experiment-with.html

So, There must be B-cutting EMF. The Paradoxe can be solved by taking in to account 2 different EMF, 1) EMF of B-cutting; 2) EMF of the increase of the surface, that is the value of inductance L.

However, there is no experimental data showing that the battery voltage equals EMF1+EMF2.

The B-cutting EMF and the increase of B-through-surface are one and the same. You get one by looking at the Lorentz law, and one by looking at Ampere's law. At the basic level, it is all Lorentz forces on individual particles and the magnetic fields defined by the biot Savart.

DeleteIf B-cutting EMF and inductance EMF were the same and one, then the energy will not balance.

DeleteThere is 2 outputs of energy: 1) For Lorentz force, F*d; 2) For Inductance, 1/2*L*I^2

This amounts for 2*EMF*I, while the electric power is EMF*I, EMF being equal to the battery voltage.

I have looked too deeply into the math, but I decided to look for a relationship between the lorentz law and maxwell's equations. Some people say that the domain of the laws does not allow them to be put together, but section 4.2.4 of this very thorough paper seems to be able to derive the lorentz force law from maxwell's equations:

Deletehttp://www.plasma.uu.se/CED/Book/EMFT_Book.pdf

regarding "There is 2 outputs of energy: 1) For Lorentz force, F*d; 2) For Inductance, 1/2*L*I^2"

You wouldn't be able to calculate the integral of F*d over your range of d because it wouldn't be a constant external force. (if the bar is massless, a constant external force would result in, at some point, a force imbalance and an infinite acceleration).

Try to approach this numerically by giving the bar some mass, and assuming constant kinetic friction, and assume the bar starts off at dx/dt>0, and assume a real resister, etc.

I would bet that if you calculate the exact forces using just mechanics, lorentzian forces and lorentzian emf, you'd see that when you look at the before and after .5LI^2 and the energy lost to heat at the resistor, inductor energy change, and heat due to friction, everything will balance out.

There is one aspect of energy lost to EM radiation, but i'm not sure how to calculate that and i think if the system moves and changes slowly, this can considered not comparable to the other losses.

I encourage you to look for a relationship between the lorentz law and maxwell's equations although I think it is impossible since Lorentz force law deal with mechanical force and Maxwell deals with E and B fields. But who knows what interesting things you will find.

DeleteI have added number after the equations in my "B-cutting" paper to facilitate discussion. In equation (7), I have proven that the work the Lorentz force does equals that done by the EMF1 in the circuit. So, the energy in the inductance comes from nowhere.