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Name: John
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Question:
If I take a coil and place a piece of steel in the center and lay the coil on its side so that the steel core is facing me, and I then take a bar magnet such that the top of the magnet is the north pole and then move the magnet from left/west to right/east, Which way will the induced magnetic field encircle the steel bar? Will it be in a clockwise direction or counter-clockwise direction?



Replies:
John -

This text sketch depicts your scenario if I understand you correctly.

|---text-sketch---use mono-spaced font------:
|     N                                     :
|     |  a --> b --> c                      :
|     S                                     :
|magnet       ||                            :
|          ___||_____coil__                 :
|         /___||_________ /    leads        :
|        //   ||        /============       :
|       //    ||       //                   :
|  ... // ... || .... //... reference_line  :
|     //      ||     //                     :
|    //_______||____//                      :
|   /_______________/                       :
|             ||                            :
|          steel_bar                        :
|                                           :
|-------------------------------------------:
Bar-magnet, orientation vertical, goes left to right through positions a,b,c.

Going a full sweep from left to right, the magnet passes through three distinct positions a, b, and c. At b, the magnet is on the axis of the steel bar. At a, it is to the left of that axis; at c, it is to the right.

I would not say the magnetic lines ever 'encircle' the bar. Rather some of the lines leaving the magnet south pole tend to focus into the nearby top end of the steel bar and go some distance down its length, because soft steel is a magnetic "short-cut", much easier than going a similar distance through empty space. Thus the steel core helps more field-lines to reach down through the inside of the coil and return around the outside of the coil, which constitutes "flux through a loop", which is good for electromagnetically induced voltage. A voltage is induced only when the sum of this loop-enclosed flux changes with time, i.e., increases or decreases. The instantaneous voltage is proportional to this rate of change.

Thinking it through:
As the magnet starts at [a], it is not very near the steel bar, so the amount of flux reaching the bar and penetrating the loop is positive but not at its highest value. As the magnet reaches [b], it is as close as it is going to get, so the captured flux is at its highest value, higher than at [a]. As the magnet starts at [c], it is again not very near the steel bar, so the amount of flux reaching the bar and penetrating the loop is once more not its highest; positive but lower than at [b].

From [a] to [b], captured flux is increasing and the induced voltage will be positive (presuming you have defined your + & - leads that way.) At [b], the captured flux gradually "rounds the top", i.e., stops increasing and starts decreasing, but at that instant its linear rate of change ( say, in flux-lines per second ) is zero. So the induced voltage is zero at that instant. From [b] to [c], the captured flux is decreasing, so the induced voltage will be negative.

It is a typical "bipolar-spike" waveform: gradually 0 to +, suddenly + to 0 to -, then gradually - to zero again. (There must be generic names for this function. It is sometimes called a "discriminator function", because the frequency-discriminator circuit in FM radios produces a similar waveform as input frequency is swept from low to high.)

In electromagnetic induction, the copper is what needs to be encircled by flux, not the steel. In fact, because the field lines are attracted to the steel, it can be difficult to get substantial encircling of it.

Cordially,

Jim Swenson



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