Does the magnetic field also change its value in the frame of a charge moving with drift velocity next to a current carrying wire? I know that if there is a charge moving next to a current carrying wire with velocity equal to that of the electrons in the conductor, then in a frame with the same velocity the charge experiences an electric force and not a magnetic one.
I have been taught that this is due to the fact that in the moving frame, the velocity of the charge=0 and thus F(magnetic)=qvB=0.
But I would like to know if the value of 'B', (i.e. magnitude of magnetic field) itself also changes in the moving frame?
 A: In the lab, where you have a (linear) proton density of $\lambda_+$ and a (linear) electron density of $\lambda_- = -\lambda_+$, the total charge density is:
$$ \lambda = \lambda_+ +\lambda_ - = 0$$
so the lab electric field is:
$$ E = 0$$
If the electrons are moving with speed $v_e=v$, then the current density is:
$$ j =(\lambda_+v_p) + (\lambda_-v_e) = -\lambda_+ v $$
That leads to a magnetic field:
$$ B_{\perp} = -\frac{\mu_0}{2\pi}\frac{\lambda v} r$$
You can transform those those fields to the frame moving at $v$ along the wire to get:
$$ E'_{\parallel}=E_{\parallel}=0$$
$$ B'_{\parallel}=B_{\parallel}=0$$
$$ E'_{\perp}=\gamma(E_{\perp}+\vec v\times\vec B)
=\frac{\gamma v^2\mu_0 \lambda_+}{2\pi r} = \frac{\gamma v^2 \lambda_+}{2\pi \epsilon_0 c^2r}$$
$$ B'_{\perp}=\gamma(B_{\perp}+\frac 1{c^2}\vec v\times\vec B)
=\frac{\gamma \mu_0v\lambda_+}{2\pi r}$$
Meanwhile, if you just transform the proton and electron densities to the moving frame, you get:
$$ \lambda'_+ = \gamma\lambda_+$$
$$ \lambda'_- = \lambda_-/\gamma = -\lambda_+/\gamma $$
so that:
$$ \lambda'=(\gamma-1/\gamma)\lambda_+=\gamma\frac{v^2}{c^2}\lambda_+$$
and the current is due to the protons:
$$ j'=\lambda'_+(-v)=-\gamma v \lambda_+=\gamma j$$
Those densities reproduce the transformed field according to the standard solutions of linear charge and current densities, as they must.
