Q 12-04-092JEE MainJEE Main 2024 (8 Apr, Shift 2)Easy
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 $\dfrac a2$ and $2a$ from the axis of the wire is:
Answer: (B) $1:1$
Inside: $B = \dfrac{\mu_0Ir}{2\pi a^2}$, so at $r = \dfrac a2$, $B_1 = \dfrac{\mu_0I}{4\pi a}$.
Outside: $B = \dfrac{\mu_0I}{2\pi r}$, so at $r = 2a$, $B_2 = \dfrac{\mu_0I}{4\pi a}$.
$B_1 : B_2 = 1 : 1$.
Solution by Sreeraj P, M.Sc Physics