Physics · Moving Charges and Magnetic Field

JEE Main 2024 — 8 April, Shift 2 — Question 37

A long straight wire of radius a carries a steady current I. The current is uniformly distributed across its cross section. The ratio of the magnetic field at $\frac{\mathrm{a}}{2}$ and 2 a from axis of the wire is :

Option A:
$1: 4$
Option B:
$4: 1$
Option C:
$1: 1$
Correct
Option D:
$3: 4$

Answer: C

Step-by-step solution

$B_{1} 2 \pi \frac{a}{2}=\mu_{\mathrm{o}} \frac{\mathrm{I}}{4}$ $B_{1}=\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{a}}$

$\mathrm{B}_{2} 2 \pi 2 \mathrm{a}=\mu_{0} \mathrm{I}$

$B_{2}=\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{a}}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 8 April, Shift 2
Subject: Physics
Chapter: Moving Charges and Magnetic Field
Topic: Magnetic Field Due to Current-Carrying Wire - Biot-Savart Law