# Why isn't Magnetic force doing any work on the rod?

Consider the situation given below : The rod (of length $$l$$ and of some resistance "r") connecting the two rails is given a velocity $$v$$ in the direction shown. Now since this motion induces an emf in the rod, so a current flows in the close circuit. Now due to this current there is a magnetic force acting on the moving rod in the opposite direction of its motion which slows down the rod.

Now my teacher told me that the loss in kinetic energy appears in the form of joule heating only which is kind of unsatisfactory to me because the magnetic field is doing negative work on the rod (right ?) , so some of the kinetic energy of the rod should be stored in the magnetic fields too . When I asked this to my teacher he said MAGNETIC FORCES CAN'T DO WORK and nothing else which left me unsatisfied.

Can someone explain why isn't some of the energy being stored in the magnetic fields ?

• The title of the question is a contradiction. How can a force not produce any work?! I am sure you meant the "magnetic field" in your title and not "magnetic force" where you actually referring to the Lorentz force generated in your example? Dec 6, 2021 at 10:56
• @Markoul11 There are plenty of examples of forces not doing work. Dec 6, 2021 at 12:14
• ...most notable in circular motion Dec 6, 2021 at 16:55
• @Ankit The answer of the provided diagram IMO is not correct for the experiment you are describing. It is wrong to say that the magnetic field here is not doing work via the induced electric dipole field in your circuit and subsequent electric current generated. The diagram is misleading. The electrons drift velocity is a few mm per hour so, Vcurrent must be neglected in your diagram, Vrod>>Vcurrent. Therefore, the Lorentz force Fmag vector is opposite to Vrod and there is no Vtotal vectror. Fmag is not vertical to Vrod. Therefore the magnetic field will slow down the rod motion. Work is done. Dec 8, 2021 at 8:34
• @Markoul11 Nope, I wasn't considering free electron beams. You can still see the effects of current and magnetism in everyday day objects and currents. I find it odd that you say the current velocity must be neglected, yet your explanation relies on the current velocity being present to produce the force component to the left. Dec 8, 2021 at 11:11

The misconception here is that only the vertical component of the charge velocity and the horizontal component of the magnetic force are considered. Certainly, if you only look at the horizontal component of the magnetic force you will see this component does work, but the entire force does no net work. A nice mechanical analogy is a block moving up an incline. The components of the normal force do work on the block, but the normal force in its entirety does no work on the block because it is perpendicular to the velocity of the block. The diagram above (not drawn to scale) shows how the magnetic force does no work on the charges in the rod. The charges have a velocity component along the rod (the current) as well as a component perpendicular to the rod (due to the charges actually moving with the rod). The magnetic force is perpendicular to the total velocity of the charges consistent with the Right Hand Rule, which in this case is up and to the left. The left component is the resistive force felt by the rod. The upward component you will find to be consistent with Lenz's law to oppose the change in magnetic flux through the current loop.

Since the magnetic force is perpendicular to the velocity of the charges, the magnetic force does no net work on the charges in the rod. This therefore means there is also no energy stored in the magnetic field.

Note that this analysis is done in the rest frame of the magnetic field as shown in the diagram (the rest of the wire loop is at rest here as well).

• so the horizontal component of force does a negative work and the vertical component of the force does a positive work (of equal magnitude ) and thus no net work on the charges . Right ?? (+1 btw) Dec 6, 2021 at 14:53
• @Ankit Yes you could look at it like that. And note that it is just no net work done by the magnetic force. Certainly there is work being done somewhere, since the charges do lose kinetic energy. Just don't go looking at the magnetic force. Dec 6, 2021 at 15:19
• @BioPhysicist Your diagram is misleading. For all proposes this question is about a practical experiment. The electrons drift velocity is a few mm per hour so, Vcurrent should be neglected in your diagram, Vrod>>Vcurrent. Therefore, the Lorentz force (Fmag) vector is opposite to Vrod and there is no Vtotal vectror. There is only Fmag opposite to Vrod. Therefore the magnetic force (Lorentz force) in this experiment is NOT VERTICAL to the velocity of the rod but antiparallel aligned. Therefore the magnetic field will slow down the rod motion thus the magnetic field is doing work in this case. Dec 8, 2021 at 8:10
• @Markoul11 You do realize that if the only velocity we are considering is the velocity of the rod then the magnetic force from this velocity component would be upwards along the rod, right? The Lorentz force you describe being opposite to the velocity of the rod arises precisely because of the current flowing through the rod; it isn't negligible. Please review the Right hand rule and cross products. Dec 8, 2021 at 8:19
• @BioPhysicist You have correctly applied the RHR with the exception that Vcurrent vector MUST be neglected. The drift velocity of the electron charges Vcurrent, in a conductor is in the scale of μm/s! Please read the WP article for reference en.wikipedia.org/wiki/Drift_velocity And also review the WP explanation, of magnetic fields not doing work on isolated monopole field charges (isolated electron) but how they CAN do work on electric dipole field charges thus the charges in the induced current in an electric circuit. en.wikipedia.org/wiki/Magnetic_field Dec 8, 2021 at 8:58

