The Unapologetic Mathematician

Mathematics for the interested outsider


Part of the reason that the Biot-Savart law isn’t usually stated in the way I did is that it’s really about currents, which are charges in motion. The point charge q moving with velocity v does give a sort of a current, but it’s so extremely localized that it doesn’t match with our usual notion of “current”. A better example is a current flowing along a curve (without boundary) c with a (constant) charge density of \lambda. It’s possible to carry through the discussion with a variable charge density, but then things get more complicated.

Anyway, just like the electric field the magnetic field obeys a superposition principle, so we can add up the contributions to the magnetic field from all the tiny differential bits of a current-carrying curve by taking an integral. The differential element of charge is again \lambda ds=\lambda\lvert c'(t)\rvert dt. The velocity is in the direction of the curve — unit vector \frac{c'(t)}{\lvert c'(t)\rvert} — and has length \lvert v\rvert. Thus the qv term near a point is \lambda\lvert v\rvert c'(t)dt=\lambda\lvert v\rvert ds. We will take the charge density \lambda as charge per unit of distance and the speed \lvert v\rvert as distance per unit of time, and combine them into the current I=\lambda\lvert v\rvert as the charge flowing through this point on the curve per unit of time. For a curve r=c(t) we have the integral:

\displaystyle B(p)=\int\limits_c\frac{\mu_0}{4\pi}\frac{Idr\times(p-r)}{\lvert p-r\rvert^3}

As an example, let’s take the infinite line of charge and set it in motion with a speed s along the z-axis. The charge density \lambda makes for a current I=\lambda s up the line. Of course, if the current is negative then the charge is just moving in the opposite direction, down the line. The obvious parameterization is c(t)=(0,0,t), so we have c'(t)=(0,0,1) and p-r=(x,y,z-t). Plugging in we find:

\displaystyle\begin{aligned}B(p)&=\int\limits_{-\infty}^\infty\frac{\mu_0\lambda s}{4\pi}\frac{(0,0,1)\times(x,y,z-t)}{\left(x^2+y^2+(z-t)^2\right)^\frac{3}{2}}dt\\&=\frac{\mu_0\lambda s}{4\pi}\int\limits_{-\infty}^\infty\frac{(-y,x,0)}{\left(x^2+y^2+(z-t)^2\right)^\frac{3}{2}}dt\\&=\frac{\mu_0\lambda s}{4\pi}(-y,x,0)\int\limits_{-\infty}^\infty\frac{dt}{\left(x^2+y^2+(z-t)^2\right)^\frac{3}{2}}\\&=\frac{\mu_0\lambda s}{4\pi}(-y,x,0)\int\limits_\infty^{-\infty}\frac{-d\tau}{\left(\rho^2+\tau^2\right)^\frac{3}{2}}\\&=\frac{\mu_0\lambda s}{4\pi}(-y,x,0)\int\limits_{-\infty}^\infty\frac{d\tau}{\left(\rho^2+\tau^2\right)^\frac{3}{2}}\end{aligned}

where I’ve used a couple convenient substitutions to put the integral into exactly the same form as last time. We can reuse all that work to continue:

\displaystyle\begin{aligned}B(p)&=\frac{\mu_0\lambda s}{4\pi}(-y,x,0)\frac{2}{\rho^2}\\&=\frac{\mu_0}{2\pi}\frac{\lambda s}{x^2+y^2}(-y,x,0)\\&=\frac{\mu_0}{2\pi}\frac{\lambda s}{\sqrt{x^2+y^2}}\frac{(-y,x,0)}{\sqrt{x^2+y^2}}\\&=\frac{\mu_0}{2\pi}\frac{\lambda s}{\lvert(-y,x,0)\rvert}\frac{(-y,x,0)}{\lvert(-y,x,0)\rvert}\end{aligned}

We find again that the magnetic field now falls off as the first power of the distance from the line current. As for the direction, it wraps around the line in accordance with the famous “right hand rule”; if you place the thumb of your right hand along the z-axis, the field curls around the line in the same direction as your fingers.

