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It's my understanding that :

  • Changing magnetic field over a coil induce current in the coil itself.

  • Current running through a coil makes it magnetic. (To put it simply)

Can these 2 rules affect each other to reach a point of equilibrium, and if so, is there a formula to get that precise point of frequency / attraction ratio ? Is there a specific name for that phenomenon ?

Ex : Would it be true to state that if i were to oscillate a magnet back & forth very quickly over a coil connected to a load, enventually the coil would emmit a magnetic field, thus attracting / repelling the magnet and slowing down it's oscillation ?

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    \$\begingroup\$ What is "frequency vs attraction" ? What frequency? How is it "vs attraction"? But it sounds like you are referring to the back-emf. \$\endgroup\$
    – Eugene Sh.
    Mar 21, 2023 at 19:35
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    \$\begingroup\$ Yes, a coil connected to a load will resist the changes in it's magnetic field, that is will resist the movement of the magnet. Because the current induced by the coil is performing a real work on the load, and this work has to be compensated by the work performed moving the magnet. \$\endgroup\$
    – Eugene Sh.
    Mar 21, 2023 at 19:38
  • \$\begingroup\$ @EugeneSh. You're right, i edited the question. What i actually meant by [frequency vs attraction] is the ratio by which the frequency of the magnet oscillation is affected by the magnetic field created by the coil. \$\endgroup\$ Mar 21, 2023 at 19:59
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    \$\begingroup\$ Frequency is mostly irrelevant. This phenomenon can be observed with one direction linear motion / DC. Have you seen these cool movies where the drop a magnet through a thick copper pipe? \$\endgroup\$
    – Eugene Sh.
    Mar 21, 2023 at 20:02
  • \$\begingroup\$ Oh, I see @vir gave this example already \$\endgroup\$
    – Eugene Sh.
    Mar 21, 2023 at 20:02

3 Answers 3

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Yes, there's a school physics demonstration you can do fairly easily: take a length (1m or so) of conductive but non-ferromagnetic pipe (copper or aluminum, say) and a strong cylindrical magnet that is slightly smaller in diameter than the inner diameter of the pipe so it can slide freely. Drop the magnet through the pipe and you'll see that it takes longer to fall through than an equivalent length of e.g. plastic pipe (or just in air) since the induced eddy voltage sets up an opposing magnetic field that slows the passage of the magnet.

See: https://www.youtube.com/shorts/o8RiIPi5OB8

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  • \$\begingroup\$ Thanks for the link ! In this case, the pipe is not connected a load, so my guess is that it can only act as some king of capacitor, otherwise, not electron flow, no magnetic field. Am i wrong ? \$\endgroup\$ Mar 21, 2023 at 20:09
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    \$\begingroup\$ The current flows in a ring around the circumference of the pipe in the area that the magnet is entering or leaving. \$\endgroup\$
    – vir
    Mar 21, 2023 at 20:10
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    \$\begingroup\$ In this example the pipe is both - the inductor and the load. You can think of it as a shorted inductor (shorted==high load). \$\endgroup\$
    – Eugene Sh.
    Mar 21, 2023 at 20:12
  • \$\begingroup\$ Awesome, that makes a lot of sense ! Tks \$\endgroup\$ Mar 21, 2023 at 20:26
  • \$\begingroup\$ There was a nice demo of this in a UK university I visited - you could drop a magnet in the top of a vertical copper tube in the stairwell, and it would take about as long to fall as for someone to walk down the stairs \$\endgroup\$
    – Chris H
    Mar 22, 2023 at 13:55
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Both principles that you describe are closely linked to the underlying fundamental principle of energy (or work).

Energy can be transformed in many ways, but it can never be destroyed in a closed system. In most energy transforming processes that we know there is some part of energy transformed into heat, which we usually call a "loss".

In the system you are describing which is changing magnetic field to current back to changing magnetic field, it's the heat losses that contradict your idea of a point of equilibrium.

But in a superconductor (without losses), your point of equilibrium can be demonstrated! (magnetic levitation over a superconductor).

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  • \$\begingroup\$ Not necessarily only heat. The hypothetical "load" can be some transducer other than a resistive element and something "useful". \$\endgroup\$
    – Eugene Sh.
    Mar 21, 2023 at 20:17
  • \$\begingroup\$ @EugeneSh. Do you have an example for what type of load or "useful thing" you mean that does not transform discussed type of energy into heat? \$\endgroup\$ Mar 21, 2023 at 20:23
  • \$\begingroup\$ Not entirely to heat. Sure. A motor, a LED, a buzzer, what not... \$\endgroup\$
    – Eugene Sh.
    Mar 21, 2023 at 20:25
  • \$\begingroup\$ @EugeneSh. Ok, now I got what you meant. \$\endgroup\$ Mar 21, 2023 at 20:28
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Changing magnetic field over a coil induce current in the coil itself.

Changing the magnetic field that's cut by a coil induces a voltage in the coil -- not a current. If you short the coil (i.e., connect its ends together) then the magnetically-induced voltage will generate a current.

Current running through a coil makes it magnetic. (To put it simply)

Correct.

You can demonstrate the moving field thing with a voltmeter and a magnet -- connect a voltmeter or oscilloscope to the coil, and move the magnet by the coil. A digital voltmeter is usually too slow, although a good old fashioned analog voltmeter works nicely.

You can demonstrate the current through a coil thing by wrapping a number (around 100) turns of insulated wire over a nail. Power that with a D-cell and you should be able to pick up small iron objects (like a nail that you didn't wrap with wire). If you're lucky, you'll be able to drop your thing when you open the circuit.

You can demonstrate how the two interact by taking a DC motor (brushed or brushless) and putting a flywheel on it. Give it a spin with the wires disconnected to anything, and note how long it takes the flywheel to spin down. Now do the same thing, only with the wires connected -- the flywheel will spin down much faster.

This is because the coils moving through the magnetic field induce a voltage; when the wires are shorted, this voltage induces a current, which makes a magnetic field, which opposes the rotation of the motor. That opposition to rotation makes the flywheel spin down.

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  • \$\begingroup\$ Re the last two paragraphs: Is this phenomenon the basis of "dynamic braking" on diesel-electric locomotives? \$\endgroup\$ Mar 22, 2023 at 15:57

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