JEE Main 202529 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 29 January
Shift: Morning Shift
Subject: Physics
Chapter: Magnetic Effects of Current

JEE Main 2025 · 29 January · Morning Shift · Physics · Magnetic Effects of Current

Consider a long straight wire of a circular cross-section (radius...

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

Options

A $[a / 4,3 a / 2]$
B $[a / 4,2 a]$
C([\mathrm{a} / 2,2 \mathrm{a}])
D $[a / 2,3 a]$

Correct answer

C. ([\mathrm{a} / 2,2 \mathrm{a}])

Step-by-step solution

Maximum possible magnetic field is at the surface

\begin{aligned} & \mathrm{B}_{\max }=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{a}} \\ & \frac{\mathrm{~B}_{\max }}{2}=\frac{\mu_0 \mathrm{I}}{4 \pi \mathrm{a}} \end{aligned}

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

\begin{aligned} & \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} \end{aligned}

For outside

\begin{aligned} & \frac{\mu_0 \mathrm{I}}{4 \pi \mathrm{a}}=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}} \\ & \Rightarrow \mathrm{r}=2 \mathrm{a} \end{aligned}

Correct answer

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

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