Imagine a long, thin current carrying conductor carrying a current $I$ and moving through space with a velocity $\mathbf v$. If there exists a magnetic field such that there is a force on the current carrying wire in a direction opposite to that of its velocity shouldn't the work done by the magnetic force on the current carrying conductor be non zero as the conductor is being displaced along the direction of its velocity?
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6 Answers
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Consider the well known "sliding rod in a magnetic field" setup:
There is an electric current 'up' (electron current 'down') through the conductor that is moving to the right, and there is a force to the left acting to slow the conductor down.
The magnetic force on the mobile electrons, due to the motion of the conductor is downward while the magnetic force due to the electric current is leftward.
- Note that the vector sum of these force components is always orthogonal the velocity vector of the mobile electrons, thus no work is done by this magnetic force on the mobile electrons.
However, due to this leftward force, the mobile electron density is greater on the trailing side of the moving conductor. The resulting electric field from right to left (within the moving conductor) produces a rightward electric force on the mobile electrons that just balances the leftward magnetic force.
But there is also a leftward electric force on the lattice ions on the leading side of the moving conductor with no balancing magnetic force.
- It is this electric force that does work, not the magnetic force.
answered Aug 21, 2019 at 20:28
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$\begingroup$ Oh thanks for that. But could you tell me why in general is it that we assume magnetic forces do not do work even before analyzing the problem? In the above problem you assumed a setup but could you prove the same for a single long conducting wire moving in a uniform magnetic field? $\endgroup$– Bzzzz..Commented Aug 22, 2019 at 2:49
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1$\begingroup$ @Bzzzz.., the moving rod above can be arbitrarily long, correct? Isn't the analysis the same? Also, I didn't assume that the magnetic force on a mobile electrons does no work, I accept that the magnetic force on a mobile electron is always orthogonal to its velocity by virtue of this equation: $\mathbf{F_B} = q(\mathbf{v}\times\mathbf{B})$ $\endgroup$ Commented Aug 22, 2019 at 10:38
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$\begingroup$ Oh ok thanks for clearing that up. $\endgroup$– Bzzzz..Commented Aug 23, 2019 at 7:01
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Adapted from my answer here:
The Lorentz force on a point charge is $$\vec{F} = q(\vec{E}+\vec{v}\times\vec{B}). $$ The force due to the magnetic field is $$ \vec{F}_{mag} = q(\vec{v}\times\vec{B}) .$$ The work done on $q$ due to the magnetic force per unit time is $$P_{mag} = \vec{F}_{mag}·\vec{v} = q(\vec{v}\times\vec{B})·\vec{v} = q(\vec{v}\times\vec{v})·\vec{B} = 0. $$ This is saying that the work done per unit time is zero because the magnetic force $\vec{F}_{mag}$ is orthogonal to the velocity $\vec{v}$.
Work done per unit time on an extended charge distribution by magnetic forces can be expressed as an integral, with the integrand (power per unit length, area or volume) again vanishing because of the magnetic force per unit length, area or volume being orthogonal to velocity. Thus, quite generally, forces due to magnetic fields do no work.
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Why is it assumed that magnetic forces arising from magnetic fields do not do work on a current carrying conductor?
The reason that it is assumed that magnetic forces do no work is not really an assumption at all; it can be proven directly from Maxwell's equations. This is known as Poynting's theorem. My favorite derivation is found at section 11.2 here: http://web.mit.edu/6.013_book/www/book.html
$$-\nabla \cdot (\mathbf E \times \mathbf H) = \frac{\partial}{\partial t} \left(\frac{1}{2}\epsilon_0 \mathbf E \cdot \mathbf E\right)+ \mathbf E \cdot \frac{\partial}{\partial t}\mathbf P + \frac{\partial}{\partial t}\left(\frac{1}{2} \mu_0 \mathbf H \cdot \mathbf H\right) + \mathbf H \cdot \frac{\partial}{\partial t} \mathbf M + \mathbf E \cdot \mathbf J$$
In your case, with just a conductor, we have $\mathbf P=0$ and $\mathbf M=0$ so the equation simplifies to a more commonly recognizable form
$$-\nabla \cdot (\mathbf E \times \mathbf H) = \frac{\partial}{\partial t} \left(\frac{1}{2}\epsilon_0 \mathbf E \cdot \mathbf E\right) + \frac{\partial}{\partial t}\left(\frac{1}{2} \mu_0 \mathbf H \cdot \mathbf H\right) + \mathbf E \cdot \mathbf J$$
Where the term on the left is the flow of energy from one region of the field to another and the first two terms on the right are the change in the energy density of the electric and magnetic fields respectively. Those are purely field terms; the only term involving an interaction with matter is the last term $\mathbf E \cdot \mathbf J$ where $\mathbf J$ is the free current. This means that the only way that work can be done on a conductor is via the $\mathbf E$ field.
