The Unapologetic Mathematician

Mathematics for the interested outsider

Currents

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 | 11 Comments

   

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