# Magnetic effects on charge

Consider a wire carrying the current $i$ and an electric charge $q$ is moving with a velocity $v$ parallel to it. The charge experiences a force due to the magnetic field, so it is accelerated. Consider an observer moving parallel to the wire with the same velocity $v$ as the charge; with respect to the observer, the charge is at rest and it doesn’t experience a magnetic field according to the formula $F=qvb$, but still it’s accelerating with respect to the observer, moving at constant velocity. How is this possible?

• Hint: electromagnetic fields are not reference-frame-independent. They'll look different (and interconvert between each other) when you're moving at a different velocity. Commented May 4, 2018 at 9:06

In the frame of the observer moving at velocity $\vec{v}$ the charge is still accelerating, yet, as you say, experiences no magnetic force ($q \vec{v} \times \vec{B}=0$). It does, though, still experience a net electric field force, $q \vec{E}$, because in this frame the electric fields due to the positive ions in the wire is not exactly equal and opposite to the electric field due to the moving free electrons. This can be regarded as due to unequal Lorentz contractions, as a detailed treatment shows.
As you might know, electromagnetic interaction is classically governed by Maxwell's field equations. These equations are invariant under the Lorentz transformations of special relativity. So if you want to compare the two viewpoints of the observer, you have to make a lorentz transformation from the inertial system of the 'lab-observer' to the inertial system of the 'charge-observer'. This can be done via lorentz transformation $\Lambda^\mu_\nu$ of the field strength tensor $F^{\mu \nu}=\partial^\mu A^\nu - \partial^\nu A^\mu$, where $\partial^\mu$ is the partial derivative in every spacetime direction and $A^\mu$ is the 4-potential. This can be done in this way: \begin{align} F^{' \alpha \beta} = \Lambda^\alpha_\mu F^{\mu \nu}\Lambda^\beta_\nu \, . \end{align} If you want to know how this is to be calculated via actual matrix multiplication, check this link. This, however, gives you the fields present in the new frame of reference via the relationships \begin{align} E_i&=c F_{0i} \\ B_i&=-\frac{1}{2}\epsilon_{ijk}F^{jk} \end{align} with the classical fields, where $\epsilon_{ijk}$ is the Levi-Civita symbol. As already mentioned in Philip Woods answer, the new configuration of fields gives you a different Lorentz force, which still leads to an acceleration in the 'charge-observer' frame.