Relative motion and magnetic field production We have Ampere's circuital law
$$\oint \vec B\cdot dl=\mu_0I+d\phi/dt \tag{1}$$
stating that every current-carrying wire produces a magnetic field. And so should a charge moving with constant velocity. But, for an observer moving with the velocity of charge, there should be no magnetic field lines, right?
Is my line of thought correct? How can it be explained intuitively?
I am a high school student, under a calculus-based physics course.
 A: It is correct that moving with the charge, there will no longer be an observed magnetic field. In fact, it hints on something much deeper.
Instead of a point charge, consider an infinite line of current. From Ampere's law, we can calculate the magnetic field
$$
\vec{B} = \frac{\mu_0 I}{2 \pi r}\hat{\phi}
$$
If we then stop the line current, we have an infinite line of charge, which has an electric field given by Gauss' law:
$$
\vec{E} = \frac{\lambda}{2 \pi \epsilon_0 r} \hat{r}
$$
From this, we can see that (at least in this case), moving with respect to electric and magnetic fields will transform them into each other!
A reasonable question to ask after making this observation is we should describe the electromagnetic field in a way that can cope with such changes in observers. We can do this by introducing the electromagnetic field strength tensor, defined by
$$
F_{\mu \nu} =
\begin{bmatrix}
0 & \frac{E_x}{c} & \frac{E_y}{c} & \frac{E_z}{c} \\
- \frac{E_x}{c} & 0 & -B_z & B_y \\
- \frac{E_y}{c} & B_z & 0 & -B_x \\
- \frac{E_z}{c} & -B_y & B_x & 0
\end{bmatrix}
$$
Each index of this tensor can then be transformed with Lorentz transformations, which will allow you to calculate the fields for an arbitrary boost:
$$
F_{\mu' \nu'} = \Lambda^{\ \ \ \mu}_{\mu'} \Lambda_{\nu'}^{\ \ \nu} F_{\mu \nu}
$$
