Physics · Moving Charges and Magnetic Field

JEE Main 2025 — 2 April, Morning Shift — Question 56

Let $B_{1}$ be the magnitude of magnetic field at centre of a circular coil or radius $R$ carrying current $I$ . Let $B_{2}$ be the magnitude of magnetic field at an axial distance ' $x$ ' from the center. For $x: R=4, \frac{B_{2}}{B_{1}}$ is :

Option A:
$16: 25$
Option B:
$25: 16$
Option C:
$64: 125$
Correct
Option D:
$4: 5$

Answer: C

Step-by-step solution

$B_{1}=\frac{\mu_{0} i}{2 R}$

$x=\frac{3 R}{4}$

$B_{2}=\frac{\mu_{0} i R^{2}}{2\left(R^{2}+x^{2}\right)^{3 / 2}}=\frac{\mu_{0} i R^{2}}{2\left(\frac{5 R}{4}\right)^{3}}$

$=\frac{64}{125}\left(\frac{\mu_{0} i}{2 R}\right)$

$\frac{B_{2}}{B_{1}}=\frac{64}{125}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 2 April, Morning Shift
Subject: Physics
Chapter: Moving Charges and Magnetic Field
Topic: Magnetic Field Due to Current-Carrying Wire - Biot-Savart Law