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Questions about electromagnetic induction

  1. May 17, 2009 #1
    Why is there an iron core in a transformer? Is it necessary?

    In a transformer, how does the changing magnetic field on the primary induce a current on the secondary? Can we think of the changing magnetic field causing a changing force on the secondary and causing it to move back and forth, cutting field lines? Or is it more like constantly switching the polarity of a magnet? Where does Lenz's law come into it?

    Say we have a coil and an ammeter in a complete circuit (ammeter will read zero because there is no power supply). When we move a magnet through the coil, why is a current registered? There is no current in the circuit so the coil does not have its own field and the magnet is not interacting with another field. Is it because the metal of the coil physically cuts the magnetic field lines?

    http://img269.imageshack.us/img269/7002/induction.jpg [Broken]
    That's a circuit with a coil and an ammeter. The magnet is moved to the right and through the coil. Lenz's law tells us that the coil will be north on the left and south on the right and these will switch around when the magnet gets past the middle of the coil. So in the first half of the motion, when the coil is north on the left and south on the right, what is the direction of the current? I don't just want the answer; I want to know how to compute it. I was thinking of Fleming's left hand rule but then what is the direction of the force? One would think that it's the force opposing the magnet, but then wouldn't it be parallel to the magnetic field?

    Thanks in advance.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 17, 2009 #2
    You have asked a lot of questions and the answers would require at least a whole book chapter.

    I will tackle the case of the basic coil - meter circuit.

    The coil does not need a pre-existing field for the magnet to interact with. The magnet interacts with the free electrons in the metal of the coil:

    When you move the magnet through the coil some percentage of the magnetic field lines, which extend into space around the magnet and move with it, are "cutting" the turns of the coil. "Cutting is a term coined by Faraday to mean crossing at right angles. The field lines penetrate the material of the coil and interact with the free electrons, forcing them to move all in the same direction according to what pole of the magnet is inserted and what direction it moves in. So long as the magnet is in motion there is, indeed, current in the circuit and the meter needle will be deflected.

    The motion of the electrons is at right angles to the motion of the magnet. If we say the magnet moves along an x-axis then the electrons move at right angles to the magnet motion; they move around the coil. If we reverse the motion of the magnet the electrons still move at right angles to that, but in the opposite direction around the coil.

    We could also hold the magnet still and move the coil and get the same effect. It is the relative motion of magnet and conductor that generates current.

    The reason metals are good conductors is that they have plenty of these free electrons which can be pushed by a magnetic field. A magnetic field has the amazing property that it can penetrate the material and selectively push these free electrons without, say, distorting the whole material. However, there is never any current unless either the magnet, or the coil, or both are moved by some outside force.
     
    Last edited by a moderator: May 4, 2017
  4. May 17, 2009 #3
    And I'll add to that with an answer about the iron core. No, a transformer doesn't need an iron core; it just works so much better with one. The core provides a high permeability path (think low resistance) for the magnetic field lines. So, if you have a high permeability core, the transformer will come very close to obeying Vs = (Ns/Np) Vp, where V is Voltage, N is number of turns, s is secondary, and p is primary. If you have a low permeability core, or even no core at all, the equation looks more like Vs = k(Ns/Np) Vp where k is a coupling coefficient that is less than one (possibly a lot less) and that is usually quite dependent on the load current.
     
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