A long straight wire of radius 'a' carries a steady current 'I'. The current is uniformly distributed across its area of cross- section. The ratio of magnitude of magnetic fields $B_1$ at a distance a/2 and $B_2$ at a distance 2a from the axis of the wire.

1 : 1

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

Magnetic field inside the wire (r < a, uniformly distributed current):

$B_\text{inside} = \frac{\mu_0 I r}{2 \pi a^2}$

Magnetic field outside the wire (r ≥ a):

$B_\text{outside} = \frac{\mu_0 I}{2 \pi r}$

Given:

$B_1$ at $r = a/2$ (inside): $B_1 = \frac{\mu_0 I (a/2)}{2 \pi a^2} = \frac{\mu_0 I}{4 \pi a}$

$B_2$ at $r = 2a$ (outside): $B_2 = \frac{\mu_0 I}{2 \pi (2a)} = \frac{\mu_0 I}{4 \pi a}$

Ratio: $\frac{B_1}{B_2} = \frac{\mu_0 I / 4 \pi a}{\mu_0 I / 4 \pi a} = 1$

Final Answer: $B_1 : B_2 = 1 : 1$