# Why does pushing a magnet inside a solenoid produce current?

If you push a bar magnet inside a solenoid, a current is produced. But why is that? I mean, the wire is being moved along the magnetic field, so taking the cross product:

$\vec{F} = I\vec{V}\times\vec{B} = I|\vec{V}||\vec{B}|\sin\theta\hat{n}$

Here, the angle between the velocity of the charge in the wires and the magnetic field is essentially zero. So there is no force on the charges in the wire and hence no current. But that does not correspond to reality. Why?

• Why ask for a solenoid rather than just across a wire ? – babou Aug 11 '13 at 8:55
• I don't know. That was the example given in my textbook. I just didn't understand it. – Gerard Aug 11 '13 at 9:04
• Because it produces a changing magnetic field. And Maxwell's equations (Faraday's law) teach us that a changing magnetic field results in an electric field, which gives rise to a current in a nearby wire. It's not the magnetic field providing the force, it's the electric field generated by the changing magnetic field. (note that the current vanishes when you stop moving the magnet wrt the solenoid, i.e. when the magnetic field stops changing) – Wouter Aug 11 '13 at 9:19
• this is related to change in magnetic flux – Dimensionless Aug 11 '13 at 9:28
• I don't want to know why this happens. My question is why does the normal approach not work here? – Gerard Aug 11 '13 at 10:10