Imagine you have an external magnetic Field $\vec{B}_{\text{ext}}$(for the sake of simplicity, it should be constant throughout the entire space), which is not generated by any current density (imagine that it's simply a solution to the setup of no current density, with apropriately choosen boundary conditions).

Now we place a current ($\vec{j}$) carrying loop somewhere, which generates a field $\vec{B}$. Initially the Field $\vec{B}$ should be antiparalell to $\vec{B}_{\text{ext}}$. For the sake of simplicity I assume that the loop has no resistance, so the current inside just flows freely.

What now happens is that the Lorentz-force will act on the charges in the loop, generating torque that will turn the loop until $\vec{B}$ and $\vec{B}_{\text{ext}}$ are paralell. Since the Lorentz-force doesn't change the energy of the charges, every bit of Energy that went into moving the loop stems from the current inside the loop: While the charges are accelerated perpendicular to the direction of the conductor (moving the loop), they are decellerated along the direction of the conductor (reducing the current).

In terms of energy conservation - so far so good, everything is fine.

My problem now is: The more both magnetic fields do align, their field energy (which is the squared sum of them, $(\vec{B}_{\text{ext}} + \vec{B})^2$ increases. Where does the energy come from, that leads to this increase of energy?


You are forgetting the field in the loop plane far from the loop, which initially points in the same direction as the external field $\vec{B}_{ext}$. This contributes positive field energy initially, but negative energy after the flip.

| cite | improve this answer | |
  • $\begingroup$ This is helpful. I'll try to calculate this, in case I fail: Do you know some resource where the calculation is done that shows the field energy actually doesn't change? $\endgroup$ – Quantumwhisp May 21 '19 at 4:03
  • $\begingroup$ To be more specific: Calculation that shows the field energy by the inside of the loop exactly cancels out the energy by the field exterior of the loop. $\endgroup$ – Quantumwhisp May 21 '19 at 4:17
  • $\begingroup$ I don't know any such source. Conservation of energy is an obvious desideratum for a consistent model of this system. However, formulating such a model in detail is not trivial, as constraint forces (necessary to keep the charges from jumping out of the conductive loop) nd the work they do need to be involved as well. One can probably assume these are just contact forces that act only when the charges are on the wire boundary or perhaps even simplify the wire to a 1D curve and use this to derive the equations of motion and then derive conservation of energy from these equations. $\endgroup$ – Ján Lalinský May 21 '19 at 5:13
  • $\begingroup$ I wouldn't have thought the calculation Needs to be that complicated. Can't I just calculate $\int \vec{B}_{\text{ext}}(\Theta) \vec{B}$ with $\vec{B}$ being an abitrary magnetic field generated by a spatially confined, sourcefree current density, and \vec{B}_{\text{ext}}(\Theta) being a constant magnetic field with strengh, pointing into the direction of $\hat{e}_{\theta}$, and then (without further specifying the current density) "somehow" Show that the result doesn't depend on $\Theta$? $\endgroup$ – Quantumwhisp May 21 '19 at 5:48
  • $\begingroup$ The complicated calculation is needed to derive the energy conservation theorem for the whole system, in particular the expression for field energy. An expression such as $\int \mathbf B\cdot\mathbf B_{ext},dV$ is a natural first guess but not obvious it is exactly accurate in this case. Exchange of energy between EM field and mechanical energy of the wire due to constraint forces is taking place and some energy may be stored in the field or charges, which is not accounted for by this expression. $\endgroup$ – Ján Lalinský May 21 '19 at 15:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.