Question

An electron enters perpendicularly to a uniform magnetic field $0.1N/amp-m$ at a speed of ${10}^{5}m/s$. Calculate the Lorentz force on the electron.

Answer 1

Given:- $B=0.1N/amp-m$

$v={10}^{5}m/s$

$q_e=1.6\times {10}^{-19}c$

$\theta ={90}^o=$Angle between velocity and magnetic field

Lorentz force on electron is given by,

$\overrightarrow{F}=\overrightarrow{F_E}+\overrightarrow{F_M}$

where $\overrightarrow{F_E}=0=$Force due to electric field ....(because there is no electric field)

$\overrightarrow{F}=0+\overrightarrow{F_M}=\overrightarrow{F_M}$

Where $F_M=$Magnetic force on electron,

and $\overrightarrow{F_M}=q_{e}v\times B=q_{e}vBsin\left(\theta \right)=q_{e}vBsin\left({90}^o\right)=1.6\times {10}^{-19}\times {10}^5\times 0.1\times 1$

$\therefore \overrightarrow{F}=\overrightarrow{F_M}=1.6\times {10}^{-15}N$