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 $$\frac{a}{2}$$ and $$2a$$ from axis of the wire is :

We need to find the ratio of the magnetic field at distances $$\frac{a}{2}$$ and $$2a$$ from the axis of a long straight wire of radius $$a$$ carrying a steady, uniformly distributed current $$I$$.

Recall that Ampere's law states: $$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}$$ where $$I_{enc}$$ is the current enclosed by the Amperian loop.

Inside the wire at a distance $$r$$ from the axis, the current is uniformly distributed over the cross-section of radius $$a$$. The enclosed current is proportional to the enclosed area: $$I_{enc} = I \cdot \frac{\pi r^2}{\pi a^2} = I \cdot \frac{r^2}{a^2}$$.

Applying Ampere's law with a circular Amperian loop of radius $$r$$ gives $$B \cdot 2\pi r = \mu_0 \cdot I \cdot \frac{r^2}{a^2}$$, so $$B_{inside} = \frac{\mu_0 I r}{2\pi a^2}$$. At $$r = \frac{a}{2}$$, this becomes $$B_1 = \frac{\mu_0 I \cdot \frac{a}{2}}{2\pi a^2} = \frac{\mu_0 I}{4\pi a}$$.

Outside the wire at a distance $$r$$ from the axis, the entire current $$I$$ is enclosed, so Ampere's law gives $$B \cdot 2\pi r = \mu_0 I$$ and hence $$B_{outside} = \frac{\mu_0 I}{2\pi r}$$. At $$r = 2a$$, this becomes $$B_2 = \frac{\mu_0 I}{2\pi \cdot 2a} = \frac{\mu_0 I}{4\pi a}$$.

The ratio of these fields is $$\frac{B_1}{B_2} = \frac{\frac{\mu_0 I}{4\pi a}}{\frac{\mu_0 I}{4\pi a}} = \frac{1}{1}$$, so $$B_1 : B_2 = 1 : 1$$.

The correct answer is Option (2): 1 : 1.