Here is a picture (from a paper by Feinberg and Yang in the Journal of the Acoustical Society of America) of an electric guitar string vibrating (mostly in the $y$ dimension) near a permanent magnet.

steel guitar string vibrating over a permanent magnet

The steel guitar string is causing the magnetic field to change in some way, which by Faraday's Law is inducing a voltage in the pickup coil. My question is how should I model the way the moving steel string is interacting with the magnetic field from the permanent magnet? I don't recall anything from my limited (first year physics) about modeling something like this. Does it have something to do with the mechanical movement of the electrons in the guitar string as the string vibrates? In that case, what difference does the permanent magnet make? I just don't know where to start thinking about this.

  • 1
    $\begingroup$ The string probably gets magnetized, and thus behaves as moving magnet, and thus induces EMF in the pickup coil. Based on my experiments, one can get some very weak signal in a coil even if no permanent magnet is inside, so either the string remains magnetized a little even without the permanent magnet, or there may be another mechanism where EMF is induced even if the string is not magnetic. $\endgroup$ Jun 20, 2023 at 14:26
  • 2
    $\begingroup$ Reference 3 in your article is to "Modeling the magnetic pickup of an electric guitar", American Journal of Physics 77, 144–150 (2009). While I can't actually access the article myself just now, Am. J. Phys. generally has stuff written at a level that's accessible to someone with an undergraduate physics education, so I suspect it'd be worth your while to check it out. $\endgroup$ Jun 20, 2023 at 14:46
  • 1
    $\begingroup$ The motion of single point on a vibrating guitar string traces an ellipse. If your model considers only the Y axis, it will not accurately represent the physical world. See Youtube for "slow motion guitar string". $\endgroup$
    – MTA
    Jun 21, 2023 at 1:03
  • $\begingroup$ With such a process, shouldn't the picked-up signal be quadratic (or at least ±y-symmetric) with respect to the y-displacement? Frequency-doubling then? Or maybe there is a built-in asymmetry that lets the coil pick a linear signal? $\endgroup$ Jun 21, 2023 at 12:20
  • $\begingroup$ This MinutePhysics video may help - The Bizarre Physics of Electric Guitars $\endgroup$
    – mmesser314
    Jun 21, 2023 at 14:12

4 Answers 4


Here's what's happening:

The permanent magnet in the pickup propagates a static magnetic field which extends out from the magnet tip. The ferromagnetic string resides within this field.

Because the string is ferromagnetic, the permanent magnet's field lines are drawn in towards it and get bunched up there. If the string is set into motion, those bunched-up field lines get dragged along with the string and they wiggle back and forth in space along with the string.

This means that the field lines close to the magnet are being wiggled back and forth a little bit too, and in so doing they "cut through" the loops of wire that surround the magnet. This generates a current flow in the loops which furnishes the electrical output of the pickup.

It is possible to artfully design the pickup in such a way as to shape the field from the magnet, so the string sits in a strong portion of the field and thereby generates a stronger signal in the coil when it vibrates.

More loops of wire around the magnet means more signal strength, as does a stronger magnet, but too many loops of wire mean too much inductance in the wire coil and too much parasitic capacitance in it, which rolls off the high-frequency response of the pickup and makes it sound muddy (although loud!).

A magnet that is too strong actually exerts a force of attraction on the string which alters its vibrations in such a way as to upset them and produce false harmonics which knock the string's intonation out of whack.

All these factors make pickup design a highly proprietary art which unfortunately contains a lot of unphysical woo-woo and mumbo-jumbo, which is manifestly obvious in their ad copy!

