$14.3.7$ 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
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when the old $\vec{E}$ is perpendicular to the velocity $\vec{\beta c}$\\
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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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### 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 $\vec{B}$.
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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:
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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 $\vec{B}$.
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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:
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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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.
$14.3.7$ 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 $\vec{B}$
$14.3.7$ 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
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 $\vec{\beta c}$\\
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
b. What magnetic field occurs when the electric field of intensity $\vec{E}$ moves at
the speed $\beta c$ if, $\beta = 1$?
the speed $\beta c$ if, $\beta = 1$?
\begin{center}
### Solution
solution
\end{center}
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 $\vec{B}$.
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 $\vec{B}$.
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:
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:
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.
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.
