# Work done in a moving conducting bar and in a Faraday disk

I deeply apologize for my ignorance, but I am asking this since I still can't seem to grasp what my teachers relayed to me a few days ago about two exercises that I solved in an intrinsically wrong manner and they both share the same mistake.

If I have a moving rod made of conducting material that is submerged in a uniform magnetic field, this will cause free charges within the rod to experience a magnetic force which is perpendicular to both the B field and the velocity, and this I believe was the reason why the rod begins polarizing on both ends which in turn establishes an electric field which will grow until charges within the rod reach an equilibrium wherein both the magnetic force and the electrostatic force are equal in magnitude.

My question is why does the book equate the emf induced (upon considering now the rod being connected to a U-shaped conductor) to the work done by the magnetic force per unit charge. As I understand it, the magnetic force cannot do work on a free-moving charge due to the direction it would point towards, yet at the same time I don't know if I can really say that in this case as am I not limiting their movement due to them being contained in the rod?

This I also have a hard time understanding for a Faraday disk such as this:

Since a teacher from my university did in fact solve it saying that the emf induced would be equal to the work done by the magnetic force upon taking a charge from the center of the disk to the edge. But once again I don't comprehend why, or why not, I would be able to say that the one doing work is the magnetic force when there is movement of the conducting material (and no change in the magnetic flux) which produces a magnetic force that forces the free charges to move.

• This is explained in Introduction to Electrodynamics by Griffiths, in example 5.3. Once the rod starts moving, the velocity of the current is no longer purely vertical. The work is done by the emf source. Commented May 25 at 20:27
• I really appreciate the example, it helped me a lot to get more perspective on the matter :D, though it also brought up a question I didn't have before: essentially even for a current loop that has a magnetic dipole moment, the torque there isn't actually due to the magnetic forces on each side but rather the electric field? Commented May 25 at 21:22
• Yes, that is correct. The agent responsible for driving the current (such as a battery) must do the work. Commented May 26 at 0:36

An electron moving freely in a magnetic field would not gain or lose speed or kinetic energy. It would be deflected and change direction.

But electrons in a wire are constrained. They can be pushed up the wire by the magnetic force. They can gain potential energy by being crowded together against their mutual repulsion. This potential energy per unit change is the EMF.

If there is resistance in the circuit, the EMF is across the ends of a resistor. It causes current to flow through the resistor. The resistance limits the current.

Any work done is due to forces produced by the interaction of the magnetic field and charged particles moving relative to the magnetic field which is abbreviated to magnetic force.
Thus in Figure 12(b) electrons in the wire are moving to the right and with a magnetic field coming out of the screen the forces $$B q v$$ on the electrons are upwards.
Those upward forces cause the free electrons in the conductor to migrate upwards which sets up an upward electric field $$E$$ in the conductor as indicated by the positive and negative charges shown at the ends of the conductor.
Eventually the downward forces on the electrons due to the electric field $$qE$$ exactly balance the upward forces on the electrons due to the motion of the elections in the magnetic field, ie $$qE = Bqv \Rightarrow E = Bv.$$

If the potential difference (work done per unit charge) across the ends of the conductor is $$V$$ and $$E = V/\ell$$ this gives $$V = B\ell v$$ and this is called the motional emf, $$\mathcal E$$.

Once this condition has been reached and if there is no external conducting path between the ends of the conductor, there is no further movement of the electrons upwards and no work needs to be done to keep the conductor moving.

If a conducting path of resistance $$R$$ is established between the ends of the conductor then a current $$I = \mathcal E/R$$ flows in the circuit and work has to be done to move the electrons round the conducting circuit at the rate of $$\mathcal E I=B^2\ell^2v^2/R$$ per unit time.

That work done to keep the conductor moving at constant speed $$v$$ does not from the magnetic field, ie the magnetic field does no work and its strength does not change.
Rather it comes from either an external force pushing the conductor or the conductor losing kinetic energy and slowing down.

In the case of the homopolar generator the speed of the conductor moving through the magnetic field varies from zero at the centre to a maximum at the outer edge so an integration has to be done to calculate the emf and again the magnetic field does no work.

• So in the case of the conducting wire really I could think of the charges making a sort of circular path relative to a fixed reference frame and thus what the magnetic force is doing isn't actually work but just bending the direction in which the charges would move as the entire wire moves to the right with a velocity V? As in, one component of it always points upward yet the vector is always normal to the trajectory that they describe in space? Also thank you very much for the answer too :D Commented May 25 at 22:33
• dannycaballero.info/phy482msu_s2020/papers/… I found the answer to the last question I made in what I believe is an appendix of a book by a fellow named Mosca. I once again thank you very much for the answer friend :D Commented May 25 at 22:50
• This may be of interest? Electricity and Magnetism: MIT 8.02 Course Notes and in particular Chapter 10 Faraday’s Law of Induction Commented May 26 at 9:32

If I have a moving rod made of conducting material that is submerged in a uniform magnetic field, this will cause free charges within the rod to experience a magnetic force which is perpendicular to both the B field and the velocity,…

A free electron that has kinetic energy (moves) with respect to a magnetic field is deflected perpendicular to its direction of movement and perpendicular to the magnetic field. How does this happen? It is known

• that the magnetic dipole of the electron and the external magnetic field interact: the magnetic dipole of the electron is aligned by the magnetic field
• that the electron emits EM radiation during this alignment
• that the electron is deflected sideways
• that the electron gradually loses its kinetic energy (slows down to a standstill).

With a little imagination, we can now assume that the emission of EM radiation causes the orientation of the electron's magnetic dipole to tilt back - which explains the cyclical process of deflection in the first place. This mechanism better explains what happens in the case of the Lorentz force, Hall effects and the generator principle.

… and this I believe was the reason why the rod begins polarizing on both ends which in turn establishes an electric field which will grow until charges within the rod reach an equilibrium wherein both the magnetic force and the electrostatic force are equal in magnitude.

Exact.

As I understand it, the magnetic force cannot do work on a free-moving charge due to the direction it would point towards, yet at the same time I don't know if I can really say that in this case as am I not limiting their movement due to them being contained in the rod?

The external magnetic field is not consumed. A permanent magnet will be just as "fresh" after the charge shift as it was before. However, the explanation has already been given above. It is the kinetic energy of the charge carriers that causes the charge shift/deflection. Even if the freely moving electrons can only move in a conductor.

This I also have a hard time understanding for a Faraday disk…

Still the same game. The electron moves in the magnetic field and is deflected inwards or outwards depending on the combination of the orientation of the magnetic field and the direction of rotation of the disk. The disk this time is slowed down faster in the magnetic field than by natural friction alone. (BTW, a moving non-conductor disk is not influenced by the magnetic field. After the immobile electrons once have been aligned by the magnetic field, they remain trapped in their position until the magnetic field is removed.)

Work done in a moving conducting bar and in a Faraday disk: Conversion of the kinetic energy of unbounded charges into EM radiation & deflection of the charges until the kinetic energy is exhausted.

• Much appreciated friend, I believe this one also allowed me to better comprehend the idea in general, I incidentally had to do an exercise that named the case of a non-conductor so this also gave me some reassurance in that regard. Thanks :D Commented May 27 at 3:31