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?

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

You did everything right, but you forgot that forces transform between frames as well.

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 $$ F = \frac{1}{\gamma} F'_E = \frac{vIq}{2\pi\epsilon_0c^2r} = \frac{\mu_0Iqv}{2\pi r}. $$


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