A particle of mass $m$ and charge $q$ is in an electric and magnetic field given by
$\overrightarrow{\mathrm{E}}=2 \hat{i}+3 \hat{j} ; \overrightarrow{\mathrm{B}}=4 \hat{j}+6 \hat{k}$
The charged particle is shifted from the origin to the point $\mathrm{P}(x=1 ; y=1)$ along a straight path. The magnitude of the total work done is :
Correct Option: , 2
(2) Resultant force on the charged particle
$=\overrightarrow{\mathrm{F}}_{\text {net }}=\mathrm{q} \overrightarrow{\mathrm{E}}+\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})$
$=(2 q \hat{i}+3 q \hat{j})+q(\vec{v} \times \vec{B})$
Also $\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})=0$
So, required work done\
$\mathrm{W}=\overrightarrow{\mathrm{F}}_{\text {net }} \overrightarrow{\mathrm{S}}=(2 \mathrm{q} \hat{\mathrm{i}}+3 \hat{\mathrm{q}}) \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}})$
$=2 q+3 q$
$=5 q$
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