Derived force on moving charge in uniform magnetic field with velocity $\overrightarrow {{v_d}} $.

Current induced from moving charge $q$ with velocity $\vec{v}$ form uniform conducting cross sectional

wire,

${l}\mathrm{I}=n \mathrm{~A} v q$

$\therefore \mathrm{I} d \vec{l}=n \mathrm{~A} v q d \vec{l}$

$\text { Magnetic force acting on } \mathrm{I} d \vec{l} \text { small current carrying element in uniform magnetic field } \overrightarrow{\mathrm{B}} \text { is }$

$\text { given as, }$

$\qquad d \overrightarrow{\mathrm{F}}=\mathrm{I} d \vec{l} \times \overrightarrow{\mathrm{B}}$

$\quad=n \mathrm{~A} \vec{v} q d l \times \overrightarrow{\mathrm{B}}$

$\therefore d \overrightarrow{\mathrm{F}}=n \mathrm{~A} q d l(\vec{v} \times \overrightarrow{\mathrm{B}})$$...(1)$

$\text { but } n \mathrm{~A} d l=\text { number density of charge in } d l \text { element. }$

$\text { Magnetic force on } q \text { electric charge, }$

$\overrightarrow{\mathrm{F}_{\mathrm{m}}}=\frac{d \overrightarrow{\mathrm{F}}}{n \mathrm{~A} d l}=\frac{n \mathrm{Adlq}(\vec{v} \times \overrightarrow{\mathrm{B}})}{n \mathrm{Adl}} \text { [From equ. (1)] }$

$\therefore \overrightarrow{\mathrm{F}_{\mathrm{m}}}=q(\vec{v} \times \overrightarrow{\mathrm{B}})$