Would Magnetic field be Zero if measured while moving along with charges? According to Einstein's relativistic interpretation magnetic field caused by charges moving inside a wire is caused only by the relative speed between charges and the observer. So, if electrons are moving inside a wire and observer is stationary, than observer will measure a magnetic field.
Now, electrons are moving relatively slowly inside a wire. Just few inches per second, or even less. 
What would happen if somebody moved the instrument measuring the magnetic field at a same speed as the movements of electrons inside a wire and in the same direction?
Would that moving instrument measure no magnetic field at all?
 A: The problem is that in a wire, you have electrons moving one way and their positive counterparts, the atoms, being stationary. When you move along with the electrons, the lattice is no longer stationary, and you replace a "negative current in + direction" with a "positive current in - direction".
What you state would be true for electrons in vacuum. But in a wire, it doesn't work.
A: To answer this question you do not need to worry about the details of how the fields are produced, if you know the relevant Lorentz transformation of the E- and B-fields. If an observer in a particular frame of reference measures a magnetic field, but no electric field (e.g. the fields of a current carrying wire), then there must be a magnetic field in all other inertial frames.
The relevant Lorentz transforms for the B-field are that:
$$\vec{B^{\prime}_{||}} = \vec{B_{||}}\ \ \ \ \ \vec{B^{\prime}_{\perp}} = \gamma (\vec{B_{\perp}} - \vec{v} \times \frac{\vec{E}}{c^2})$$
Thus if there is no E-field in one frame, but a non-zero B-field, there will always be a non-zero transformed B-field in any other frame of reference moving with velocity $\vec{v}$ (and Lorentz factor $\gamma$) with respect to the original frame.
The situation is different if your current was just caused by an electron beam, because then there would also be an E-field associated with them. If you move with the electrons in this case, the B-field will be zero in the moving frame, because the second term inside the bracket is exactly large enough to cancel with the first term.
