0
$\begingroup$
  • If a magnet is moving through a solenoid, why is there no electromotive force being induced while the magnet is at the centre of the solenoid? (Is it because emf is induced in both directions?)

  • Can a magnet induce an electromotive force in a single strand of wire? (If so, how does Lenz's Law apply? Is the emf negligible because it's only a single strand of wire?)

Thanks in advance!

$\endgroup$
  • $\begingroup$ Suppose the solenoid is a single loop of wire. There would be no EMF if the rate of change of magnetic flux through the loop is zero. $\endgroup$ – Keith McClary Oct 8 '15 at 3:52
1
$\begingroup$

Indeed, the emf in one half of the solenoid will cancel that in the other half. This is most easily visualized if you split the solenoid in half, now you have one solenoid that is being entered, and one that is being exited so their EMF will be equal and opposite.

Or if you're looking at the center only, all of the flux lines in the middle are parallel to the motion, so there will be no change in flux.

Magnetic field lines

So if you look at a plot of number a field lines encircled by a ring at a given position (which is proportional to the flux) you'll get something that looks like:

flux vs. position

As you can see there is a maximum amount of flux right in the middle, and on either side it is symmetrically decreasing and increasing. The EMF produced by a coil is proportional to the change in flux with time. If the magnet/coil is moving at a constant velocity, that means the EMF will be proportional to the change in flux with position.

derivative of flux with respect to position

Here is the derivative of the flux with respect to position. This would be proportional to the EMF that an infinitely thin coil would experience as a magnet passed through it.

If the coil had some width to it, we would want the average change in flux over the whole coil. To take the average one can integrate and divide by the length, but since we'd be integrating the derivative of the flux, we can just take the difference between the endpoints.

For relatively thin coils this ends up looking almost identical to the infinitely thin coil, but as the coil length increases the graph does begin to look different. However, the middle must always pass through zero because the graph of the flux is symmetric (because the magnetic field is symmetric, so however much the EMF is positive on one side it will exactly cancel the negative on the other side).

thin coil

medium coil

long coil

$$ $$

All charged particles feel a force when passing through a magnetic field. When the wire doesn't form a loop it's necessary to calculate the EMF via the Maxwell–Faraday equation instead of Lenz's Law.

$\endgroup$
  • $\begingroup$ So if you drop a magnet through a solenoid, would it be the second scenario? $\endgroup$ – Cr4zyM4tt Oct 9 '15 at 15:34
  • $\begingroup$ @Cr4zyM4tt if you drop a magnet through a solenoid, it would create an EMF as it approached the solenoid, then the emf would drop to zero as it passed through the middle and the emf would go the opposite direction as the magnet continued on it's path. $\endgroup$ – Rick Oct 9 '15 at 15:37
  • $\begingroup$ Will the emf drop to zero be because of the first scenario or the second scenario given in your answer? $\endgroup$ – Cr4zyM4tt Oct 12 '15 at 18:50
  • $\begingroup$ @Cr4zyM4tt yes. They aren't separate phenomenon, just different ways of looking at the same thing. The fact the all the flux lines are parallel in the middle is justification/ and explanation for why that location is where the maximum flux is. Thus coils that the magnet are approaching are gaining flux and coils that the magnet is receding from are loosing flux. $\endgroup$ – Rick Oct 13 '15 at 18:06
  • $\begingroup$ But if we look at the second way of looking at it, where the flux is parallel to the motion of the magnet hence no emf is produced, doesn't it contradict the first way? The first way states emf is induced in both directions and hence cancelled out. $\endgroup$ – Cr4zyM4tt Oct 17 '15 at 10:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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