Q 12-04-063JEE MainJEE Main 2025 (29 Jan, Shift 1)Medium
Consider a long straight wire of 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 anywhere due to the wire will be
Answer: (C) $[a/2,\ 2a]$
Inside: $B = \dfrac{\mu_0Ir}{2\pi a^2}$; outside: $B = \dfrac{\mu_0I}{2\pi r}$. The maximum is at the surface: $B_{max} = \dfrac{\mu_0I}{2\pi a}$.
- Inside, $B \propto r$: half the maximum at $r = a/2$.
- Outside, $B \propto 1/r$: half the maximum at $r = 2a$.
$[a/2,\ 2a]$
Solution by Sreeraj P, M.Sc Physics