Physics · Moving Charges and Magnetic Field

JEE Main 2025 — 28 January, Evening Shift — Question 59

An infinite wire has a circular bend of radius a, and carrying a current $I$ as shown in figure. The magnitude of magnetic field at the origin O of the arc is given by :

Option A:
$\frac{\mu_{0}}{4 \pi} \frac{I}{a}\left[\frac{\pi}{2}+1\right]$
Option B:
$\frac{\mu_{0}}{4 \pi} \frac{\mathrm{I}}{\mathrm{a}}\left[\frac{3 \pi}{2}+1\right]$
Correct
Option C:
$\frac{\mu_{0}}{2 \pi} \frac{I}{a}\left[\frac{\pi}{2}+2\right]$
Option D:
$\frac{\mu_{0}}{4 \pi} \frac{\mathrm{I}}{\mathrm{a}}\left[\frac{3 \pi}{2}+2\right]$

Answer: B

Step-by-step solution

(1)

$\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{a}} \otimes$

$\mathrm{B}_{2}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{i}}{\mathrm{a}}\left(\frac{3 \pi}{2}\right) \otimes$

$B_{3}=0$

$B=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{i}}{\mathrm{a}}\left(1 \frac{3 \pi}{2}\right) \otimes$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 28 January, Evening Shift
Subject: Physics
Chapter: Moving Charges and Magnetic Field
Topic: Magnetic Field Due to Current-Carrying Wire - Biot-Savart Law