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In the presence of magnetic field ' $B$ ' and electric field ' $E$ ', the total force on a moving charged particle is
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Verified Answer
The correct answer is:
$\overrightarrow{\mathrm{F}}=\mathrm{q}[(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})+\overrightarrow{\mathrm{E}}]$
Lorentz force on a charged particle in presence of magnetic and electic field is
$\begin{array}{l}
\overrightarrow{\mathrm{F}}=\overrightarrow{\mathrm{F}}_{\mathrm{e}}+\overrightarrow{\mathrm{F}}_{\mathrm{m}} \\
\Rightarrow \overrightarrow{\mathrm{F}}=\mathrm{q} \overrightarrow{\mathrm{E}}+\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})
\end{array}$
$\begin{array}{l}
\overrightarrow{\mathrm{F}}=\overrightarrow{\mathrm{F}}_{\mathrm{e}}+\overrightarrow{\mathrm{F}}_{\mathrm{m}} \\
\Rightarrow \overrightarrow{\mathrm{F}}=\mathrm{q} \overrightarrow{\mathrm{E}}+\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})
\end{array}$
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