Maximum possible magnetic field is at the surface $\begin{aligned}
& \mathrm{B}{\max }=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{a}} \
& \frac{\mathrm{~B}{\max }}{2}=\frac{\mu_0 \mathrm{I}}{4 \pi \mathrm{a}}
\end{aligned}ItcanbeobtainedinsideaswellasoutsidethewireForinside,\begin{aligned}
& \frac{\mu_0 \mathrm{I}}{4 \pi \mathrm{a}}=\frac{\mu_0 \mathrm{Ir}}{2 \pi \mathrm{a}^2} \
& \Rightarrow \mathrm{r}=\frac{\mathrm{a}}{2}
\end{aligned}Foroutside\begin{aligned}
& \frac{\mu_0 \mathrm{I}}{4 \pi \mathrm{a}}=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}} \
& \Rightarrow \mathrm{r}=2 \mathrm{a}
\end{aligned}Correctanswer\left[\frac{\mathrm{a}}{2}, 2 \mathrm{a}\right]$
Consider a long straight wire of a circular cross-section (radius a) carrying a steady current I. The current is uniformly distributed across this…
JEE Main 2025 — Physics Electromagnetism
2025mcqmedium
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
Official previous-year question
Held on 29 Jan 2025 · Verified 6 Jul 2026.
Options
- A
[a/4,3a/2]
- B
[a/4,2a]
- C
[a/2,2a]
- D
[a/2,3a]
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