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I'm trying to understand the reasons why an electric current should occur in the following circumstances:

Given a magnetic field in space, and a neutral (not charged) body moving at some speed, in a direction so the magnetic force is not zero upon its charged particles:

Will the electric current be caused by:

1.Electromagnetic induction (since the flux of the magnetic field via the body is changing).

2.The magnetic force upon the particles, will cause them to move from each other, thus charging the body which will create an electric field, and a current.

And what about a constant magnetic field in space?

As the body moves along a constant field, the magnetic flux will not change.

But the magnetic field will cause particles from different signs to move away from each other, so an electric field will appear, and a current also.

But this is in contradiction to Faraday law since $\epsilon = -\frac{\partial\Phi}{\partial t} = 0$ and therefore no current should be created.

What is my misunderstanding?

Thank you.

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1 Answer 1

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You can view the force as being due to the Lorentz force $q{\bf v}\times {\bf B}$ or you can move to a frame of reference moving with the body. In that frame the Lorentz transformation of the magnetic field will give rise to a uniform electric field.

Faraday's law involving flux applies to closed loops with $\Phi$ the flux through the loop. A changing flux gives rise to a non-zero ${\rm curl}\,{\bf E}$. In the constant field case, seen from the moving frame, the flux is not changing, so the moving body experiences a curl-free electric field -- i.e one given by the gradent of a scalar potential. The constant ${\bf E}$ field is indeed the gradient of a potential.

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  • $\begingroup$ Yes, but what I didn't understand is whether when calculating the electric current through a loop, one should consider both the Faraday law and both the effect of magnetic field upon charges in the loop (that will cause + and - charges to go to different directions and then cause a potential etc...). Does the Faraday law "contain" the second effect I have mentioned? so one should consider the Faraday law only? Thank you. $\endgroup$
    – Taru
    Commented Dec 31, 2016 at 8:56

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