A thick long straight wire of radius 'a' is carrying steady current I…

The correct answer is Option (3) → 3:4

Given:

Radius of wire $= a$

Current $= I$ (uniform distribution)

Magnetic field inside the wire ($r < a$):

$B = \frac{\mu_0 I r}{2 \pi a^2}$

At $r = \frac{a}{2}$:

$B_{in} = \frac{\mu_0 I \cdot (a/2)}{2 \pi a^2} = \frac{\mu_0 I}{4 \pi a}$

Magnetic field outside the wire ($r > a$):

$B = \frac{\mu_0 I}{2 \pi r}$

At $r = \frac{3a}{2}$:

$B_{out} = \frac{\mu_0 I}{2 \pi (3a/2)} = \frac{\mu_0 I}{3 \pi a}$

Ratio:

$\frac{B_{in}}{B_{out}} = \frac{\frac{\mu_0 I}{4 \pi a}}{\frac{\mu_0 I}{3 \pi a}} = \frac{3}{4}$

Final Answer: Ratio $= \frac{3}{4}$