  • 1
    $\begingroup$ Do you know if pickups are more sensitive to vertical string displacement than horizontal? $\endgroup$ Jun 20, 2023 at 20:23
  • 1
    $\begingroup$ They are far more sensitive to side-to-side movement, but the pickup designers can and do play with shaping the field to increase their sensitivity to z-axis movement as well. $\endgroup$ Jun 20, 2023 at 22:37
  • 1
    $\begingroup$ @StevanV.Saban, I used to own a stringed instrument with two piezoelectric pickups built-in to the bridge in such a way that one supposedly responded to horizontal vibration, and the other, to vertical vibration. A simple toggle switch allowed me to choose one or the other, but not both. I don't know what they really responded to, but the end result was two distinctly different timbres. (Neither of which was pleasant. I sold that instrument after owning it for less than half a year.) $\endgroup$ Jun 22, 2023 at 18:17
  • 2
    $\begingroup$ @SolomonSlow, yes, they do that by cutting one crystal so it responds to compression (vertical displacement) and the other to shear (sideways displacement). remind me sometime to tell you stories about how a satellite manufacturer used acoustic guitar pickups in their products! $\endgroup$ Jun 22, 2023 at 20:43
  • $\begingroup$ @SolomonSlow Acoustic and classic guitars prefer the vibrations along the plane of the soundboard (vertical displacement) for maximum amplification and best tone. If the piezo pickups are built into the bridge, then they are designed for vertical displacement. None of the struts and bracings for an acoustic guitar are designed with horizontal displacement in mind. Personally I've had the best success with piezos that I can mount on the underside of the soundboard. Position of the piezo can create drastic differences in timbre alone without the need to select for vibration planes. $\endgroup$ Jun 23, 2023 at 3:02

Although niels nielsen's answer contains valuable information (the last 3 paragraphs), I think user1079505's answer is closer to the truth, and the article (that I don't have access to) hinted at by Michael Seifert is probably spot on. My explanation would be:

Magnet and string and surrounding air form a magnetic circuit. The magnetic flux through this circuit depends on the air gap between magnet and string, smaller gap means smaller 'resistance' hence bigger flux. Moving string thus causes changing flux. The coil reacts to (only) changes in flux.

Additional notes: (1) The coil doesn't have to be anywhere near the string, as long as it surrounds the magnet (or more correctly, the flux). 'Wiggling field lines' play no role. (2) The device can be hugely improved by almost completing the flux circuit with more iron, the casing around the magnet-and-coil, to dramatically decrease the second gap between the other pole of the magnet and the string. Effective electromagnets are built this way.


Electric guitar strings are ferromagnetic. In the presence of magnetic field from the magnets in the pickup, the strings become magnetized. As they move (vibrate), the magnetic field they generate also changes. Change of magnetic field generates electrical current in the coil conductor.

As acoustic guitar strings are often less ferromagnetic, they may not work well on electric guitars.


Great question. The first part of the answer by niels nielsen is quite inaccurate, talking about "dragging" and "wiggling" of field lines to make the concept more intuitive, but in my opinion it obscures things much more. Zaaikort's answer is great, I only want to add an answer with more detail, and quotes from the reference mentioned in the comments and in Zaaikort's answer.

This explanation is from "Modeling the magnetic pickup of an electric guitar", American Journal of Physics 77, 144–150 (2009). Link.

A simple magnetic pickup (see Fig. 1) is comprised of a permanent magnet surrounded by a coil of wire with typically several thousand turns. The guitar strings consist of wires made of a ferromagnetic material and are parallel to the face of the magnet. The magnetic field of the pickup/wire system, and thus the magnetic flux through the coil, depends critically on the position of the wire. Therefore, moving the wire changes the magnetic flux through the coil. According to Faraday’s law the current induced inside the coil is proportional to the time rate of change of the magnetic flux through the coil. As the string moves through the magnetic field a time-varying current is produced in the coil. This current is used to produce a potential drop across a resistor, which is then amplified and sent to a speaker.

This verifies what you already describe, but then you ask

...how should I model the way the moving steel string is interacting with the magnetic field from the permanent magnet? ... Does it have something to do with the mechanical movement of the electrons in the guitar string as the string vibrates?

First, a tip: Unless you are dealing with quantum effects, you can almost always ignore the role of electrons in electromagnetics. Continuous current distributions are the language of classical E&M, like a material continuum is the language of fluid dynamics. That's at least my opinion, others may disagree.

About the modeling: In short, a ferromagnetic material (the guitar string) has high permeability $\mu$, and thus the resulting field is more intense inside the volume. This is magnetization. The magnetic field around the magnet is not uniform, so changing the position of the string alters the induced magnetic field in the ferromagnetic material, and therefore the total magnetic field is changed (superimpose the field of the magnet and the induced field of the ferromagnetic material).

