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In the classical example of the slidewire generator where the rod slides on a U-shaped conductor in a magnetic field, we get a charge separation due to the Lorentz force. The way the induced emf is usually derived is the following: $\mathcal{E}=W_l/q=vBL$ where $W_l$ is the work done by the magnetic (Lorentz) force. But we know that work is defined as a scalar product of force and displacement. In case of Lorentz force, which is always perpendicular to the velocity vector, the scalar product should always be zero. So why then we allowed to use the above expression?

It even confuses me more, because later on the book states:

The Joule heating produced in the slidewire generator is: $\Delta Q=R I_{induced}^2 \Delta t=\frac{(vBL)^2}{R} \Delta t$

....

But the Lorentz force does no work, so where does this energy come from? It turns out that there is another Lorentz force acting in the system due to the relative motion of the charges...

So, on the one hand it derives motional emf by assuming $F_l$ does work $W_l$ and on the other hand, it says later that it does no work (which can cause a confusion because heat is still being produced in the circuit).

What's going on here?

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There are two things that lead to your confusion:

1) the integral $\oint_C q(\mathbf v\times \mathbf B)\cdot d\mathbf s = vBl$ along the circuit $C$ is not $work$ of magnetic force; it is called electromotive force of the circuit (emf). This is because $\mathbf v$ in this expression is not the velocity of the charged particle, but that of the element of conductor. The charged particle moves in a more complicated way.

The work of magnetic force per unit charge per displacement $d\mathbf r$ would be $$ (\mathbf u\times \mathbf B)\cdot d\mathbf r $$ where $d\mathbf r$ is displacement of the charged particle and $\mathbf u$ is its velocity. Since $\mathbf u = \frac{d\mathbf r}{dt}$ is parallel to $d\mathbf r$, the above product vanishes and the magnetic force does not work on the charged particle.

2) if magnetic force does not work on the charges, then why there is current and evolution of heat in the conductor in the first place?

In macroscopic theory, the description of such situations where current is due to nonelectrostatic force (such as current in the battery, or due to thermal gradients, or due to magnetic field, as in the rod) is via general notion of electromotive intensity $\mathbf E^*$. This is not electric field, but has the same units and its meaning is that the total force acting on the macroscopic charge is

$$ q(\mathbf E + \mathbf E^*). $$

One can derive approximate condition for $\mathbf E^*$ for the case of perfect conductor; the total force acting on the charge has to be zero and

$$ \mathbf E^* = - \mathbf E. $$

Various circumstances lead to various expressions for $\mathbf E^*$ in terms of other measurable quantities, for example in the Seebeck effect $\mathbf E^* = -k \nabla T$. In the case of magnetic forces, experiments are well explained by assuming that

$$ \mathbf E^* = \mathbf v \times \mathbf B, $$ but explaining why this works from the point of view of observer in the laboratory frame seems hard. It would require going into some microscopic theory of conduction and taking into account both the velocity of the rod $\mathbf v$ and the velocities of the individual charge carriers $\mathbf u_i$.

However, the above choice for $\mathbf E^*$ is well explained by the theory of relativity, if we look into the conduction from the point of view of obsever on the rod. There the electric field is composed of two parts, the electric field due to charges of the rod $\mathbf E_0'$ which from the theory of relativity is almost equal to $\mathbf E$, and the electric field due to external bodies $\mathbf E'_{ext}$, which from the theory of relativity is approximately equal to $\mathbf v \times \mathbf B$. So the total electric field in the frame of the rod is approximately $\mathbf E' = \mathbf E + \mathbf v\times \mathbf B$ and putting $\mathbf E' = \mathbf 0$ for ideal conductor of the rod and using the Ohm law for the rest of the circuit (resistance $R$), we can derive the required current $vBl/R$ and can motivate why $\mathbf E^* = \mathbf v\times\mathbf B$ works in the original frame.

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    $\begingroup$ Thanks for the answer. As far as I understand, the particles are moving with the speed of the rod (approximately of course), so there is at least one component of the Lorentz force which is $q \mathbf{v} \times \mathbf{B}$ where $\mathbf{v}$ is the velocity of the conducting rod and hence the particles. The other component is due to the effect from the previous force: the particles are being separated i.e. electrons are moving downwards (in addition to their movement together with the rod). $\endgroup$ – user37127 Jan 10 '14 at 18:18
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If you could cite the reference more completely, it would have been much clearer. However, as far as I understood, there is only one Lorentz force which is $ F = q(E+\bar vX\bar B)$. The heating/work that you are referring to is called Joule heating and is caused by the motion of electrons in the wire.