As it happens, despite how popular the rule is it’s purely conventional, with no actual physical significance. It’s hard to explain just why that is right now, but it will become clear later. For now, I can justify that it makes no difference in the effect of currents on moving charges, since the Biot-Savart law involves a triple vector product which can be rewritten:

\displaystyle a\times(b\times c)=(a\cdot c)b-(a\cdot b)c

which formula involves no cross products and no choice of right-hand or left-hand rules.

January 7, 2012 - Posted by | Electromagnetism, Mathematical Physics


  1. John, this post is showing up with the wrong date for some reason… It says December 31 when I’m looking at it right now.

    Comment by Gilbert Bernstein | January 8, 2012 | Reply

  2. Sorry, this is a bug in WordPress when I write posts in advance. I usually catch it, but sometimes it slips through.

    Comment by John Armstrong | January 8, 2012 | Reply

  3. […] charge density and a velocity vector , each of which is a function of space. In our example of an infinite line current, this density was concentrated along the -axis, where the velocity was vertical. But it could exist […]

    Pingback by Gauss’ Law for Magnetism « The Unapologetic Mathematician | January 12, 2012 | Reply

  4. […] let’s see where we are. There is such a thing as charge, and there is such a thing as current, which often — but not always — arises from charges moving […]

    Pingback by Maxwell’s Equations « The Unapologetic Mathematician | February 1, 2012 | Reply

  5. […] we can compare this to the last time we computed the magnetic field of the straight infinite current by integrating the Biot-Savart law […]

    Pingback by Deriving Physics from Maxwell’s Equations « The Unapologetic Mathematician | February 3, 2012 | Reply

  6. […] to list! A very small sample chosen mostly at random: Coulomb’s Law, The Biot-Savart Law, Currents, Gauss’ Law for Magnetism, Faraday’s Law, Conservation of Charge, Deriving Physics from […]

    Pingback by Fourteenth Linkfest | February 7, 2012 | Reply

  7. Very frustrating, trying to understand the basics of Maxwell’s equations and I do not understand what current is, aside from the motion of charge. Why exactly is current needed in the equations, why can’t it just be defined in terms of charge? You say a current is not always a motion of charge. What else is it? Where is a good exposition of this from first principles?

    Comment by Benson Bear | February 14, 2012 | Reply

  8. Well, as I mention in Maxwell’s correction to Ampère’s law, a changing magnetic field also counts as a current without any actual charge moving around. As for first principles, you’d really have to ask a physicist; I’m mostly doing this all as a mathematical model, where there just are these things called currents and that’s that.

    Comment by John Armstrong | February 14, 2012 | Reply

  9. In the ME version of AL, the J refers to real honest to goodness current, does it not? Don’t you mean in AL that a changing electric field (not a magnetic field) “counts as a current” because it induces a magnetic field as well as genuine currents (J) themselves do? That is the “correction term” based on E. I know it has units of current, but it seems to be linguistically violent and confusing revisionary to call it a current. In any case, what I was confused about and hope you can confirm, is that the J term really is nothing but “genuine” current.

    Comment by Benson Bear | February 14, 2012 | Reply

  10. Yes, in Maxwell’s equations J is a current density due to actual matter moving around. That said, it doesn’t necessarily correspond to a change in charge.

    Consider two uniform charge distributions along a line; one is positive and stationary, while the other is negative and moving in one direction. When we add the two together we get no net charge distribution, but we do have a current!

    Comment by John Armstrong | February 14, 2012 | Reply

  11. Yes, I didn’t mean to suggest if I did that current density corresponds to change in charge. As I understand what ME are supposed to be about, it is about particles moving around, causing currents, electric and magnetic fields, which the in turn affect how the particles move. The underlying basis for all of this is the particles. So both charge density and current come from the particles. If you have a huge current, you may not have any change in charge at any location. The current is the particles in motion not the charge in motion. A given particle carries a charge along, but if it is in a current sustained at each point, it is instantaneously replaced by another particle coming in: no net change in charge density or current (an idealisation to be sure). Thanks.

    Comment by Benson Bear | February 14, 2012 | Reply

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