There are two infinitely long straight current-carrying conductors and they are held at right angles to each other so that their common ends meet at the origin as shown in the figure given below. The ratio of current in both conductors is $1:1$. The magnetic field at point $P$ is$_________.$ ![JEE Main 2021 Physics, Electromagnetic Waves — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Electromagnetic_Waves/648b5a69417cc3fb48d66e1b/question_1__q_648b5a69417cc3fb48d66e1b__cdn-question-pool.getmarks.app__90babf89-cd02-40fb-bdec-78468dd7066d-image__95e5fc549d_final_ppt_sync.png)

There are two infinitely long… - JEE Main 2021 Physics PYQ

JEE Main 2021 — Physics Electromagnetism

2021mcqeasy

There are two infinitely long straight current-carrying conductors and they are held at right angles to each other so that their common ends meet at the origin as shown in the figure given below. The ratio of current in both conductors is $1:1$ . The magnetic field at point $P$ is $_________.$

JEE Main 2021 Physics, Electromagnetic Waves — question figure

Official previous-year question

Held on 1 Sept 2021 · Verified 6 Jul 2026.

Options

A
$\frac{{\mu }_{0}I}{4\pi xy}[\sqrt{{x}^{2}+{y}^{2}}-(x+y)]$
B
$\frac{{\mu }_{0}Ixy}{4\pi }[\sqrt{{x}^{2}+{y}^{2}}-(x+y)]$
C
$\frac{{\mu }_{0}I}{4\pi xy}[\sqrt{{x}^{2}+{y}^{2}}+(x+y)]$
D
$\frac{{\mu }_{0}Ixy}{4\pi }[\sqrt{{x}^{2}+{y}^{2}}+(x+y)]$

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Solution

$B=\frac{{\mu }_{0}I}{4\pi y}(1+\mathrm{sin}{\theta }_{1})+\frac{{\mu }_{0}I}{4\pi x}(1+\mathrm{sin}{\theta }_{2})$

$B=\frac{{\mu }_{0}I}{4\pi y}(1+\frac{x}{\sqrt{{x}^{2}+{y}^{2}}})+\frac{{\mu }_{0}I}{4\pi x}(1+\frac{y}{\sqrt{{x}^{2}+{y}^{2}}})$

$B=\frac{{\mu }_{0}I}{4\pi x}+\frac{{\mu }_{0}I}{4\pi y}+\frac{{\mu }_{0}I}{4\pi }[\frac{x}{y\sqrt{{x}^{2}+{y}^{2}}}+\frac{y}{x\sqrt{{x}^{2}+{y}^{2}}}] \Rightarrow B=\frac{{\mu }_{0}I}{4\pi }[\frac{1}{x}+\frac{1}{y}]+\frac{{\mu }_{0}I}{4\pi }\frac{\sqrt{{x}^{2}+{y}^{2}}}{xy}=\frac{{\mu }_{0}I}{4\pi }[\frac{x+y+\sqrt{{x}^{2}+{y}^{2}}}{xy}]$

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Options

$\frac{{\mu }_{0}I}{4\pi xy}[\sqrt{{x}^{2}+{y}^{2}}-(x+y)]$

$\frac{{\mu }_{0}Ixy}{4\pi }[\sqrt{{x}^{2}+{y}^{2}}-(x+y)]$

$\frac{{\mu }_{0}I}{4\pi xy}[\sqrt{{x}^{2}+{y}^{2}}+(x+y)]$

$\frac{{\mu }_{0}Ixy}{4\pi }[\sqrt{{x}^{2}+{y}^{2}}+(x+y)]$

Solution

$B=\frac{{\mu }_{0}I}{4\pi y}(1+\mathrm{sin}{\theta }_{1})+\frac{{\mu }_{0}I}{4\pi x}(1+\mathrm{sin}{\theta }_{2})$

$B=\frac{{\mu }_{0}I}{4\pi y}(1+\frac{x}{\sqrt{{x}^{2}+{y}^{2}}})+\frac{{\mu }_{0}I}{4\pi x}(1+\frac{y}{\sqrt{{x}^{2}+{y}^{2}}})$

Options

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Solution

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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