Lorentz Force
Category : JEE Main & Advanced
When the moving charged particle is subjected simultaneously to both electric field \[\overrightarrow{E}\] and magnetic field \[\overrightarrow{B}\], the moving charged particle will experience electric force \[\overrightarrow{{{F}_{e}}}=q\overrightarrow{E}\] and magnetic force \[\overrightarrow{{{F}_{m}}}=q(\overrightarrow{v}\times \overrightarrow{B})\]; so the net force on it will be \[\overrightarrow{F}=q\mathbf{[}\overrightarrow{E}+\mathbf{(}\overrightarrow{v\,}\times \overrightarrow{B}\mathbf{)]}\]. Which is the famous 'Lorentz-force equation'.
Depending on the directions of \[\overrightarrow{v},\,E\] and \[\overrightarrow{B}\] following situations are possible
(i) When \[\overrightarrow{v},\,\overrightarrow{E}\] and \[\overrightarrow{B}\] all the three are collinear : In this situation the magnetic force on it will be zero and only electric force will act and so \[\vec{a}=\frac{{\vec{F}}}{m}=\frac{q\vec{E}}{m}\]
(ii) The particle will pass through the field following a straight-line path (parallel field) with change in its speed. So in this situation speed, velocity, momentum and kinetic energy all will change without change in direction of motion as shown
(iii) \[\overrightarrow{v\,},\,\overrightarrow{E}\] and \[\overrightarrow{B}\] are mutually perpendicular : In this situation if \[\overrightarrow{E}\] and \[\overrightarrow{B}\] are such that \[\overrightarrow{F}=\overrightarrow{{{F}_{e}}}+\overrightarrow{{{F}_{m}}}=0\] i.e., \[\overrightarrow{a}=(\overrightarrow{F}/m)=0\]
as shown in figure, the particle will pass through the field with same velocity, without any deviation in path.
And in this situation, as \[{{F}_{e}}={{F}_{m}}\] i.e., \[qE=qvB\] \[v=E/B\]
This principle is used in 'velocity-selector' to get a charged beam having a specific velocity.
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