Magnetic field generated by a current in a wire - special relativity

[Frostman]

Homework Statement

Calculate the magnetic field $\underline B$ generated by a current $I$ in an infinitely long straight wire, knowing the shape of the electric field generated by a straight-line and uniform distribution of static electric charges, the transformation laws of electromagnetic fields for Lorentz boost, and appropriately using the superposition principle for electromagnetic fields.

Relevant Equations

Gauss theorem
Lorentz transformations for current density and electromagnetic fields

First I wrote in $S'$, by using Gauss theorem
$$
\int_{\Sigma} \underline E' \cdot \hat n d\Sigma = \frac Q {\varepsilon_0} \rightarrow E'(r)2\pi rH=\frac{\lambda'H}{\varepsilon_0}
$$
$$
\underline E'(\underline r)=\frac{\lambda'}{2\pi\varepsilon_0r}\hat r
$$
Its components are:

$E_x'=\frac{\lambda'}{2\pi\varepsilon_0r}\cos\theta$
$E_y'=\frac{\lambda'}{2\pi\varepsilon_0r}\sin\theta$
$E_z'=0$

Now I notice that $Q=\rho'HA$ and $\lambda'=\frac{Q}{H}$, so $\rho'=\frac{\lambda'}{A}$

With this information, I can made $J'\mu=(\rho', 0, 0, 0)$ and consider a $S$ frame moving respect $S'$ with a speed $\underline v=- v\hat z$. For a Lorentz boost we have:

$\rho = \gamma(\rho'-vJ_z')=\gamma\rho'$
$J_x=J_x'=0$
$J_y=J_y'=0$
$J_z=\gamma(J_z'-v\rho')=-\gamma v\rho'$

Now we have $J\mu=(\gamma\rho', 0, 0, -\gamma v \rho')$, it appears a current density in $S$ frame. I evaluate the new fields:

$E_x=\gamma(E_x'+vB_y')=\gamma E_x' = \gamma \frac{\lambda'}{2\pi\varepsilon_0r}\cos\theta$
$E_x=\gamma(E_y'-vB_x')=\gamma E_y' = \gamma \frac{\lambda'}{2\pi\varepsilon_0r}\sin\theta$
$E_z=E_z'=0$
$B_x=\gamma(B_x'-vE_y')=-\gamma v E_y' = \gamma v \frac{\lambda'}{2\pi\varepsilon_0r}\sin\theta = \frac{J_zA}{2\pi\varepsilon_0r}\sin\theta=\frac{I}{2\pi\varepsilon_0r}\sin\theta$
$B_y=\gamma(B_y'-vE_x')=\gamma v E_x' = \gamma v \frac{\lambda'}{2\pi\varepsilon_0r}\cos\theta = \frac{-J_zA}{2\pi\varepsilon_0r}\cos\theta=- \frac{I}{2\pi\varepsilon_0r}\cos\theta$
$B_z=B_z'=0$

I observe that it shows an electric field in $S$ frame that we should delete, and we do this by using superposition principle for electromagnetic fields introducing a linear charge density that delete $E_x$ and $E_y$, and its value is $\tilde{\lambda} = -\gamma \lambda'$.

I hope that everything is right also at the level of signs given the reference systems I used, I shouldn't have mistaken the Lorentz transformations.

I have a doubt, beyond what I have done. Surface $A$ does not undergo any contraction since the motion is orthogonal to the surface, right?

The moment I introduce a new linear charge density $\tilde{\lambda}$, the magnetic field becomes more intense due to the fact that more current is added, especially it should double right?

 

[TSny]

With this information, I can made $J'\mu=(\rho', 0, 0, 0)$ and consider a $S$ frame moving respect $S'$ with a speed $\underline v=- v\hat z$. For a Lorentz boost we have:

$\rho = \gamma(\rho' - vJ_z')=\gamma\rho'$
$J_x=J_x'=0$
$J_y=J_y'=0$
$J_z=\gamma(J_z'-v\rho')=-\gamma v\rho'$

I believe the minus signs in the first and last equations above should be plus signs. You are transforming from the primed frame to the unprimed frame. Note that if $\rho'$ is positive, you would expect $j_z$ to also be positive since the positively charged cylinder is moving in the positive z-direction relative to the unprimed frame.

Now we have $J\mu=(\gamma\rho', 0, 0, -\gamma v \rho')$, it appears a current density in $S$ frame.

$j_z$ will not have the minus sign

I evaluate the new fields:

$E_x=\gamma(E_x'+vB_y')=\gamma E_x' = \gamma \frac{\lambda'}{2\pi\varepsilon_0r}\cos\theta$

OK

$E_x=\gamma(E_y'-vB_x')=\gamma E_y' = \gamma \frac{\lambda'}{2\pi\varepsilon_0r}\sin\theta$

$E_x$ on the far left should be $E_y$. Otherwise OK.

$B_x=\gamma(B_x'-vE_y')=-\gamma v E_y' = \gamma v \frac{\lambda'}{2\pi\varepsilon_0r}\sin\theta = \frac{J_zA}{2\pi\varepsilon_0r}\sin\theta=\frac{I}{2\pi\varepsilon_0r}\sin\theta$

The minus sign after the second "$=$" should carry over to the expression after the third "$=$". Check to see if this will change the sign of the final expression in terms of $I$.

$B_y=\gamma(B_y'-vE_x')=\gamma v E_x' = \gamma v \frac{\lambda'}{2\pi\varepsilon_0r}\cos\theta = \frac{-J_zA}{2\pi\varepsilon_0r}\cos\theta=- \frac{I}{2\pi\varepsilon_0r}\cos\theta$

I'll let you double-check the signs for this equation.

I observe that it shows an electric field in $S$ frame that we should delete, and we do this by using superposition principle for electromagnetic fields introducing a linear charge density that delete $E_x$ and $E_y$, and its value is $\tilde{\lambda} = -\gamma \lambda'$.

Yes, you can superpose this static negative charge distribution in the unprimed frame to get rid of the electric field in the unprimed frame.

I have a doubt, beyond what I have done. Surface $A$ does not undergo any contraction since the motion is orthogonal to the surface, right?

Right

The moment I introduce a new linear charge density $\tilde{\lambda}$, the magnetic field becomes more intense due to the fact that more current is added, especially it should double right?

No. you are superposing a static negative charge distribution in the unprimed frame.
It will not affect the B-field that you have calculated in the unprimed frame.