A long straight wire of circular cross-section (radius a) carries a uniformly distributed steady current I. For a point inside the wire at distance r < a, the magnetic field varies as:

1. For $r < a$, the enclosed current is the fraction in area $\pi r^2$: $I_e = I\dfrac{r^2}{a^2}$. 2. Ampère's law: $B(2\pi r) = \mu_0 I_e = \mu_0 I\dfrac{r^2}{a^2}$. 3. Solve: $B = \dfrac{\mu_0 I}{2\pi a^2}\,r$. 4. With $a$ fixed, $B \propto r$ (the $1/r$ behaviour applies only outside, for $r > a$). _Source: NCERT Class 12 Physics Ch 4 "Moving Charges and Magnetism", p.151_