Problem 14.3.19

Solution

To solve this problem we can use the Lorentz transformation for current density and charge density.
Also, we need to keep in mind that the current density in the Earth frame is related to the charge density in the Earth frame as follows:

\begin{equation} j = -\rho \beta c \end{equation}
The negative sign is due to the sign of the charge, because it is the current density of the electrons.
Now, the Lorentz transformations for current density and charge density are:

\begin{equation} j' = \gamma (j - \rho \beta c) \qquad \text{and} \qquad \rho' = \gamma (\rho - \frac{j \beta_1}{c}) \quad \text{where} \quad \gamma = \frac{1}{\sqrt{1 - \beta_1^2}} \end{equation}

Using the equation for the charge density we can calculate it in the moving frame:

\begin{equation} \rho' = \gamma (\rho - \frac{j \beta_1}{c}) = \gamma (\rho + \frac{\rho \beta c \beta_1}{c}) =\rho \frac{(1 + \beta \beta_1)}{\sqrt{1-\beta_1^2}} \end{equation}