JEE Main 2022 — Physics Electromagnetism

2022mcqeasy

Two long parallel conductors ${S}_{1}$ and ${S}_{2}$ are separated by a distance $10\mathrm{cm}$ 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 $\vec{v}=(2\hat{i}+3\hat{j})m{s}^{-1}$ ; where $\hat{i}$ & $\hat{j}$ represents unit vector along $x$ & $y$ axis respectively.

The force acting on the charge particle is $4\pi \times {10}^{-5}(-x\hat{i}+2\hat{j})N$ . The value of $x$ is

JEE Main 2022 Physics, Magnetic Effects of Current — question figure

Official previous-year question

Held on 27 Jun 2022 · Verified 6 Jul 2026.

Options

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

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Solution

JEE Main 2022 Physics, Magnetic Effects of Current — solution figure

Net magnetic field at $P$ is ${B}_{\mathrm{Net}}=(\frac{{\mu }_{0}\times {i}_{1}}{2\pi {r}_{1}}-\frac{{\mu }_{0}{i}_{2}}{2\pi {r}_{2}})(-\hat{k})$

${B}_{\mathrm{Net}}=\frac{{\mu }_{0}4}{2\pi \times \frac{4}{100}}-\frac{{\mu }_{0}2}{2\pi \times \frac{6}{100}}$

$=\frac{{\mu }_{0}}{\pi }\times 100[\frac{1}{2}-\frac{1}{6}]$

$=\frac{{\mu }_{0}}{\pi }\times 100[\frac{4}{2\times 6}]$

$=\frac{{\mu }_{0}\times 100}{3\pi }(-\hat{k})$

So, force acting on charge particle is $\vec{F}=q(\vec{v}\times \vec{B})=3\pi [2\hat{i}+3\hat{j}]\frac{{\mu }_{0}\times 100}{3\pi }(-\hat{k})$

$=4\pi \times {10}^{-7}[2\hat{i}+3\hat{j}]\times 100(-\hat{k})$

$=4\pi \times {10}^{-5}[2\hat{j}-3\hat{i}]$

$\therefore x=3$

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Solution

JEE Main 2022 Physics, Magnetic Effects of Current — solution figure

Net magnetic field at $P$ is ${B}_{\mathrm{Net}}=(\frac{{\mu }_{0}\times {i}_{1}}{2\pi {r}_{1}}-\frac{{\mu }_{0}{i}_{2}}{2\pi {r}_{2}})(-\hat{k})$

${B}_{\mathrm{Net}}=\frac{{\mu }_{0}4}{2\pi \times \frac{4}{100}}-\frac{{\mu }_{0}2}{2\pi \times \frac{6}{100}}$

$=\frac{{\mu }_{0}}{\pi }\times 100[\frac{1}{2}-\frac{1}{6}]$

$=\frac{{\mu }_{0}}{\pi }\times 100[\frac{4}{2\times 6}]$

$=\frac{{\mu }_{0}\times 100}{3\pi }(-\hat{k})$

So, force acting on charge particle is $\vec{F}=q(\vec{v}\times \vec{B})=3\pi [2\hat{i}+3\hat{j}]\frac{{\mu }_{0}\times 100}{3\pi }(-\hat{k})$

$=4\pi \times {10}^{-7}[2\hat{i}+3\hat{j}]\times 100(-\hat{k})$

$=4\pi \times {10}^{-5}[2\hat{j}-3\hat{i}]$

$\therefore x=3$

JEE Main 2022 — Physics Electromagnetism

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JEE Main 2022 — Physics Electromagnetism

Options

Did you get this right?

Solution

Your note

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Electromagnetism