Magnetic Effects & Magnetism

JEE Main level, single correct, medium

161 of 276

Question 161 of 276

A particle of mass mm and charge qq is in an electric and magnetic field given by

E→=2i^+3j^;B→=4j^+6k^\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 P(x=1;y=1)\mathrm{P}(x=1 ; y=1) along a straight path. The magnitude of the total work done is :

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Answer: (B) 5q5 \mathrm{q}

Resultant force on the charged particle

=F→net =qE→+q(V→×B→)=(2qi^+3qj^)+q(v→×B→)\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 q(V→×B→)=0\mathrm{q}(\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}})=0

So, required work done

W=F→net⋅S→=(2qi^+3qj^)⋅(i^+j^)=2q+3q=5q\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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