From Ampere's circuital law for a long straight wire of circular cross-section carrying a steady current, the variation of magnetic field in the inside and outside region of the wire is
Answer: (D) A linearly increasing function of distance $r$ upto the boundary of the wire and then decreasing one with $\dfrac{1}{r}$ dependence for the outside region.
Take a circular Amperian loop of radius $r$ for a wire of radius $R$ carrying current $I$ uniformly.
Inside ($r < R$): enclosed current $I\dfrac{r^2}{R^2}$, so $B \cdot 2\pi r = \mu_0 I\dfrac{r^2}{R^2} \Rightarrow B = \dfrac{\mu_0 I r}{2\pi R^2} \propto r$.
Outside ($r > R$): $B \cdot 2\pi r = \mu_0 I \Rightarrow B = \dfrac{\mu_0 I}{2\pi r} \propto \dfrac{1}{r}$.
Solution by Sreeraj P, M.Sc Physics