Physics · Moving Charges and Magnetic Field

JEE Main 2025 — 29 January, Morning Shift — Question 30

Consider a long straight wire of a circular cross-section (radius a) carrying a steady current I. The current is uniformly distributed across this cross-section. The distances from the centre of the wire's cross-section at which the magnetic field [inside the wire, outside the wire] is half of the maximum possible magnetic field, any where due to the wire, will be

Option A:
$[a / 4,3 a / 2]$
Option B:
$[\mathrm{a} / 2,2 \mathrm{a}]$
Correct
Option C:
$[a / 2,3 a]$
Option D:
$[a / 2,3 a]$

Answer: B

Step-by-step solution

Maximum possible magnetic field is at the surface

$B_{\max }=\frac{\mu_{0} I}{2 \pi a}$ $\frac{\mathrm{B}_{\text {max }}}{2}=\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{a}}$

It can be obtained inside as well as outside the wire For inside,

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

$\Rightarrow \mathrm{r}=\frac{\mathrm{a}}{2}$ For outside

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

$\Rightarrow \mathrm{r}=2 \mathrm{a}$ Correct answer

$\left[\frac{\mathrm{a}}{2}, 2 \mathrm{a}\right]$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 29 January, Morning Shift
Subject: Physics
Chapter: Moving Charges and Magnetic Field
Topic: Magnetic Field Due to Current-Carrying Wire - Biot-Savart Law