That's the gist. From an engineering or physics perspective, this situation can be modeled as either (a) a magnetostatic problem, (b) a magnetic circuit, or (c) a problem of (fictitious) magnetic charge density. Each of the representations are suitable, they represent the same phenomena, just at different levels of generality and "physicality". To develop a model, first begin with a description of the magnetic field due to the permanent magnet, without the guitar string. Most models of permanent magnets are too complicated to be useful outside of computation, but using the fictitious concept of magnetic charge yields a good approximation with simpler expressions, and this is the route taken by the above-cited paper. Then, determine the effect of the nearby ferromagnetic material. This might involve assuming the wire to be a series of parallel plate magnets with no width and a height equal to the diameter of the wire, above the permanent magnet. From the paper:

... We model the wire as a series of infinitesimally wide magnets whose strength is linearly proportional to the local field at the position of the wire due to the permanent magnet, and the height of these infinitesimal magnets is equal to the diameter of the wire.

We assume that the coercitive field of the permanent magnet is sufficiently large so that the presence of the wire does not affect the magnetic charge density on the surface. Therefore, the only change in the magnetic field at $(x_p,y_p,z_p)$ due to the presence of the wire is caused by the magnetization of the wire itself. We further assume that the magnetic intensity at the wire is less than is required for saturation, and we ignore the effects of hysteresis. Therefore, the magnetic field of the permanent magnet results in a linear change in magnetization at the wire.

Finally, the model itself is calculated from what is essentially Coulomb's law, but for magnetic charge. It is calculated in three parts. First, the field due to the permanent magnet; then, the magnitude of the magnetic field at a given position is calculated; finally, the z-component of the magnetic field is used to calculate flux.

I will reproduce the reasoning in the linked paper for completeness. Let the radius of the permanent magnet be $\psi$, with center $(x_0,y_0,z_0)$, with its face perpendicular to the $z$-axis, and plate separation equal to its length. Denote the magnetic charge density by $\sigma$, and assume this density is uniform. Then the $z$-component of the magnetic field at $(x',y',z')$ is $$ B_{z}(x',y',z') = \int_{0}^{2\pi} \int_{0}^{\psi} \frac{\sigma(z'-z_0)\rho}{\left[ (x'-[x_0-\rho \cos \phi])^{2} + (y' - [y_0-\rho \sin \phi]) ^{2} + (z'-z_0)^{2} \right]^{3/2} } d\rho d\phi, $$ Next, modeling the wire as an infinitesimally wide disk located at $(x',y',z')$, the magnitude of the magnetic field at this disk, before taking the permeability into account, is: $$ |\mathbf {B}_w (x',y',z')| = \int_{0}^{2\pi} \int_{0}^{\psi} \frac{\sigma\rho}{\left[ (x'-[x_0-\rho \cos \phi])^{2} + (y' - [y_0-\rho \sin \phi]) ^{2} + (z'-z_0)^{2} \right] } d\rho d\phi, $$ (Note: subscript $w$ is for "wire") Finally, the z-component of this field is calculated by multiplying the above expression with a constant of proportionality related to the permeability/susceptibility of the wire, and resolving it into its z-component, for an arbitrary position $(x,y,z)$, which is the location of interest for the total magnetic field, $$ B_{w,z}(x,y,z) = \gamma |\mathbf B_w(x',y',z')| \frac{ (z'-z) }{\left[(x'-x)^2 + (y'-y)^2 +(z'-z)^2\right]^{3/2}}, $$ and this is summed for each segment of the wire. In other words, splitting the wire into $N$ pieces, with positions $(x_i,y_i,z_i)$ for $i=1,2,\cdots,N$, the field at $(x,y,z)$ is

$$ B_{w,z}^{tot}(x,y,z) = \gamma \sum_i |\mathbf B_w(x_i,y_i,z_i)| \frac{ (z_i-z) }{\left[(x_i-x)^2 + (y_i-y)^2 +(z_i-z)^2\right]^{3/2}}. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.