0
$\begingroup$

If a magnet is introduced speedily in a coil there is voltage by EMI Faraday's law induced in the coil.

We try to reverse the roles of a coil and the magnet unconventionally but still cutting the magnetic lines.

If a copper wire is wound around a non-magnetic or non conducting long plastic rod from pole to pole (circuit contains an indicator bulb) and dropped fast into a stationary hollow cylinder magnet hole, would there be current in the windings?

EDIT1:

The doubt arises because magnetic field lines outside the magnet are not cut by the moving rod wound with coils, may generate reduced EMF.

enter image description here

$\endgroup$

1 Answer 1

0
$\begingroup$

Yes there would be an induced EMF due to the falling magnet.

The strength of the magnetic field of a permanent magnet is inversely related to the distance from the magnet so the change in flux and therefore the induced EMF is related to the change in distance from the magnet to the coil (its speed). Additionally, during the moments when the center of the magnet passes through the coil the direction of the magnetic field abruptly switches sign and the induced EMF spikes.

Also something notable is that the magnetic field created by the coil due to the the induced current will oppose the motion of the magnet and slow it down.

$\endgroup$
2
  • $\begingroup$ Here it is the rod that is falling into or rising out of the fixed hollow magnet. $\endgroup$
    – Narasimham
    Nov 6, 2023 at 21:02
  • $\begingroup$ @ Er Jio: Trying to understand the electromagnetic field action. Can the sketch be taken to represent a crude linear armature / inside a magnetic stator as a linear induction motor under certain reciprocation inducing conditions? Have toroidal rotors ever been designed and operated? $\endgroup$
    – Narasimham
    Nov 16, 2023 at 15:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.