# Deriving the magnetic field around a current from Coulomb's law and SR

The magnetic force which a current-carrying wire exerts on a moving charge can be derived from Coulomb's Law and special relativity. It's easy to show that if a straight wire has current density $\lambda$ and a particle with charge $q$ is a distance $r$ away from the wire, then the particle experiences a force perpendicular to the wire of magnitude $$F = \frac{\lambda q}{2\pi\epsilon_0r}$$ Further, we expect $$F_B = \frac{\mu_0 Iqv}{2\pi r}$$ Where $v$ is the velocity of the particle. For simplicity's sake, assume the wire is neutrally charged in the given reference frame, and the wire has a density $\phi = \frac{I}{v}$ of negative moving charges (which are also moving at speed $v$ for simplicity). Letting $F=F_B$, we expect the apperent charge density of the wire in the particle's inertial frame to be $\lambda = \mu_0\epsilon_0 I v=\frac{Iv}{c^2}$.

Shifting to the particle's inertial frame, The density of negative charges decreases by a factor of $\gamma$, and the density of positive charges increases by a factor of $\gamma$. Therefore, $$\lambda = \phi\gamma - \phi/\gamma = \phi(\gamma-1/\gamma) = \gamma\frac{Iv}{c^2}$$ Which is off from what we expect by a factor of $\gamma$. This can be fixed if we assume $v/c\approx0$, but does this mean that the law used to derive $F_B$ is only valid for slowly moving currents (which seems strange, since that formula follows from Ampere's law), or have I made a mistake in my reasoning?

• You can't "derive" any of this. It's been measured and the theory is a fit to the measurements. – CuriousOne Aug 7 '16 at 18:13

Starting where you did, we have the net force in the lab frame being the magnetic one: $$F_B = \frac{\mu_0Iqv}{2\pi r}.$$ The net charge density in the particle's frame is then $$\lambda' = \lambda'_+ - \lambda'_- = \gamma\lambda_+ - \frac{1}{\gamma} \lambda_- = \bigg(\gamma-\frac{1}{\gamma}\bigg) \lambda = \frac{\gamma v^2\lambda}{c^2} = \frac{\gamma Iv}{c^2},$$ where $\lambda$ is the density of both charges in the lab frame. In the particle's frame, then, the net force is the electrostatic one: $$F'_E = \frac{\lambda'q}{2\pi\epsilon_0c^2r} = \frac{\gamma vIq}{2\pi\epsilon_0c^2r}.$$ If we fix the $x$-direction to be that of the wire, and the $y$-direction orthogonal to this, then the $4$-force has particle-frame components $$f^{\mu'} = \gamma' \bigg(\frac{1}{c} \vec{F}'\cdot\vec{v}', \vec{F}'\bigg) = (0, 0, F'_E).$$ Lorentz-transforming this back tells us the $4$-force in the lab frame has components $$f^\mu = (0, 0, F'_E)$$ (that's right -- the four-force components stay the same, in this particular case), which means the net $3$-force is in the $y$-direction with magnitude $$F = \frac{1}{\gamma} F'_E = \frac{vIq}{2\pi\epsilon_0c^2r} = \frac{\mu_0Iqv}{2\pi r}.$$