Joule heating is caused by interactions between the moving particles that form the current (usually, but not always, electrons) and the atomic ions that make up the body of the conductor. Charged particles in an electric circuit are accelerated by an electric field but give up some of their kinetic energy each time they collide with an ion. The increase in the kinetic or vibrational energy of the ions manifests itself as heat and a rise in the temperature of the conductor. Hence energy is transferred from the electrical power supply to the conductor and any materials with which it is in thermal contact.

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  • $\begingroup$ The charges are not only moving to the right, but are also moving upwards, so there are two components of the Lorentz force. I can't provide the source because it is non-English. I've attached an image though. Thanks. $\endgroup$ – user37127 Jan 10 '14 at 14:16
  • $\begingroup$ Yes, but the force is to the left. Now, the electrons in the wire cannot jump out of the wire as they are obstructed by the "atomic ions of the conductor" as mentioned in the second paragraph. This force doesn't give a $Fds$ term. $\endgroup$ – Torsten Hĕrculĕ Cärlemän Jan 10 '14 at 14:38
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Just for clarity, if I say positive y (direction) then I mean up, and if I say positive x (direction) I mean to the right.

We will follow the life of an electron, we can call it El.

So, our conduction electron El is happily moving on the top wire, to the right. It's guided by electrostatic forces in the sense that when it bumps into something that slows it down, the conductions electrons in front keep going (so postiive charge imbalance there, attracting it) and the electrons behind it keep building up (negative charge imbalance there, pushing it forward). Same if something pushes it extra. It's like traffic, there are consequences for stepping out of line, in this case electrostatic consequences.

OK, so this electron El will go into the wire on the right for the same reason, electrostatically the electrons in front of El went that way, so there is an opening (will be a charge imbalance), but turning that corner all those protons are moving at speed v to the right. So even though there isn't an earlier electron to "follow" there is a charge imbalance there, so electrostatically it gets pulled that way, and will continue until it also is moving at speed v to the right. But now something new is happening.

Now El has a velocity to the right, and so can experience an magnetic force downwards, and the electrons in front there are experiencing the same kinds of forces, so everyone can have some force per unit charge in the direction of the circuit element. We can measure this, and it's called EMF. So the conduction electron gets an EMF because of the rightwards component of the velocity.

This is not work because force is orthogonal to the motion (a more precise analysis would note that the average velocity of the particle needs to get it from the top at one time to a point below it ... and to the right since by the time it gets to the bottom the wire is farther to the right, but there will still be no work in this more careful analysis, and it's only the part of v orthogonal to the circuit element and the magnetic field that contributes to the EMF, so it's not an entire fraud. An even more careful analysis would have a sea of random velocities with a net bias for a drift velocity, but again it's the average velocity that contributes a net effect, and again only the rightwards component of the velocity that matters to the EMF).

Now the downwards motion of the electrons also contributes to a magnetic force. And there is the normal electromagnetic pull and push downwards (traffic opening up in front and honking behind) and the magnetic force helping now too (because of the rightwards velocity). This has a magnetic effect to send the electrons onthe left of the wire out (farther left) and those on the right deeper into the wire), and either way electrostatic forces respond to that charge imbalance by pushing in the opposite direction to keep the charge from building up an imbalance on the right (magnetism pushes charges on the right into the wire, but then electrostatic push back to the right, so the right side is a tiny bit charged, exactly enough to counter the magnetic force) and the charge imbalance on the left (magnetism pushes charges on the left out of the wire, but then electrostatic forces push back to the right, so the left side is a tiny bit charged, exactly enough to counter the magnetic force).

Basically, you have neutral wire and a pretty steady current (not changing fast relative to the speed of light) so magnetic forces do push the charges around, but always orthogonal to the velocity of any fixed charge and small charge imbalances counteract any deviation from the steady flow of charge.

Now, there is a current, and there is energy lost due to the current. Where does that energy come from? When the electrostatic forces smooth out the charges, the charges exert equal and opposite forces on those protons. So when you pull an electron on the right edge to the right, the protons get pulled left. And when you pull an electron on the left edge of the wire to the right, the protons get pulled to the left. The net effect is that either the bar is going to get pulled to the left, or else something else also needs to pull the bar to the right to keep it going at a steady speed.

So to summarize. The bar moving to the right means there will be increased flow in the direction of the circuit element, so we get an EMF. This flow is smoothed out by electrostatic forces so that it doesn't build up or get away from the wire. These electrostatic forces on the conduction electrons individually have equal and opposite forces, so the net total electrostatic force on all the conduction electrons is equal and opposite to the net electrostatic force on the other charges (the protons, bound electrons, etc.), and this either slows that moving rod down, or is opposed by a totally new force, the force pulling that rod over to the right.

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