Question

An electron is moving with speed $2\times {10}^{5}m/s$ along the positive $x$-direction in the presence of magnetic induction $\overline{B}=(\overline{i}+4\overline{j}-3\overline{k})T$. The magnitude of the force experienced by the electron in newtons is $(e=1.6\times {10}^{-19}C)$

• A. $18\times {10}^{-13}$
• B. $1.28\times {10}^{-13}$
• C. $1.6\times {10}^{-13}$
• D. $1.73\times {10}^{-13}$

Answer 1

The correct option is C $1.6\times {10}^{-13}$

The force experience by a electron of charge $q$ and velocity $v$ moving in a magnetic field $B$ is $\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$. So the magnitude of the force is $1.6\times {10}^{-13}$.