This is a relatively long question however I think I hit on a fundamental issue regarding the conservation of energy during magnetic attraction and how it is taught in standard texts. I'd really appreciate it if you'd stick through for the ride. The situation I will describe and question is from Griffiths' famous Intro to Electrodynamics 4th Ed, pg 373-377.

Griffiths illustrates an example of magnetic attraction wherein a very large current loop ($I_b$) of radius $b$ is placed horizontally on a table with current flowing in the anticlockwise direction when viewed from above. A similarly oriented but much smaller current loop ($I_a$) of radius $a$ (with $a\ll b$) is placed on the floor beneath the table. Both current loops have fixed values of current at all times ($I_a$ and $I_b$ for each loop) which are sustained by two external power supplies $A_{ext}$ and $B_{ext}$ respectively. This setup clearly constitutes two parallel loops of current so that there is an attractive force between the loops which will cause the smaller loop $I_a$ to rise. The larger loop $I_b$ remains fixed in position though, because of the constraint forces of the table.

As the smaller loop $I_a$ rises by a height $dz$, Griffiths claims that its gravitational potential energy rises by a value $ dU$. It is clear that as the smaller current loop rises and changes position, a motional emf is induced in it which tends to reduce its current. This emf is overcome by the power supply $A_{ext}$ which does an amount of work exactly equal to $dU$ according to Griffiths. Similarly, there is an emf induced in the upper loop $I_b$ that tends to oppose its current which needs to be overcome by power supply $B_{ext}$, which in doing so, ends up doing an amount of work exactly equal to $dU$ once more. Thus, after the rise in height equal to $dz$, The total work done by the external power supplies is equal to $W_{ext}=2\cdot dU$ whilst the gravitational potential energy of the lower loop has only increased by $W_{grav}=dU$. Seemingly, we have violated energy conservation as an amount of energy equal to $dU$ has vanished ; external work on the system being equal to $2\cdot dU$ while the systems energy only increases by $dU$. Griffiths "solves" the apparent violation by stating that the extra energy $dU$ simply went into increasing the energy of the magnetic fields. He does this by stating that the energy stored in the fields of two current carrying wires is equal to $$U_B=\frac{1}{2}L_aI_a^2+\frac{1}{2}L_bI_b^2+MI_aI_b \tag{1}$$ As the lower loop rises, both loops maintain their current and shape so that the first two terms in eq (1) do not change. However, the mutual inductance of the two loops increases as the lower loops rises. Hence the field energy increases due to the final term in eq (1). This increase is shown to be exactly $dU_B=dU$ so that the systems energy increase is actually exactly equal to $2\cdot dU$ (the gravitational potential energy increasing by $dU$ and the magnetic field energy increasing by $dU$).

Something is fundamentally wrong here though. As the lower loop rises, its kinetic energy will rise as well (we've all seen two appropriately aligned magnets attract). So as the lower loop rises towards the table, its gravitational potential energy clearly rises, so too does its kinetic energy (the force of attraction getting stronger the closer the lower loop gets to the attracting upper loop), and according to Griffiths so too does the field energy. If the external power supplies provide enough energy to increase the field energy and the gravitational potential energy, then what provides the work needed for the increase in kinetic energy?? This situation is meant to be analogous to the situation of a large permanent magnet attracting a smaller one. But if two permanent magnets attract, we know that there is a conversion of energy from something (presumably magnetic field energy) into kinetic energy. But according to this analogy, the magnetic field energy actually increases as well! How can the kinetic energy increase as they attract whilst the field energy increases as well? The only possible way for the field energy to decrease according to equation (1) is if the mutual inductance decreases as the two magnets (or current loops) approach each other. But this is never the case as far as I am aware. Not in the situation illustrated by Griffiths and not in any other situation I've come across.

So what is going on here? What provides the necessary work for the increase in kinetic energy in Griffiths' setup and how does the field energy of two current loops increase as they attract each other but decrease if the loops are replaced by permanent magnets?


