# 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

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.