Relative motion + electromagnetism

We say that Force exerted by a current carrying wire on charge $q$ moving with velocity $v$ is: $F=qvB\tag*{}$ (Where $B$ is the magnetic field , and electric Force is 0 because wire is neutral.)

But an observer moving in the direction of charge, with velocity $v$, would see the charge at rest. So, in this frame, no Force should act on the charge?

Please tell me where am I wrong?

If charge $q$ in a system $S$ moves with velocity $v$ in a direction perpendicular to a magnetic field $B$ (and there is no electric field in $S$), it experiences in $S$ a Lorentz force $$F=qvB$$ If you are in an inertial system $S'$ moving with the charge at constant velocity $v$ then you see an electrical force on the charge $$F'=qvB'$$ where $$B'=\frac{B}{\sqrt{1-v^2/c^2}}$$ is the magnetic field in $S'$ in the same direction as in $S$. For small velocities $v<<c$, the magnetic field in $S'$ is approximately $B'=B$. This means that, in this approximation, the Lorentz force $F$ on $q$ in $S$ is equal to the electrical force $F'$ on $q$ in $S'$ $$F=F'$$ Thus what is a magnetic force on the moving charge in $S$ appears as a purely electrical force in $S'$. This follows from the Lorentz transformation of electric and magnetic fields in inertial systems moving with velocity $v$ relative to each other. See, e.g., Chapter 12.3 Relativistic Electrodynamics, in D.J. Griffiths, Introduction to Electrodynamics, 3rd edition, 1999.

• If you change to $S'$, why do you use $v$, not $v'$ in the expression for force? In the question exactly this point is not understood. – queezz Mar 2 '18 at 5:18
• @queezz - Because it was assumed that the charge doesn't move in $S'$. See the text above.. – freecharly Mar 2 '18 at 17:32
• I just think it could be much more obvious if you explicitly state the transformation of the electromagnetic forces. – queezz Mar 2 '18 at 19:18
• @queezz - You are right, I just wanted to keep the answer short. – freecharly Mar 2 '18 at 20:27

Let us have two frames, $S$ where charge is moving and $S'$ where it is at rest. Then for $S'$ we can wright down $$\mathbf{E_\parallel}' = \mathbf{E_\parallel} \\ \mathbf{B_\parallel}' = \mathbf{B_\parallel} \\ \mathbf{E_\bot}' = \gamma \left( \mathbf{E}_\bot + \mathbf{v} \times \mathbf{B} \right) \\ \mathbf{B_\bot}' = \gamma \left( \mathbf{B}_\bot - \frac{1}{c^2} \mathbf{v} \times \mathbf{E} \right)$$ where $\gamma \ \overset{\underset{\mathrm{def}}{}}{=} \ \frac{1}{\sqrt{1 - v^2/c^2}}$, see Wiki for more details.
Important part now is $\mathbf{E_\bot}' = \gamma \left( \mathbf{E}_\bot + \mathbf{v} \times \mathbf{B} \right)$, even if $\mathbf{E} = 0$ in the frame $S$, in the frame $S'$ electric field would not be zero. Assuming velocity of the charge $v<<c$, in your case $\mathbf{E}' = \mathbf{v}\times \mathbf{B}$.
So if $\mathbf{v} \perp \mathbf{B}$, the force in $S'$ will be defined as follows: $$F' = qE = qvB = F$$
Force would not change, though the electromagnetic field in frame $S'$ will. And it will have an electric component now, which will act on the charge at rest.
Quote from undergraduate physics textbook from Knight: Basically, when you trying to observe in a "moving" reference frame, $E$ and $B$ field started to rotate by an imaginary angle. In this case, $B$ field became $E$ field, thus in stead of the magnetic force, there ought to be a equivalent electronic statics force.