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Describing Electromagnetic Induction of a Guitar Pickup

  1. Nov 2, 2012 #1
    First Post!

    So I was curious about how I could relate my new found knowledge of electromagnetic inductance on guitar pickups and I came across this on the internet:


    Which says, "All of these magnets are surrounded by a coil of very fine magnet wire containing thousands of wraps. When a steel string on the guitar vibrates, it perturbs the field of the magnet just beneath it. This changing magnetic field induces a current in the coil of wire. With a properly designed pickup, the perturbations in both the magnetic field and the current in the coil are synchronized with the vibration of the string. When the signal from the pickup is amplified, it produces the sound of the electric guitar."

    I find conflict with this. Here is how i would describe a pickups mechanism. Please fill in any holes in my understanding:

    So what I think is happening is that the magnetic field is pointing perpendicular to the plane that the strings are vibrating in. This creates a current in the guitar string. A guitar string then has it's own magnetic field created by it's current. This brings us to Faraday's law stating that the Change in the Magnetic Flux over time is proportional and opposite in direction to the induced EMF/current. When this string moves back and forth over thousands of coils of copper wire, it's small magnetic field generates a current big enough to be sent to an amplifier to amplification. So I think that the string's vibrating source of a changing magnetic field is what creates the change in the flux in the coils. There is no CHANGE in flux given by the magnetic field generated by the Neodymium magnet, or so I believe.
    A bit more in depth, I think that not only the fact that the string has it's own magnetic field determines the sound, but it also has to do with the mechanical wave propagating through it. When the string moves to the right over a upward magnetic field, the current in the wire propagates backward by the right hand rule. When going to the left, the string creates a current going forward through the string. This creates clockwise and counter-clockwise magnetic fields. So, I postulate that the induced current in the pickup is Alternating Current! The frequencies of the mechanical waves propagating through the string are reflected in the frequencies of the induced current, which allow it to retain its musical characteristics like frequency. Also, I've noticed that if you pluck a string harder on an electric guitar, it sounds louder on an amplifier. I think this is because the velocity of each infinitesimal part of the string is greater, which increases the current, thus increasing the strings magnetic field, thus making a stronger induced EMF.

    Something I'm blurry on:
    - how does the magnetic field cause a current in the string in the first place? I'm not sure if F = qv x B or more appropriately F = Idl x B.

    The law of Biot and Savart may apply if I knew if/why current was going through the strings.

    - people on the internet keep saying that since the string is a conductor, it distorts the magnetic field of the natural magnet thus creating a flux in the coil. How much truth is there to this?
  2. jcsd
  3. Nov 3, 2012 #2

    Andrew Mason

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    Welcome to PF!

    There are many kinds of electro-magnetic devices that convert sound into electrical signals. A dynamic microphone or a telephone uses a diaphram that vibrates and causes relative motion of a coil and magnet. The guitar pickup is a bit different. Instead of using relative motion of a magnet and coil to induce a current, it uses a vibrating steel string to change the effective permeability of the core of a magnet.

    As I understand it, the pickup consists of a coil of wire wound around a magnetized iron core. The string is made of iron (steel). When the string's position relative to the coil/core changes, the permeability changes so the magnetic flux enclosed by the coil changes slightly. The rate of change of the magnetic flux enclosed by the coil determines the induced emf in the coil (Faraday's law).

  4. Nov 3, 2012 #3
    So there is truth to what is being said on the link I include in my previous post? So, the vibration of the string changes the magnetic field of the natural magnet.
    This means there is no current running through the string creating it's own magnetic field?
    Or if there is, it is too minuscule to induce a voltage in the coil all by itself?
    Could someone refer me to where I could read more about this effect where a conductor moving through a magnetic field distorts the magnetic field?
  5. Nov 4, 2012 #4

    Andrew Mason

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    The distance between the iron atoms in the string and the iron atoms in the magnetized core of the pickup affects the overall permeability, which affects the strength of the magnetic field lines enclosed by the coil.

    There is no induced current and associated magnetic field in the string itself. It is the changes in proximity of the iron in the string to the magnetic pickup that causes the magnetic field enclosed by the pickup coil to change.

    It is not about a conductor moving through a magnetic field. It is about a string made with a with a substance having high permeability vibrating in close proximity to a magnet. If you used aluminum or copper strings, the pickup would not work.

  6. Nov 4, 2012 #5

    Andrew Mason

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    Again, it is not about a conductor moving through a magnetic field. It is about a material with high magnetic permeability vibrating close to a magnet (one that has a coil wrapped around it). See "www.physics.princeton.edu/~mcdonald/examples/guitar.pdf" [Broken]

    Last edited by a moderator: May 6, 2017
  7. Feb 5, 2014 #6
    Hi, I joined the Forum just for this subject, other internet searches were wrong or incomplete. Yes, string permeability seems to be key.
    Question- Strings are plucked in a "lateral" direction relative to the pickup. It seems the output signal would be twice the string frequency because of symmetry. String motion to the left would produce the same voltage polarity it does going to the right. Which doesn't seem to be the case.

    Allen Storaasli
  8. Feb 5, 2014 #7

    Philip Wood

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    Forgive me if someone's already pointed this out, but although there will be a small emf induced in the 'string', because it is moving in the field due to the permanent magnet, there won't be an induced current, because the wire is not part of a closed loop circuit: its ends aren't connected to anything. Try swapping the steel 'string' for a brass one; my guess is that the pickup won't work any longer, though it would still work if the induced current explanation were the right one.

    If I may say so, your first post was a model of clarity.
  9. Nov 19, 2015 #8
    In any AC generator, the movement in the direction of the coil in relation to the magnetic field induces a voltage in the brushes. In the case of a guitar pickup, the string is the coil and the coil of wire around the magnets are the brushes. The movement of the string through the magnetic field causes the magnetic field to induce an alternating current in the brushes, or coils of wire around the magnets. Movement in one direction induces a positive current, and a negative current is induced when the string moves in the opposite direction. If you look at the pickup as six (or four or however many strings the guitar has) a generator with six rotating coils connected to one set of brushes, each rotating at different speeds, it makes perfect sense. The physical relationship between the components is what throws most non-electronic technician/engineers off because it does not look like what it is...a simple ac generator.
    Last edited: Nov 19, 2015
  10. Nov 19, 2015 #9
    The strings are not coils and the brushes in a real motor are just contacts. There is no emf induced in the brushes.
    Why force a bad analogy when the mechanism has a good explanation already?
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