1 Answer 1


The system that Griffiths describes here makes heavy use of static phenomena (static magnetic fields , generated by a constant current). Strictly speaking they are only valid when the velocity of the loops is also zero (or rather infinitely small). So you can imagine the system to go through a quasistatic process, in which the velocity of the lower loop is kept at a very low value. If you would realize an experiment like this, then the calculations in question would be an accurate model.

If you allow the experiment to be conducted at finite speeds, then you have a whole can of worms from accelerated charges (which we will ignore here). But that aside, the increase in kinetic energy will come from the current in the small loop: When you move the small loop $a$ through the magnetic field of $b$, then charges in it will be subject to the lorenz-force. The component along the direction of the wire will be bigger if the velocity is greater. Any increase of the kinetic energy of the wire can therefore be accounted for by an decrease in the kinetic energy of the current. The calculation is the same, no matter wether the small loop is moving in the big field at finite speeds, or at (almost) zero speed.

In Detail: If you want to employ the formalism of a back electromotive force (and faradays law), keep in mind that this electromotive force is proportional to the flux change in the loop. If the loop moves faster (has kinetic energy), then the EMF will also be greater, because flux changes faster.

As for the permanent magnets: You are right, the energy has to come from somewhere. It can't come from the energy stored in the magnetic fields, so it has to be drained (again) from the system that generates the magnetic field in the first place.

I'll now employ a VERY SIMPLE model of magnetism (there are countless more difficult ones): The magnet being made up from electron states with a certain spin (and with certain energy). As the magnets move together, and the field energy increases, those spin states will be subjected to an increasing magnetic field aligned with their direction. And in that case, they will lose energy (which can be seen for example in the Zeeman effect).

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    $\begingroup$ @SalahTheGoat to the first comment: Yes, that is right (although I wouldn't employ the BAck EMF at all, in the end it's another part of the same lorentz force anyway). To the 2nd comment: Yes, it is true that the field energy increases as well. Magnets align so that their magnetic fields are parallel, you can calculate that in that case the field-energy increases. Yes, you identified the right term that decreases the energy of each individual dipole moment. $\endgroup$ Commented Sep 13, 2022 at 9:19
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    $\begingroup$ Faradays law is the statement that a change of magnetic flux induces an electromotive force. If you look into faradays law closely, it can either be that the magnetic field changes (which induces a rotation in the E field - this is maxwell's law) - This is not the case in our example. The other option for the flux to change is by the loop moving - In that case, the EMF is caused by the Lorentz force that the charges experience. So in our case, the Lorentz force is the solely underlying mechanism of the BACK EMF. $\endgroup$ Commented Sep 13, 2022 at 9:47
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    $\begingroup$ I have made an edit. If you want to you can stick to the back emf, in that case the back emf accounts for both the increase in potential energy and in kinetic energy. Faradays law is applicable, independent of the lab frame. But depending on the frame, the underlying reason will either be rotation induced in the E field, or lorentz force because of the moving loop. Both cases amount for the same decrease of energy of the current. $\endgroup$ Commented Sep 13, 2022 at 10:17
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    $\begingroup$ Awesome, pretty much all clear. It ends up coming down to the Quasistatic approximation. Griffiths assumes the ring rises with a negligible velocity so that there is no gain in kinetic energy AND so that he may use the formulas for the quasistatic fields. If we assume that the loop rises with a non-negligible speed, then the fields are no longer quasistatic. Using the true field equaitons, we would obtain an external work done equal to $2dU+dK$ where $dK$ is just enough to compensate for the increase in kinetic energy? $\endgroup$ Commented Sep 13, 2022 at 10:22
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    $\begingroup$ @BenVoigt you integrate about the whole space. Both current loops maintain their magnetic field ($\vec{B}_a$ and $\vec{B}_b$, and the integral over their squares stays the same. But you also have the cross term $2 \vec{B}_a \vec{B}_b$, and if the two fields align, then this integral will grow. $\endgroup$ Commented Sep 13, 2022 at 17:38

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