Question

An electron travels with $\overrightarrow{v}=5.0\times {10}^6\hat{i}\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$through a point in space where $\overrightarrow{E}=(2.0\times {10}^5\hat{i}-2.0\times {10}^5\hat{j})\raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{$m$}\right.$and $\overrightarrow{B}=-0.10\hat{k}T$. What is the force on the electron?

Short answer

The force on the electron will be$27\times {10}^{-17}N$.

Step-by-step solution

Given information

We have been given that an electron travels with $\overrightarrow{v}=5.0\times {10}^6\hat{i}\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$through a point in space where $\overrightarrow{E}=(2.0\times {10}^5\hat{i}-2.0\times {10}^5\hat{j})\raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{$m$}\right.$and localid="1650176818409" $\overrightarrow{B}=-0.10\hat{k}T$.

We need to calculate the force on the electron.

Simplify

According to Lorentz force :$F=q(E+v\times B)$

Firstly we have to calculate the value of $v\times B$.

$$\begin{gathered} v\times B=5\times {10}^6\hat{i}\times (-0.10\hat{k}) \\ v\times B=-5\times {10}^5\hat{j} \end{gathered}$$

Now,

$$\begin{gathered} F=q(E+v\times B) \\ F=q(2\times {10}^5\hat{i}-2\times {10}^5\hat{j}-5.0\times {10}^5\hat{j}) \\ F=1.6\times {10}^{-19}(2\times {10}^5\hat{i}-7\times {10}^5\hat{j}) \\ F=1.6\times {10}^{-19}\times 1700 \\ F=27\times {10}^{-17}N \end{gathered}$$