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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 :
Options:
$$
\overrightarrow{\mathrm{E}}=2 \hat{i}+3 \hat{j} ; \overrightarrow{\mathrm{B}}=4 \hat{j}+6 \hat{k}
$$
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 :
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
$$
\begin{aligned}
\mathrm{W} &=\overrightarrow{\mathrm{F}}_{\mathrm{net}} \cdot \overrightarrow{\mathrm{S}}=(2 \mathrm{q} \hat{\mathrm{i}}+3 \mathrm{q} \hat{\mathrm{j}}) \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}) \\
&=2 \mathrm{q}+3 \mathrm{q} \\
&=5 \mathrm{q}
\end{aligned}
$$
$$
\begin{array}{l}
=\overrightarrow{\mathrm{F}}_{\text {net }}=\mathrm{q} \overrightarrow{\mathrm{E}}+\mathrm{q}(\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}}) \\
=(2 \mathrm{q} \hat{\mathrm{i}}+3 \mathrm{q} \hat{\mathrm{j}})+\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})
\end{array}
$$
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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