Does a collapsing and re-establishing magnetic field impart a force on a stationary charged particle? Does the charge particle get repelled and or attracted? Does it move or spin?


3 Answers 3


Yes, it will create a force. The force is directed solenoidally around the change in a magnetic field.

To see this, look at Maxwell's equation $\nabla \times \mathbf{E} = -\partial_t \mathbf{B}$. This is analogous to the equation from magnetostatics: $\nabla \times \mathbf{B} = \mu_0 \mathbf{J}$. Thus a changing magnetic field sources an electric field the same way a current sources a magnetic field.

So for a concrete example, suppose you have a solenoid and you turn on a current so the magnetic field strength increases at a constant rate. Then $\partial_t \mathbf{B}$ is constant in the solenoid, and the electric field you get will be the same as the magnetic field you would get from a constant $\mathbf{J}$ in the region of the solenoid. That is, the electric field you get will look like the magnetic field from a wire. So outside the solenoid, you will get an electric field wrapping around the axis of the solenoid. This electric field will cause a force on the charge. Note, the force is directed in a circle, but it will not cause circular motion. Instead, the charge will eventually spiral away from the solenoid.

Notice that in some sense you would say the force is caused directly by the electric field, and it is only indirectly caused by the magnetic field. However, I am still going to say that a "yes" answer is more appropriate in this case, and anyway I think this indirect effect is what you were trying to get at in the first place.

  • $\begingroup$ Thanks, so it is the electric field that imparts the force on the charged particle not the magnetic field. Also once the charge is in motion does the magnetic field impart a force. And do we have to account for both fields at this point? $\endgroup$
    – StarDrop9
    Oct 8, 2015 at 14:41
  • $\begingroup$ Yes, once the charge starts moving it will experience a $\mathbf{v} \times \mathbf{B}$ force. $\endgroup$ Oct 8, 2015 at 18:50
  • $\begingroup$ As fan of en.wikipedia.org/wiki/Jefimenko%27s_equations , I would not say B-dot sources E. Rather, the rho, J, and J-dot required to create B-dot also, by law, create an E with curl E proportional to B-dot. $\endgroup$
    – JEB
    Jul 16, 2021 at 15:37
  • $\begingroup$ So this assumes that the charge is situated outside the changing B-field? I agree with your example where the charge is outside the changing B-field but e.g. if the charge were placed in a unform but time-dependent B-field, are you saying that the charge would accelerate? $\endgroup$
    – ProfRob
    Mar 10, 2022 at 17:54

In the absence of any charge motion and current density then a changing magnetic field is accompanied by a changing electric field. From Ampere's law we can say that

$$ \vec{E} = \frac{1}{\epsilon \mu} \int \nabla \times \vec{B}(t)\ dt$$

This electric field will exert a force $q\vec{E}$ on the particle.

The above equation though is not fully determined because there is a time-independent constant of integration. The question cannot therefore be answered without specifying additional boundary conditions on the problem.

  • $\begingroup$ The above formula, even if differentiated wrt time, doesn't look right to me. In any case Ampère's law is about how a current induces a magnetic field, which is unrelated to the question. The matter here is Farady's law of induction. $\endgroup$
    – bhbr
    Mar 10, 2022 at 16:03
  • $\begingroup$ @bhbr How are you proposing to rewrite the Ampere-Maxwell law if you think the above equation is wrong? On the contrary, Faraday's law only tells you what the curl of the electric field is. Knowing the curl of the E-field does not enable you to answer the question except in certain symmetrical cases - e.g. that $dB/dt$ is in a uniform, unchanging direction. so that E can be deduced from its curl. The equation above gives the E-field directly. $\endgroup$
    – ProfRob
    Mar 10, 2022 at 17:38
  • $\begingroup$ Other than that there can be a time-independent E-field present... $\endgroup$
    – ProfRob
    Mar 10, 2022 at 18:21
  • $\begingroup$ At face value, your equation would seem to imply that so long as the magnetic field is spatially uniform, then $\nabla \times \vec{B} = 0$ and there is no electric field. $\endgroup$ Mar 10, 2022 at 18:27
  • $\begingroup$ Michael, I don't think that this equation under the circumstances $\nabla × \vec{B}=0$ would predict $\vec{E} = 0$, Remember we are partially differentiating, meaning any function independant of time could be present, aka a time independant $E_{0}(r)$ field. Then obtained via boundary conditions surrounding Gauss law $\endgroup$ Mar 10, 2022 at 19:06

that curlE=dB/dt basically comes from faraday's flux law. this flux law doesn't work in all case unlike lorentz force. When a loop is moving, the flux law and lorentz force argument, both will lead to the same result. But in the case when a loop is static, it feels like lorentz force law doesn't work here but flux rule certainly gives the explicit answer. this situation makes the farady's flux rule in case when the loop is stationary and field is changing a fundamental law. just imagine a charged particle (with some initial velocity) moving a time varying magnetic field. Now if there's an induced electric field in the region, there must be the loss of kinetic energy of the particle.even if i go into relativistic way, no new equations show the presence of induced electric field. so the point worth musing here is that faraday's law is something invariant and fundamental but why it is so.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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