Question

Scientists in the laboratory create a uniform electric field $\overrightarrow{E}$=$1.0\times {10}^6$$\hat{k}$V/m in a region of space where$\overrightarrow{B}=\overrightarrow{0}$. What are the fields in the reference frame of a rocket travelling in the positive x-direction at $1.0\times {10}^6$m/s?

Short answer

Required field ${\overrightarrow{B}}_B=1.11\times {10}^{-5}\mathrm{T}$

Step-by-step solution

Given information:

We have been given that (a)$\overrightarrow{E}=1.0\times {10}^6\hat{R}\mathrm{V}/\mathrm{m}$(b)$\overrightarrow{B}=0$(c)$\overrightarrow{v_B}=1.0\times {10}^{6}i\mathrm{m}/\mathrm{s}$

Simplify

The equation for the Electric field is as:

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

But it is known that:

${\overrightarrow{E}}_B={\overrightarrow{E}}_A+{\overrightarrow{v}}_B\times {\overrightarrow{B}}_A={\overrightarrow{E}}_A=1.0\times {10}^6\hat{R}\mathrm{V}/\mathrm{m}$

Therefore:

${\overrightarrow{B}}_B={\overrightarrow{B}}_A-\frac{1}{C^2}{\overrightarrow{V}}_B\times {\overrightarrow{E}}_A$Equation 1

Putting the known values in Equation 1, We get:

${\overrightarrow{B}}_B=\frac{1.0\times {10}^{12}\hat{j}}{{\left(3\times {10}^8\right)}^2}$

${\overrightarrow{B}}_B=1.11\times {10}^{-5}\mathrm{T}$