Physics · Moving Charges and Magnetic Field

JEE Main 2025 — 8 April, Evening Shift — Question 51

Figure shows a current carrying square loop $A B C D$ of edge length is 'a' lying in a plane. If the resistance of the $A B C$ part is $r$ and that of $A D C$ part is $2 r$ , then the magnitude of the resultant magnetic field at centre of the square loop is

Option A:
$\frac{\sqrt{2} \mu_{0} l}{3 \pi a}$
Correct
Option B:
$\frac{\mu_{0} l}{2 \pi a}$
Option C:
$\frac{2 \mu_{0} I}{3 \pi a}$
Option D:
$\frac{3 \pi \mu_{0} l}{\sqrt{2} a}$

Answer: A

Step-by-step solution

$R_{A B C}=r$ $R_{A D C}=2 r$ $i_{1}=\frac{2 I}{3}$

$i_{2}=\frac{l}{3}$

$B_{\text {centre }}=\frac{2\left(\mu_{0} \sqrt{2}\right)}{4 \pi\left(\frac{a}{2}\right)}\left[\frac{2 I}{3}-\frac{I}{3}\right]=\sqrt{2} \frac{\mu_{0} l}{3 \pi a}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 8 April, Evening Shift
Subject: Physics
Chapter: Moving Charges and Magnetic Field
Topic: Magnetic Field Due to Current-Carrying Wire - Biot-Savart Law