A long straight wire with a circular cross-section having radius $$R$$, is carrying a steady current $$I$$. The current $$I$$ is uniformly distributed across this cross-section. Then the variation of magnetic field due to current $$I$$ with distance $$r$$ ($$r < R$$) from its centre will be

We need to find how the magnetic field $$B$$ varies with distance $$r$$ from the centre of a long straight wire of radius $$R$$ carrying a uniformly distributed current $$I$$, for $$r < R$$.

By applying Ampere’s circuital law to a circular Amperian loop of radius $$r$$ inside the wire, one writes $$\oint \vec{B}\cdot d\vec{l} = \mu_0 I_{enc}$$. Because the current is uniformly distributed over the cross section, the enclosed current is $$I_{enc} = I \frac{\pi r^2}{\pi R^2} = I \frac{r^2}{R^2}$$. By symmetry, $$B$$ is tangential and constant along the loop, so $$B\,(2\pi r) = \mu_0 I \frac{r^2}{R^2}$$, which yields $$B = \frac{\mu_0 I r}{2\pi R^2}$$.

Hence, inside the wire, the magnetic field varies as $$B \propto r$$. The correct answer is Option A: $$B \propto r$$.