This debate is a bit of "much ado about nothing". The basis is as follows. If you look at the Lorentz force we see $$\vec F = q \vec E + q \vec v \times \vec B$$ which leads immediately to $$P = \vec F \cdot \vec v = q\vec E \cdot \vec v + (q \vec v \times \vec B) \cdot \vec v = q \vec E \cdot \vec v$$ So from the Lorentz force it seems that the B field does no work.

Similarly, if we look at Poynting's theorem we see: $$\frac{\partial u}{\partial t} + \nabla \cdot \vec S + \vec J \cdot \vec E = 0$$ where $$u$$ is the energy density of the EM field and $$\vec S$$ is the flux of EM field energy from one location to another. The only term that involves matter is $$\vec J \cdot \vec E$$, which is energy that is leaving the EM field and going into the matter, hence it is the work done on matter. So from Poynting's theorem it also seems that the B field does no work.

However, the reason that this is "much ado about nothing" is that the E and B fields are not independent of one another. While you can write the work as $$\vec E \cdot \vec J$$ we also have Maxwell's equations which describe relationships between $$\vec E$$ and $$\vec J$$ and $$\vec B$$, especially via Faraday's law and Ampere's law. $$\vec E \cdot \vec J$$ is a function of $$\vec B$$.

Suppose, in particular, that we wish to investigate further the specific claim:

the loss in kinetic energy appears in the form of joule heating only

The problem is that the term $$\vec E \cdot \vec J$$ includes all of the work done on the conductor. That is, it includes both the joule heating and also any other form of work done on matter. For instance, in a battery it will include joule heating and also chemical work increasing the chemical potential. Similarly, in a conductor it will include joule heating and also mechanical work.

Let us make the assumption that in a conductor there is no other form of energy transfer besides joule heating and mechanical work. With that assumption, if the conductor is at rest then the mechanical work is 0 and $$\vec E \cdot \vec J$$ is indeed the Joule heating. However, in this case the conductor is not at rest. To determine the Joule heating we need to transform into the reference frame where the conductor is at rest.

Assuming that $$v \ll c$$ we have the following transformation equations where the primes represent quantities in the reference frame where the conductor is at rest:

$$\vec E' = \vec E + v \times \vec B$$ $$\vec J' = \vec J - \rho \vec v$$

Substituting those in we have: $$\vec E \cdot \vec J = (\vec E' - \vec v \times \vec B) \cdot (\vec J' + \rho \vec v)$$ $$=\vec E' \cdot \vec J' + \vec E' \cdot \rho \vec v - (\vec v \times \vec B) \cdot \vec J' - (\vec v \times \vec B) \cdot \rho \vec v$$ $$=\vec E' \cdot \vec J' + \rho \vec v \cdot (\vec E+ \vec v \times \vec B) - (\vec J-\rho \vec v) \cdot (\vec v \times \vec B)$$ $$=\vec E' \cdot \vec J' + \rho \vec v \cdot \vec E - \vec J \cdot (\vec v \times \vec B)$$

So finally we end with $$\vec E \cdot \vec J=\vec E' \cdot \vec J' + \vec v \cdot (\rho \vec E + \vec J \times \vec B)$$ which says that the total work $$\vec E \cdot \vec J$$ on a moving conductor is equal to the Joule heating $$\vec E' \cdot \vec J'$$ plus the mechanical work $$\vec v \cdot (\rho \vec E + \vec J \times \vec B)$$ where the mechanical work has the expected form which includes mechanical work done by the B field.

So while it is correct to say that $$\vec E \cdot \vec J$$ means that all of the work is due to the E field, it is also correct to say that $$\vec E \cdot \vec J=\vec E' \cdot \vec J' + \vec v \cdot (\rho \vec E + \vec J \times \vec B)$$ and therefore the B field does some mechanical work as expected. What is definitely incorrect to say is that the work is "in the form of joule heating only" because $$\vec E \cdot \vec J$$ is the total work and not only joule heating.

