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Question: Answered & Verified by Expert
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 :
PhysicsMagnetic Effects of CurrentJEE MainJEE Main 2019 (11 Jan Shift 2)
Options:
  • A $(0.35) \mathrm{q}$
  • B $5 \mathrm{q}$
  • C (2.5) $\mathrm{q}$
  • D $(0.15) \mathrm{q}$
Solution:
2511 Upvotes Verified Answer
The correct answer is: $5 \mathrm{q}$
Resultant force on the charged particle

$$

\begin{array}{l}

=\overrightarrow{\mathrm{F}}_{\text {net }}=\mathrm{q} \overrightarrow{\mathrm{E}}+\mathrm{q}(\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}}) \\

=(2 \mathrm{q} \hat{\mathrm{i}}+3 \mathrm{q} \hat{\mathrm{j}})+\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})

\end{array}

$$

Also $\mathrm{q}(\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}})=0$

So, required work done

$$

\begin{aligned}

\mathrm{W} &=\overrightarrow{\mathrm{F}}_{\mathrm{net}} \cdot \overrightarrow{\mathrm{S}}=(2 \mathrm{q} \hat{\mathrm{i}}+3 \mathrm{q} \hat{\mathrm{j}}) \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}) \\

&=2 \mathrm{q}+3 \mathrm{q} \\

&=5 \mathrm{q}

\end{aligned}

$$

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