# 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?

• In the moving frame, the electrons are stationary, but the Lorentz contracted protons appear as a current (in the same direction as the lab current, but stronger by $\gamma$).
– JEB
Commented Jul 27, 2021 at 22:48

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.