If the solenoid covers the whole length of a shake flashlight, will it light up?

[Spock2230]

In a Faraday flashlight, a magnet is shaken to and fro in a solenoid to induce an EMF and thus current. Something like this:
http://express.howstuffworks.com/gif/autopsy-flashlight-tube.jpg

If the solenoid were to cover the entire length of the interior, and thus the magnet is only moving within the solenoid, will there still be an effective magnetic flux linkage to induce a current? Why or why not?

I understand that Lenz law states that a current is produced because the solenoid opposes the motion of the magnet, by setting up an like pole to repel it away. So if the magnet is contained entirely within a solenoid, what will happen? Is it still able to generate an EMF?

Thanks for your help. :)

 

[Q-reeus]

If the solenoid were to cover the entire length of the interior, and thus the magnet is only moving within the solenoid, will there still be an effective magnetic flux linkage to induce a current? Why or why not?

Yes there will still be an induced emf and thus current - owing to the finite length of that solenoid a differential flux linkage always exists for two-an-fro magnet motion. But will lessen as the solenoid grows ever longer. [That assumes solenoid length extends equally beyond the motion range of magnet. If magnet motion range continues to match solenoid spatial extent, the net emf will be roughly unchanged in maximum magnitude (bounce at ends unchanged), but is overall less efficient since the period will increase hence less voltage pulses per second.]

I understand that Lenz law states that a current is produced because the solenoid opposes the motion of the magnet, by setting up an like pole to repel it away. So if the magnet is contained entirely within a solenoid, what will happen? Is it still able to generate an EMF?

Answer is the same since the question is the same!

 

[Spock2230]

Hi Q-reeus, thanks for your reply.

I'm just a high school student, so I'm not very sure what a 'differential flux linkage' is. Could you please tell me more?

You see, I was discussing with my physics teacher about this, and he's contention is that, since the magnet is entirely contained in the solenoid, the solenoid cannot set up a like pole to oppose the magnet's motion, and hence no magnetic work can be done to induce an EMF, so the LED won't light up.

I'd like to think of one long solenoid as multiple short solenoids, and so each part of the solenoid is still able to set up a repelling pole to oppose the motion for work to be done. I'm not really sure about this.

Thanks. :)

 

[Q-reeus]

Hi Q-reeus, thanks for your reply.

A pleasure.

I'm just a high school student, so I'm not very sure what a 'differential flux linkage' is. Could you please tell me more?

It simply means that, at each furtherest excursion of that sliding magnet, the number of magnet flux lines linking through all the coil turns of the solenoid, is less than when that magnet is at the center of solenoid. A little difficult to calculate exactly, but unless that solenoid is either infinitely long, or wraps around on itself to form a toroid, there will be such a differential.

You see, I was discussing with my physics teacher about this, and he's contention is that, since the magnet is entirely contained in the solenoid, the solenoid cannot set up a like pole to oppose the magnet's motion, and hence no magnetic work can be done to induce an EMF, so the LED won't light up.

He is using an energy argument in effect - based on an infinitely long solenoid and finite excursion magnet motion. And that's valid if, as I have also earlier done, we neglect possible phase delay effects and assume 'instantaneous' flux linkage - a very good approximation for shaking-magnet device.

I'd like to think of one long solenoid as multiple short solenoids, and so each part of the solenoid is still able to set up a repelling pole to oppose the motion for work to be done. I'm not really sure about this.

No that's not valid - intermediate 'poles' will exactly cancel leaving only the end 'poles' to consider.
[Edit: Just a clarification from your #1: "I understand that Lenz law states that a current is produced because the solenoid opposes the motion of the magnet, by setting up an like pole to repel it away."
That is not actually correct although there is such a correspondence in respect of currents opposing motion. It is the Maxwell-Faraday law that determines whether an emf in circuit is produced]

 

[Spock2230]

Hi Q-reeus, thank you very much for your time and the detailed explanation.

I understood the differential equations, but I don't get why d(B-field) cannot be 0. Could you take a look at this picture please?

This is what my physics teacher drew, to explain. The EMF induced by either sides are supposed to cancel out, thus d(B-field) = 0, and no EMF or current induced.
Well, it does seems quite reasonable, could you please explain what is wrong with it?

Thanks again for your patience and time.
You have been very friendly and helpful, leaving me with an excellent impression of this forum (I'm new here), and I'm encouraged to participate here more actively in the future.

 

[Q-reeus]

Hi Q-reeus, thank you very much for your time and the detailed explanation.

I understood the differential equations, but I don't get why d(B-field) cannot be 0.

You are probably referring to the ∅_B term in integral form expression. That is not B field but magnetic flux: the integrated amount of B field threading through a given area which the circuit encloses. No time-varying flux linkage = no emf. On the other hand, there is a finite ∇×E 'at a point' because it relates to rates-of-change not absolute value around a circuit.

Could you take a look at this picture please?

This is what my physics teacher drew, to explain. The EMF induced by either sides are supposed to cancel out, thus d(B-field) = 0, and no EMF or current induced.
Well, it does seems quite reasonable, could you please explain what is wrong with it?

[STRIKE]Just guessing for now. I'd say it involves having a closed circuit lying just outside of a region containing a magnetic field (say a very long magnetized rod). There is no appreciable threading flux through circuit, but if the flux in the rod or whatever is time-varying, it generates a circular E field that acts on the circuit. If the circuit has a sector shape, it becomes clear why cancellation applies to the inner and outer curving sides - E varies as 1/r thus voltage induced in each curved side are equal in strength but cancel direction wise as far as circulation around the loop. Straight radial sides intersect E field normal to their direction so no voltage acts in them. Thus net emf is zero, in keeping with Faraday's law of induction. Is that about right? [No - but leaving it now anyway!][/STRIKE]

OK - guessing over. Tried three different browsers and finally had luck with Google Chrome - others didn't show any image! So basically it gets back to scenario in #1. It's a fair/crude approximation depending on pov. It's true the net flux-linkage will be fairly small when both magnet ends are well inside of a long solenoid. However if that magnet is moving there will always be some net change in flux-linkage except at dead-center. You can see this by noting magnet approximates to a dipole and since the axial field of a dipole drops off as 1/r³, there is a difference in threading flux through solenoid ends - and thus through solenoid as a whole in general. Voltage in coil will be fairly 'spiky' as large changes in threading flux will occur only when magnet end(s) exits/re-enters the coil. In that sense he is right.

Thanks again for your patience and time.
You have been very friendly and helpful, leaving me with an excellent impression of this forum (I'm new here), and I'm encouraged to participate here more actively in the future.

Glad you feel that way Spock2230 and hope your stay is pleasant and beneficial!