Note, this is a derivation based on the macroscopic Maxwell's equations. A similar derivation can be done based on the microscopic Maxwell's equations, and again the only term involving an interaction with the matter of a conductor is $\mathbf E \cdot \mathbf J$. So regardless of if you are talking about the macroscopic or microscopic Maxwell's equations, for a conductor, the conclusion is the same: all work is done by the $\mathbf E$ field and the amount of work is given by $\mathbf E \cdot \mathbf J$.
It is not an assumption, it is a theorem. It holds as long as Maxwell's equations hold.
As @Alfred Centauri described in his answer whenever you have a situation that looks like there is a magnetic field doing work, you can always "dig deeper" and find where it is actually the $\mathbf E$ field doing the work.
However, instead of digging deeper, let's say that we want to step back a bit. Is there anything else we can learn? The term $\mathbf E \cdot \mathbf J$ includes not only the mechanical work, but also the non-mechanical work. Usually we want to maximize the mechanical work and we want to minimize the non-mechanical work. So suppose, instead of trying to find where the $\mathbf E$ field is, we try to separate out the non-mechanical work from the mechanical work.
To do so, we will transform to the rest-frame of the conductor, since in that frame there is no mechanical work so all of the work in that frame is non-mechanical. Assuming that $v<<c$ the transformation equations are: $$\mathbf E' = \mathbf E + \mathbf v \times \mathbf B$$ $$\mathbf J' = \mathbf J - \rho \mathbf v$$ where the primed quantities are quantities in the rest frame of the conductor. Substituting those into $\mathbf E \cdot \mathbf J$ and simplifying, we get: $$\mathbf E \cdot \mathbf J = \mathbf E' \cdot \mathbf J' + \mathbf v \cdot (\rho \mathbf E + \mathbf J \times \mathbf B)$$
So, that means that the $\mathbf E \cdot \mathbf J$ term itself contains within it the mechanical work due to the magnetic field that you are interested in. If you pull out the non-mechanical work, then the mechanical work is exactly what you would expect including a term from the magnetic field.
This result may seem a little surprising or confusing, as it appears to contradict the above, but it does not. The thing is that all of the fields in electromagnetism are closely related to each other. You can often express the same thing in multiple ways, or dig out hidden dependencies. So although Poynting's theorem holds and although it clearly states that the total work is always $\mathbf E \cdot \mathbf J$, it is not a mere coincidence that formulas describing only the mechanical work correctly include the $\mathbf B$ field and show that it does mechanical work.
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$\begingroup$ You seem to believe Poynting's equation as derived from Maxwell's equations by itself implies that magnitude of work per unit time and unit volume is $\mathbf E\cdot \mathbf J$. That is mistaken; that statement is an assumption independent of Maxwell's equations and independent of Poynting's equation. This assumption is necessary to interpret Poynting's equation in terms of work (as law of local conservation of energy) as is customary. Without this assumption, Poynting's equation is merely a single scalar constraining equation between field vectors and source densities $\rho,\mathbf j$. $\endgroup$ Commented Aug 23, 2019 at 10:16
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$\begingroup$ Yes, work per unit time on the current carrying mobile charges can be expressed as $\int \mathbf E \cdot \mathbf J~dV$, but this does not prove or explain how magnetic forces do not work on the conductor and all work is done by electric forces. It merely shows (in case of perfect conductor) how the work of EM field on the mobile charges forming the electric current equals numerically to work of macroscopic magnetic force on the conductor. But these are two different works! Because different forces act on different bodies (mobile charges vs. the conductor body). $\endgroup$ Commented Aug 23, 2019 at 10:24
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$\begingroup$ “This assumption is necessary to interpret Poynting's equation in terms of work (as law of local conservation of energy) as is customary”. Maybe. I have heard that claim before and not been terribly convinced. It is trivially true that any equation in physics needs to be interpreted in terms of physics. I don’t think this objection goes beyond the trivial. The other terms have the general form of a local conservation law for the field and the units for conservation of energy. Also, it can be derived via Noether’s theorem from time translation symmetry. Given that, it is an easy interpretation $\endgroup$– DaleCommented Aug 23, 2019 at 11:24
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$\begingroup$ > "I don’t think this objection goes beyond the trivial." I do, that's why I took the time to explain the problem with your post. The idea that Maxwell's equations or Poynting's theorem imply by themselves magnitude of work being done has no grounds and is bound to confuse people into putting Poynting's theorem on a mental pedestal that it doesn't belong to. Not all energy exchange is taken into account by Poynting's theorem, for example, there are things such as electrochemical and thermoelectric phenomena where other forces from those of electric field do work on mobile charges. $\endgroup$ Commented Aug 23, 2019 at 14:25
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$\begingroup$ In simple cases like ohmic conductor $\mathbf E\cdot \mathbf J$ is enough, so the Poynting theorem is applicable, but you need to know that $\mathbf E\cdot \mathbf J$ is enough, based on some idea of what is going on in the conductor! Inside battery, for example, there is another force per unit charge $\mathbf E*$ - electrochemical electromotive force - that has opposite direction from the electric force due to field $\mathbf E$, so $\mathbf E\cdot \mathbf J$ does not give complete picture of work done on the mobile charges, $\mathbf E*$ has a say too. $\endgroup$ Commented Aug 23, 2019 at 14:30
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Yes. This is how generators work. You move a wire through a magnetic field. The field generates an EMF, i.e. an electric field parallel to the wire. The EMF/field causes a current to flow in wire. The current together with the magnetic field produces a force directed opposite to the motion of wire. Consequently whatever is pushing the wire has to do work to move the wire. The work-rate is equal to the product of the current and the total EMF, and this is the power that the generator is supplying to the external user.
