Consider a point charge $q$ situated at the origin, and a uniform magnetic field, covering all of space, pointing in the $z$ direction $\mathbf{B}=B_0\hat{\mathbf{k}}$. What happens when you turn off the magnetic field? Which way will the charge go? There is obviously a changing magnetic field, and this will in turn produce an electric field, thereby exerting a force on $q$. But in which direction?

This is a textbook problem (Griffiths) and the solutions say that there are insufficient boundary conditions. OK, but what kind of boundary conditions are missing?


This seems to be an unphysical problem in that whilst you can hypothesise a uniform magnetic field everywhere, there is no way it can be changed everywhere uniformly.

However in this case I don't think that is the issue. Faraday's law is that $$\nabla \times \vec{E} = - \frac{\partial \vec{B}}{\partial t},$$ so knowing the rate of change of the magnetic field (as a vector) only gives you the curl of the electric field. To get the Lorentz force you need to know the electric field itself and this can only be calculated from the curl of the field with additional boundary conditions, for instance the value of the E-field at a particular position or some other symmetry arguments. Without this, you could add any stationary, curl-free E-field to your solution and it could be consistent with the changing magnetic field.

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  • $\begingroup$ But we ought to be able to find the direction the charge would move, right? $\endgroup$ – Michael Angelo Mar 31 '16 at 12:25
  • $\begingroup$ @fawningflagellum OK, so if I tell you that $\nabla \times \vec{E} = -t B_0 \exp(-t)\ \hat{k}$ (for $t \geq 0$), which way does the charge move? $\endgroup$ – Rob Jeffries Mar 31 '16 at 12:30
  • $\begingroup$ $\vec{E} = 1/2 \frac{d\vec{B_0}}{ dt} (y \vec{i} − x \vec{j})$ would do the job, in which case the force would be zero, but we could add any constant vector to this, and make the force anything we like... Is troubles me... Edit: Oh is the point that you could, in the real world situation, take zero at infinity as boundary coundition, but that this is not possible here because it's an artificial problem? $\endgroup$ – Michael Angelo Mar 31 '16 at 12:34
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    $\begingroup$ @fawningflagellum This is only one possibility. $\vec{E}$ depends on a boundary condition You cannot uniquely invert a curl (essentially perform an integration) without a boundary condition. $\endgroup$ – Rob Jeffries Mar 31 '16 at 12:42
  • $\begingroup$ To be specific you could add any curl-free E-field to it. $\endgroup$ – Rob Jeffries Mar 31 '16 at 12:51

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