Special theory of relativity > The transformation of electric and magnetic fields > Problem 14.3.7

Statement

$14.3.7$ a. When moving at the speed $\vec{\beta c}$ of a state in which there was only an electric field, a magnetic field with induction $\vec{B}$
arises , associated with the new electric field $\vec{E}$ by the relation $\vec{B} = [\vec{\beta} \times \vec{E}]$. Prove this relation in the case
when the old $\vec{E}$ is perpendicular to the velocity $\vec{\beta c}$

b. What magnetic field occurs when the electric field of intensity $\vec{E}$ moves at
the speed $\beta c$ if, $\beta = 1$?

Solution

a) In the moving frame there is no magnetic field, only an electric field. But in the Earth frame there are an electric field $\vec{E}$ and a magnetic field $\vec{B}$.
To prove the relation we will use the Lorentz transformation of the magnetic field to calculate the magnetic field in the moving frame; this is zero:

\begin{equation}
\vec{B}' = 0 = \frac{\vec{B} - [\vec{\beta} \times \vec{E}]}{\sqrt{1-\beta^2}} \rightarrow \vec{B} = [\vec{\beta} \times \vec{E}]
\end{equation}

In the equations above I set $c=1$ to simplify the calculations. Restoring the equation to include $c$:

\begin{equation}
\vec{B} = \frac{[\vec{\beta} \times \vec{E}]}{c}
\end{equation}

b) Using the previous equation we can calculate the magnetic field in each case:

\begin{equation}
\vec{B} = \frac{[\vec{\beta} \times \vec{E}]}{c}
\end{equation}

\begin{equation}
\vec{B_1} = \frac{[\vec{\beta_1} \times \vec{E}]}{c}
\end{equation}

\begin{equation}
\vec{B_c} = \frac{[\hat{\beta} \times \vec{E}]}{c}
\end{equation}

Where $\hat{\beta}$ is the unit vector in the direction of velocity. The magnetic field will be zero when the velocity is parallel to the electric field.

Answer

\begin{equation}
\vec{B} = \frac{[\vec{\beta} \times \vec{E}]}{c}
\end{equation}

\begin{equation}
\vec{B_1} = \frac{[\vec{\beta_1} \times \vec{E}]}{c}
\end{equation}

\begin{equation}
\vec{B_c} = \frac{[\hat{\beta} \times \vec{E}]}{c}
\end{equation}