The inverse-square law of magnetism


The observation that opposite charges attract while like charges repel, with a force proportional to the inverse square of distance, motivates the study of electrostatics. Although we often don’t solve problems in electrostatics using Coulomb’s law directly, relying instead on Gauss’s law or on techniques for solving Poisson’s equation for the electric scalar potential, there is no physical information contained in those techniques that isn’t already in Coulomb’s law. Therefore, the entire field of electrostatics, including the introduction of the electric field E itself, may be viewed as an exploration of the consequences of this basic fact about stationary electric charges (and continuous distributions thereof).

At the beginning of the chapter on magnetostatics in Introduction to Electrodynamics, Griffiths notes that parallel currents attract while antiparallel currents repel. You might think that he would go on to write down a law similar to Coulomb’s law that described this force between currents directly. However, this approach is not pursued. Instead, the magnetic field B and the Lorentz force law are introduced right away. In a subsequent section, the Biot–Savart law is used to compute the magnetic field of a long straight wire, which is then immediately used to find the force between two wires.

There is a good reason for this, which I’ll get to at the end. There’s just one small problem, which is that it’s not clear that magnetism actually obeys an inverse-square law and that the force that wire 1 exerts on wire 2 is equal and opposite to the force that wire 2 exerts on wire 1. Students are also confused by being asked to use the right-hand rule to assign an apparently arbitrary direction to the magnetic field. It may appear that the universe prefers one handedness over the other for some strange reason (which, for what it’s worth, is true for the weak interactions, but not the electromagnetic).

If there were an analogue to Coulomb’s law for magnetostatics, it would become clear that the magnetic force obeys an inverse-square law, obeys Newton’s third law, and does not involve an arbitrary choice of handedness. Obviously I don’t mean the Biot–Savart law; I mean a law that gives the force between currents directly. In fact, we can derive such a law using basic knowledge of vector calculus. Recall that the Biot–Savart law gives the magnetic field at a point due to some given source distribution:
\displaystyle B(x) = \frac{\mu_0}{4\pi} \int \frac{J(x') \times (x - x')}{\|x - x'\|^3} \, \mathrm{d}^3x'
The force this exerts on some other current distribution J(x) is given by applying the Lorentz force law and integrating again over x:
\displaystyle F = \frac{\mu_0}{4\pi} \iint J(x) \times \frac{J(x') \times (x - x')}{\|x - x'\|^3} \, \mathrm{d}^3x' \, \mathrm{d}^3 x
We can simplify this using the vector identity A \times (B \times C) = B(A \cdot C) - C(A \cdot B):
\displaystyle F = \frac{\mu_0}{4\pi} \iint J(x') \left(J(x) \cdot \frac{x - x'}{\|x - x'\|^3}\right) - (J(x) \cdot J(x')) \frac{x - x'}{\|x - x'\|^3} \, \mathrm{d}^3x' \, \mathrm{d}^3 x
To see that the integral over the first term vanishes, we can first exchange the order of integration and pull J(x') out of the inner integral, then observe that (x - x')/\|x - x'\|^3 is the gradient of -1/\|x - x'\|:
\displaystyle \iint J(x') \left(J(x) \cdot \frac{x - x'}{\|x - x'\|^3}\right) \, \mathrm{d}^3 x' \, \mathrm{d}^3 x = \int J(x') \int J(x) \cdot \nabla(-\|x - x'\|^{-1}) \, \mathrm{d}^3 x \, \mathrm{d}^3 x'
This allows us to perform integration by parts on the inner integral:
\displaystyle \int J(x) \cdot \nabla(-\|x - x'\|^{-1}) \, \mathrm{d}^3 x = \\ -\int \|x - x'\|^{-1} J(x) \cdot \mathrm{d}a + \int (\nabla \cdot J(x))\|x - x'\|^{-1} \, \mathrm{d}^3 x
where the surface integral is taken over some sufficiently large surface that completely encloses the current distribution J(x) acted on and therefore vanishes, and the volume integral vanishes because the divergence is zero for a steady current. We conclude that:
\displaystyle F = -\frac{\mu_0}{4\pi} \iint (J(x) \cdot J(x')) \frac{x - x'}{\|x - x'\|^3} \, \mathrm{d}^3x' \, \mathrm{d}^3 x
This now reads very much like Coulomb’s law:
\displaystyle F = \frac{1}{4\pi\epsilon_0} \iint \rho(x) \rho(x') \frac{x - x'}{\|x - x'\|^3} \, \mathrm{d}^3x' \, \mathrm{d}^3 x
Both laws describe inverse-square forces that act along the line joining the two elements; the difference in signs is because like charges repel while like (parallel) currents attract. It is now easy to see that the force exerted on J(x) by J(x') is equal and opposite to the force exerted on J(x') by J(x). The right-hand rule does not appear anywhere. Parallel currents attract and antiparallel currents repel, period.

