1. Magnetic field inside a long current-carrying wire

For a wire of radius $a$ carrying total current $I$ uniformly distributed over its cross-section:

• Current density, $J = \frac{I}{\pi a^2}$

At a distance $r<a$ from the axis, the enclosed current is $I_{\text{enc}} = J \cdot \pi r^2 = I\frac{r^2}{a^2}$

Using Ampere’s law, $B(2\pi r)=\mu_0 I_{\text{enc}}$

So, $B=\frac{\mu_0 I_{\text{enc}}}{2\pi r}=\frac{\mu_0 I}{2\pi r}\cdot \frac{r^2}{a^2} = \frac{\mu_0 I r}{2\pi a^2}$

At $r=\frac{a}{2}$, $B_{a/2}=\frac{\mu_0 I}{2\pi a^2}\cdot \frac{a}{2}=\frac{\mu_0 I}{4\pi a}$

2. Magnetic field outside the wire

For $r>a$, the entire current $I$ is enclosed, so $B=\frac{\mu_0 I}{2\pi r}$

At $r=2a$, $B_{2a}=\frac{\mu_0 I}{2\pi(2a)}=\frac{\mu_0 I}{4\pi a}$

3. Find the ratio

$B_{a/2}:B_{2a} = \frac{\mu_0 I}{4\pi a} : \frac{\mu_0 I}{4\pi a} = 1:1$

4. Option check

• A: $4:1$ ❌
• B: $3:4$ ❌
• C: $1:1$ ✅
• D: $1:4$ ❌

Therefore, the correct answer is C.