• The two parts of your answer are taken from different reference frames, right? The latter part is looking at the rest frame of the rod. So isn't it a little more complicated than just saying in that frame the magnetic field does work? The total work in that frame depends on the magnetic field, but at the end of the day where the energy comes from can't depend on the frame, right? Dependence on $B$ isn't the same thing as saying the magnetic field does work Dec 8, 2021 at 12:36
• @BioPhysicist only the primed quantities, E' and J', are in the conductor's reference frame. All of the unprimed quantities are in the original frame. The only purpose of the primed quantities is to write the joule heating compactly. It could be written in terms of the unprimed quantities, but then it wouldn't be clear what it represents.
– Dale
Dec 8, 2021 at 13:22
• Right, I understand that, but how can where the energy comes from be frame dependent? Saying the magnetic field does no work in one frame but does do work in another frame seems odd to me. Certainly you have terms that depend on $B$ in the latter case, but how do we make the jump to then say the magnetic field is actually doing work? Dec 8, 2021 at 13:33
• @BioPhysicist where energy comes from is indeed frame dependent even in normal Newtonian mechanics, but that is besides the point here. The other frame is only used to pull out the joule heating from the mechanical work. The mechanical work is expressed in the original frame and the joule heating is frame independent (for v << c).
– Dale
Dec 8, 2021 at 13:51
• How can the actual sources/sinks of energy be frame dependent? e.g. if you had some sort of meter that showed how much energy left/entered something wouldn't they need to read the same thing at the end of the process? Maybe I'm using imprecise terminology here. Dec 8, 2021 at 14:15

Consider what would happen to your rod without the application of an external force on the moving rod.
Due to the interaction of the current passing through the rod and the magnetic field there is a force on the rod to the left.
That force does negative work on the rod with the result that the kinetic energy of the rod would decrease.
There is no change in the strength of the external magnetic field acting on the rod and so there is no change to the energy stored in that external magnetic field.

To keep the rod moving at constant speed an external force must be applied to the rod to the right and then that external force will do positive work on the rod. An external force of just the right magnitude will result in no net work being done on the rod and so the rod will move at constant speed to the right.

The current in the rod is produced because of the interaction between the charges in the rod (all being made to move to the right) and the external magnetic field which has the effect of applying forces on the charges in the rod resulting in the mobile charge carriers moving along the rod (and around the rest of the complete circuit).
The forces on the charges in the rod are at right angles to the direction of motion of the rod and the direction of the external magnetic field.
This is the force on the charges in the rod (the magnetic force) which is at right angle to the direction of motion of the rod so in terms of that motion, the magnetic force does no work.

If the rod had no mobile charge carriers, ie was an insulator, then the steady state condition would be that the rod would have no force acting on it as there is no current passing though it, and the rod would move with a constant speed.

• and why doesn't the Magnetic field strength change ? Dec 6, 2021 at 10:11
• The magnetic field does not change because it is produced outside the system being considered, ie from a horseshoe permanent magnet. Dec 6, 2021 at 10:35
• So why the $-1$ downgrade? Dec 8, 2021 at 18:22

Οf course the magnetic field is doing work in your example and applying the magnetic Lorentz force on your moving target charges (force applied only to the segments in your drawing which are perpendicular to the magnetic field vectors).

(Notice, because the moving part of your conductive loop frame is elevated in relation from the rest of the frame as shown in your figure, your total frame is asymmetric and expect also a net torque to be applied to your frame).

This is a common misconception and wrong generalization about magnetic fields never doing any work, thus applying a force aligned to the motion direction and is true only for isolated charges (i.e. an electron inside a magnetic field, monopole electric field).

Because the magnetic force is always perpendicular to the motion of isolated charges, a static magnetic field cannot do work on an isolated charge. However, it can do work indirectly, via the electric potential dipole field generated in a conductor by a changing magnetic field. Magnetic field WP

In your case shown in your provided circuit diagram, the rod moving to the right and cutting the magnetic flux lines will change the magnetic flux in the area of the loop, thus a changing magnetic field inside the loop that will induce an electric field. The resulting vector field for the moving rod inside the magnetic field is shown below (right hand rule applied with the direction of the induced current in the conductor $$I$$ assumed to point upwards): As shown above, the changing magnetic field is generated by initial mechanical force on the rod which starts to move to the right and cuts the magnetic flux of the static B field and increases the conductor (circuit) loop area. Thus a changing magnetic field inside the conductive loop, field B, that induces a current $$I$$, in the circuit thus the Lorentz force $$F_\text{mag}$$ is applied on the moving rod segment antiparallel to its velocity $$V_\text{rod}$$. Therefore, $$F_\text{mag}$$ is not vertical to $$V_\text{rod}$$ but antiparallel and the magnetic field via Faraday's induction is actually doing work here in this experiment. $$F_\text{mag}$$ is slowing down the velocity of the rod $$V_\text{rod}$$ to the right.

The electrons drift velocity inside the conductor rod is a few mm per hour so, $$V_\text{current}$$ therefore inside a conductor is practically zero and must be neglected $$V_\text{rod}\gg V_\text{current}$$. Drift velocity

So yes indirectly magnetic fields can do work on electric dipole field charges like that inside an electric current circuit but not on electric monopole field charges like that of an isolated electron.