Solution

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$14.3.7$ 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, $\beta = 1$?

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solution
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a) In the moving frame there is no magnetic field, only a electric field. But in the Earth frame there are a electric field $\vec{E}$ and a magnetic field $\vec{B}$.
For prove the relation we will use the loretnz transformation of the magnetic field for calculate the magnetic field in the moving field, 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 of before I set $c=1$ for simplify the calculations, fixing the equation of before:

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

b) Using the equation of before we can calculate the magnetic fields 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 unitary vector in the direction of velocity. The magnetic field will be zero when the velocity were parallel to the electric field.

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