A rocket zooms past the earth at $v=2.0\times {10}^{6}m/s.$a. Scientists on the rocket have created the electric and magnetic fields shown in $FIGUREEX31.5.$What are the fields measured by an earthbound scientist?

We have been given that a rocket zooms past the earth at $v_{AB}=2.0\times {10}^6\hat{i}\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$. The Scientists on the rocket have created the magnetic field${\overrightarrow{B}}_A=1.0\hat{j}T$and electric field${\overrightarrow{E}}_A=1.0\times {10}^6\raisebox{1ex}{$\hat{R}V$}\!\left/ \!\raisebox{-1ex}{$m$}\right.$shown in $FIGUREEX31.5$.

We need to calculate the electric field, ${\overrightarrow{E}}_B$measured by an earthbound scientist.

Given values :

The electric field created by the scientists on the rocket is ${\overrightarrow{E}}_A=1.0\times {10}^6\hat{R}\raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{$m$}\right.$.

The magnetic field created by the scientists on the rocket is${\overrightarrow{B}}_A=1.0\hat{j}T$

The relative velocity of the rocket with suspect to earth is $v_{AB}=2.0\times {10}^6\hat{i}\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

The relative velocity of the earth with suspect to rocket will be opposite i.e. $v_{BA}=-2.0\times {10}^6\hat{i}\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

We know the formula,

${\overrightarrow{E}}_B={\overrightarrow{E}}_A+{\overrightarrow{v}}_{BA}+{\overrightarrow{B}}_A$

Substituting the values of ${\overrightarrow{E}}_A,{\overrightarrow{v}}_{BA},{\overrightarrow{B}}_A$, we get:

${\overrightarrow{E}}_B=1.0\times {10}^6\hat{R}+(-2.0\times {10}^6\hat{i})\times 1.0\hat{j}$

role="math" localid="1650186338852" ${\overrightarrow{E}}_B=-1.0\times {10}^6\hat{R}\raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{$m$}\right.$

The electric field measured by an earthbound scientist is role="math" localid="1650186356029" ${\overrightarrow{E}}_B=-1.0\times {10}^6\hat{R}\raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{$m$}\right.$.

We have been given that a rocket zooms past the earth at $v_{AB}=2.0\times {10}^6\hat{i}\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$. The Scientists on the rocket have created the magnetic field ${\overrightarrow{B}}_A=1.0\hat{j}T$and electric field ${\overrightarrow{E}}_A=1.0\times {10}^6\hat{R}\raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{$m$}\right.$shown in $FIGUREEX31.5$.

We need to calculate the magnetic field ${\overrightarrow{B}}_B$measured by an earthbound scientist.

Given values:

The electric field created by the scientists on the rocket is${\overrightarrow{E}}_A=1.0\times {10}^6\hat{R}\raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{$m$}\right.$.

The magnetic field created by the scientists on the rocket is${\overrightarrow{B}}_A=1.0\hat{j}T$.

The relative velocity of the rocket with suspect to earth is$v_{AB}=2.0\times {10}^6\hat{i}\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

The relative velocity of the earth with suspect to rocket will be opposite i.e.$v_{BA}=-2.0\times {10}^6\hat{i}\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

We know the formula,

${\overrightarrow{B}}_B={\overrightarrow{B}}_A-\frac{1}{l^2}\left({\overrightarrow{v}}_{BA}\times {\overrightarrow{E}}_A\right)$

Substituting the values of ${\overrightarrow{B}}_A,l,{\overrightarrow{v}}_{BA},{\overrightarrow{E}}_A$, we get:

${\overrightarrow{B}}_B=1.0\hat{j}-\frac{1}{{(3.0\times {10}^6)}^2}(-2.0\times {10}^6\hat{j}\times 1.0\times {10}^6\hat{R})$

${\overrightarrow{B}}_B=0.99\hat{j}T$

The magnetic field measured by an earthbound scientist is ${\overrightarrow{B}}_B=0.99\hat{j}T$.