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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?

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

To see this, look at maxwell's equation $\nabla \times \mathbf{E} = -\partial_t \mathbf{B}$. This is analagous 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.

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  • $\begingroup$ 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 '15 at 14:41
  • $\begingroup$ Yes, once the charge starts moving it will experience a $\mathbf{v} \times \mathbf{B}$ force. $\endgroup$ – Brian Moths Oct 8 '15 at 18:50
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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.

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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.

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