We can directly use this law to compute the force between two long, straight, parallel current-carrying wires. For such a problem we would use the version of the law expressed in terms of currents, I, rather than current densities, J. (I derived the law using current densities because it’s more general that way, and we can easily argue that we can go from J to I but it’s less obvious why the form involving J‘s should be true given the form involving I‘s.)
\displaystyle F = -\frac{\mu_0 I_1 I_2 }{4\pi} \iint \frac{x - x'}{\|x - x'\|^3} \, \mathrm{d}\ell \cdot \mathrm{d}\ell'
To get the force per unit length, we just compute the inner integral at x = 0. We know by symmetry that the force will be perpendicular to the wires, so we can just throw away the longitudinal component. And the two wires are everywhere parallel so \mathrm{d}\ell \cdot \mathrm{d}\ell' = 1. It’s then clear that
\displaystyle \|F\|/L = \frac{\mu_0 I_1 I_2 }{4\pi} \int_{-\infty}^\infty \frac{d}{(d + x)^{3/2}} \, \mathrm{d}x = \frac{\mu_0 I_1 I_2 }{2\pi d}
This is basically the same calculation that we would do with the Biot–Savart law and the Lorentz force law; it’s simply a bit more direct this way.

Still, if magnetism were introduced in this way, it would be necessary to get from the inverse square law to the field abstraction somehow. With electrostatics this is easy: we just pull out \rho(x) and call everything else E(x). With the inverse-square law for the magnetic force, we can try to do something like this. It’s a little bit trickier because the force also depends on the direction of the current element J(x) (while charge density, of course, doesn’t have a direction). To represent an arbitrary linear function of J(x) that gives the force density f on the current element, we need to use a matrix field rather than a vector field:
\displaystyle F = -\frac{\mu_0}{4\pi} \iint \left[\frac{x - x'}{\|x - x'\|^3} J^t(x')\right] J(x) \, \mathrm{d}^3x' \, \mathrm{d}^3 x = \int B(x)J(x) \, \mathrm{d}^3 x
where the matrix field B evidently satisfies
\displaystyle B(x) = -\frac{\mu_0}{4\pi} \int \frac{x - x'}{\|x - x'\|^3} J^t(x')  \, \mathrm{d}^3x'
The problem is that what we’ve done here is, in some sense, misleading. The local law f = BJ we may be tempted to conclude is not really true, in that B(x)J(x)\, \mathrm{d}^3 x is not the real force density on the infinitesimal current element J(x). In performing an integration by parts earlier, we essentially averaged out the term we wanted to get rid of over the entire target distribution. But this destroyed the information about what the true distribution of force is over the target current loop, with observable consequences unless the loop is rigid. In other words, we can only compute the total force on the loop, but not the stress tending to deform the loop. The true Lorentz force law that has the correct local information also has the form f = BJ, but the matrix B is antisymmetric; we also need an antisymmetrization (wedge product) in the equation for B. It’s not clear to me whether it’s possible to get from the inverse-square law to the true Lorentz force law and the true Biot–Savart law, or whether the loss of local information was irreversible. So, as I said, there is a very good reason why we don’t just postulate the inverse-square law and take that as the basis for magnetostatics: it is (probably) actually incomplete in a way that Coulomb’s law is not. Still, I feel that it contains some useful intuition.

Advertisements

About Brian

Hi! I'm Brian Bi. As of November 2014 I live in Sunnyvale, California, USA and I'm a software engineer at Google. Besides code, I also like math, physics, chemistry, and some other miscellaneous things.
This entry was posted in Uncategorized. Bookmark the permalink.

1 Response to The inverse-square law of magnetism

  1. I do believe magnetism does conform to the inverse square law. every magnet i have tasted along with electromagnets have fallowed the inverse square law.
    test;
    1; tested the detection of a may very powerful neodymium magnets out as far as it could be detected, in every test the magnetic field was only detectable the actual length of the magnet it’s self. ie the powerful magnet was two inches in length and the field was only detectable out two inches.
    2. would numerous electromagnets and tested each one and again i came up with the very same results. each electromagnets magnetic field only projected out the actual length of the core it’s self. on the last test i performed it was on a three inch long electromagnet would the entire length. i used two amp of excitation and measured with same results. i them saturated the electromagnet with 10 amps of excitation and only then the magnetic field was detectable out 3.2 to 3.2 inches. even with saturation the magnetic field still abide by the inverse square law and the fact that a magnetic field will only project out the length of the core it’s self.
    I proved in thew Figuera patent that the primaries has to be twice as long as the secondaries with Physics.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s