Why is mechanical work equal to Joule heating?

This question came about when I was doing a problem related to a conducting loop with resistance $$R$$ moving with velocity $$v$$ in a magnetic field. It asks you to verify an expression for the force acting on the loop. Essentially, in the solution, it uses (and states) the following: $$Fv = \frac{V^2}{R}$$, where the electromotive force $$V$$ is obtained in another part of the question.

I don't understand this equivalence. On the one hand, we have the mechanical work due to the loop moving in the magnetic field, and on the other hand, we have work due to the motions of the electrons in the wire? What is the physical reason that these are equal?

• I don't quite understand the deeper point of this question. A quick intuitive answer as written would be conservation of energy but I have a feeling that's not what you're after.
– noah
Mar 29 at 18:59
• Why would conservation of energy imply this equality? Are you just saying that the work done to move the rod "allowed" the induced current which resulted in the Joule heating? Perhaps the wording isn't clear but my question is about the connection between the macroscopic motion of the loop in the field, and the microscopic random motions of the electrons within the loop. Mar 29 at 19:07

On the one hand, we have the mechanical work due to the loop moving in the magnetic field, and on the other hand, we have work due to the motions of the electrons in the wire? What is the physical reason that these are equal?

They aren't, in general. The loop is also subjected to air drag, for example. No material is perfectly ohmic. Material properties and applied fields aren't constant. Various other energy transfers are potentially occurring, such as heat transfer, for instance.

But the reader is expected to notice that the problem is a textbook-type problem about certain underlying concepts of limited scope (generated EMF, work, energy, and power) and to notice that no other information is provided.

The reasonable conclusion is that only one work input is relevant in this context: the force needed to move the loop at constant velocity. One one energy output is relevant in this context: Joule heating.

The reader is expected to set $$Fv$$ equal to $$\frac{V^2}{R}$$ because of pedagogical reasons, not any fundamental law (although the equivalence is consistent with a simplified, ideal system following our understanding of physical law).

V=IR is derived under the assumption that the current is constant, the potential in this equation is the potential required to maintain a constant current under the influence of a resistive term.

This resistive term is responsible for negative work being done on electrons in the wire. And is also the amount of energy being transfered as Heat.

Ohms law is also derived as a steady state solution. Aka, the amount of work required to actually get a current going initially is negligible, the only amount of energy needed is to overcome those resistive forces

For a constant current to exist. The amount of work that the induced emf is doing, (as a result of macroscopic motion) must be equal to the amount of negative work that the resistive term is doing on electons. This is because there is a net 0 gain of KE for a constant current.

Thus work that is being lost to due the resistive term is thermal motion, aka the energy that s being transfered to the object through collisions with the electrons that make up the current

Because we have equated the amount of work done by the emf and the amount of energy lost as thermal motion, we can the use

P=IV

to find the rate at which energy is being lost as heat.

As this equation represents the amount of work that is being done on charges in a volume through a potential V. So when we input the potential as V= IR, we know this is going to be the same thing as the energy lost through joule heating.

• Hi, thanks for the answer, but my question is about how the macroscopic motion of the wire itself in the field is related to the microscopic motions of the electrons in the wire, not about what the EMF is. Mar 29 at 19:23
• I know. My answer wasn't trying to teach you what EMF is. I Was saying that the work being done by the induced emf, can be equated to the resistive terms in the drude model of conductivity( energy being lost to heat) because there is a constant current, so the amount of energy I put in, is equal to the amount of work being stolen through heat, as the amount of work to get a current going is miniscule ( which is how ohms law is derived, ignoring this energy) Mar 29 at 19:26
• OK, I understand the part about the work put in being equal to the amount dissipated through heat. But are we then essentially saying that we know that the work done by moving the loop in the field is the "work put in"? Mar 29 at 19:34
• The work "put in" would be the work done by the induced EMF. If I were to actually forcefully rotate a loop, the total amount of work that I put in, would be the work lost as heat , and the rotational KE of the loop. In either case, the work we are concerned with is the work done by the induced EMF. Mar 29 at 19:37
• I've edited it to be more clear Mar 29 at 19:43