# Can $F=qvB$ be applied to a solenoid and a parallel moving magnet?

If you take a scenario like this, Charge gathers at either end, and if you could somehow form a complete circuit without there being a canceling effect, there would be a current flowing. This makes sense to me, the charges are moving in a B field with constant velocity, so F=qvB the electrons are forced down the rod.

But then, my textbook states for working out the emf across the conductor can be modelled using Faradays law, this doesn't make sense to me because there is not changing magnitude of flux in the conductor.

The derivation they use is as follows:

Displacement, $$s=v\Delta t$$

Area of flux cut, $$A=lv\Delta t$$

So $$\Delta \Phi = BA = Blv\Delta t$$

Faraday's law gives $$\epsilon = \frac{\Delta \Phi}{\Delta t} = \frac{Blv\Delta t}{\Delta t} = Blv$$

In then the converse fashion, in the situation where there is a solenoid moving relative to a permanent magnet: Then here, Faradays law makes sense to me, as an emf is only induced when there is some change of flux over time, not constant relative motion, but then how can I relate this F=qvB Because the electrons are moving parallel to the magnetic field (roughly) so I assume there would be no significant motion of electrons.

My overall question is how are these two processes related? The first scenario makes sense to me physically, but not in relation to Faraday's law but for the second the converse is true. I understand the relation to Faraday's law but not how the magnetic field is moving electrons in the wire. I fear the second is to do with special relativity effects, and I will not understand the calculus but if that is the case, at least I know it is.

• Have you heard of the Biot-Savart law? It might help. – Gareth Meredith Apr 7 at 14:41

## 1 Answer

This question certainly made me think a lot about magnetic fields, the magnetic force $$\vec{F}=q\vec{v}x\vec{B}$$, coils moving in (uniform and non-uniform) magnetic fields and relativity!

I think relativity is not so important here (maybe "a bit" in the second example) because it only tells us the $$\vec{B}$$-field is actually an electrical field.

In your first example, if we transform the rod in a circular form (I think that this is what you mean by "completing the circuit"), there will certainly not flow a current. All the electrons will accumulate on the bottom of the ring and the positive charge on the top. This happens because of the magnetic force acting on the electrons, as you understand. I can understand why you think how this can't be described by referring to Faraday's law because there is no change in the flux. But in the case of the rod (not the ring, which has a different geometry) there is no flux going through a closed circuit and Faraday's law can be used as your book stated.

In the second example, the coil is not moving through a uniform magnetic field and so the electrons are not moving parallel to the field lines (and "no significant motion of electrons", as you write is not true). The electrons on top of the coil experience a magnetic force opposite to the electrons on the bottom and two opposite current will flow towards the two ends of the coil which causes the emf. So both Faraday's law and the magnetic force exerted on the electrons can be used to see what happens. We can do this because relativity (still, there it is) says that moving the magnets (which is done in the two pictures) is equivalent to moving the coils (when you move the magnets then the changing magnetic force in the coils, considered at rest, will produce an electric field, according to Maxwell's equations.

A coil moving in a uniform magnetic field will develop opposite charges on the top and bottom, and if it moves parallel to the field, nothing will happen at all.

I hope it made things a bit clearer!