Question 211 of 276

Two long parallel conductors ${S}_{1}$ and ${S}_{2}$ are separated by a distance $10cm$ and carrying currents of $4A$ and $2A$ respectively. The conductors are placed along $x$-axis in $X-Y$ plane. There is a point $P$ located between the conductors (as shown in figure).
A charge particle of $3\pi$ coulomb is passing through the point $P$ with velocity $\vec{v}=\left(2\hat{i}+3\hat{j}\right)m{s}^{-1}$; where $\hat{i}$ & $\hat{j}$ represents unit vector along $x$ & $y$axis respectively.
The force acting on the charge particle is $4\pi \times {10}^{-5}\left(-x\hat{i}+2\hat{j}\right)N$. The value of $x$ is

A$2$B$1$C$3$D$-3$

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Answer: (C) $3$

Net magnetic field at $P$ is ${B}_{Net}=\left(\dfrac{{\mu }_{0}\times {i}_{1}}{2\pi {r}_{1}}-\dfrac{{\mu }_{0}{i}_{2}}{2\pi {r}_{2}}\right)\left(-\hat{k}\right)$
${B}_{Net}=\dfrac{{\mu }_{0}4}{2\pi \times \dfrac{4}{100}}-\dfrac{{\mu }_{0}2}{2\pi \times \dfrac{6}{100}}$
$=\dfrac{{\mu }_{0}}{\pi }\times 100\left[\dfrac{1}{2}-\dfrac{1}{6}\right]$
$=\dfrac{{\mu }_{0}}{\pi }\times 100\left[\dfrac{4}{2\times 6}\right]$
$=\dfrac{{\mu }_{0}\times 100}{3\pi }\left(-\hat{k}\right)$
So, force acting on charge particle is$\vec{F}=q\left(\vec{v}\times \vec{B}\right)=3\pi \left[2\hat{i}+3\hat{j}\right]\dfrac{{\mu }_{0}\times 100}{3\pi }\left(-\hat{k}\right)$
$=4\pi \times {10}^{-7}\left[2\hat{i}+3\hat{j}\right]\times 100\left(-\hat{k}\right)$
$=4\pi \times {10}^{-5}\left[2\hat{j}-3\hat{i}\right]$
$∴x=3$