answered Aug 21, 2019 at 18:52
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A stationary magnetic field does not perform work but a time dependent one does, by virtue of the Maxwell-Faraday equation.
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1$\begingroup$ Doesn't this seem more of a comment than an answer? $\endgroup$ Commented Aug 21, 2019 at 21:30
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Yes, the macroscopic magnetic force on wire, given by familiar formula $BIL$, can do often non-zero work (this is how DC electric motor gets spinning due to magnetic forces acting on current-carrying wires). The idea that magnetic force cannot do work comes from the microscopic theory, where magnetic part of Lorentz force on charged particle indeed does not work, but this does not translate into macroscopic theory, because there magnetic force means something different.
In macroscopic theory, magnetic force means usually the macroscopic force due to external magnetic field acting on the body as a whole, not just on current forming mobile charges, or individual charged particles. This macroscopic force, in terms of microscopic theory as Alfred described in his answer, is actually internal (possibly electric, but that is not important) force due to electrons pushing on the rest of the wire. This push occurs because the electrons are pushed by the magnetic part of the Lorentz force towards the wire boundary but since they are bound to the wire, they cannot jump out, so they translate the push on the rest of the wire.
answered Aug 22, 2019 at 0:17
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$\begingroup$ Poynting’s theorem can be derived from the macroscopic Maxwell’s equations without much additional difficulty. In the absence of magnetizable media the usual term arises macroscopically also. So I don’t think that the distinction between microscopic and macroscopic is a solid explanation. Can you add some math showing the general form of electromagnetic work in the macroscopic theory? $\endgroup$– DaleCommented Aug 23, 2019 at 11:30
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$\begingroup$ The issue is partly terminological and partly understanding about which work we are talking about. Macroscopic magnetic force does work on the conductor, because it is acting on the conductor body, not merely on mobile charge carriers, and this macroscopic magnetic force has component along conductor's velocity. Microscopic magnetic force does not work on the charged particle, because it is perpendicular to velocity of the particle. I hope you see now that this is not really a mathematical issue that warrants "adding some math". $\endgroup$ Commented Aug 23, 2019 at 14:35
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$\begingroup$ "this is not really a mathematical issue that warrants 'adding some math'". That is a very disappointing attitude to see expressed by someone with as much physics knowledge as you have. You imply in your answer that the correct energy conservation law differs between the microscopic and macroscopic cases. I would think that claim would warrant showing the correct general conservation law for each case and explicitly pointing out the difference. But I guess that is just me. By the way, there is a mathematical difference between the two, but it goes to zero for the scenario considered here. $\endgroup$– DaleCommented Aug 23, 2019 at 18:17
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$\begingroup$ No, you are missing the point. The point is not that the conservation law is different. It may be different, depending on the kind of microscopic model we are thinking about. The point was that the usual claim about magnetic force not doing any work applies only to microscopic magnetic forces that act on charged particles, force being expressed by the Lorentz formula. The claim does not apply to macroscopic magnetic forces, also called Laplace forces, or Ampere forces, or ponderomotive forces, which act on the conductor. These are two very different concepts of magnetic force. $\endgroup$ Commented Aug 23, 2019 at 20:46
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$\begingroup$ If you wonder how that could be given Maxwell's equations hold both in macro and micro theory, the reason is matter is modelled differently in microscopic theory; the EM forces have much simpler expression (Lorentz formula) and if we want to have a simple model, those are the only forces present.In macroscopic theory, there are usually various kinds of forces on different classes of matter (mobile charges vs liquid body vs solid body) that don't always have such simple expression in macroscopic terms.Macroscopic theory Poynting is a more crude description of energy exchanges that misses a lot. $\endgroup$ Commented Aug 23, 2019 at 21:09