1. Magnetic field due to a long current-carrying wire

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

• Inside the wire $(r<a)$, $B_{\text{in}}(r)=\frac{\mu_0 I r}{2\pi a^2}$ since enclosed current is proportional to area.

• Outside the wire $(r\ge a)$, $B_{\text{out}}(r)=\frac{\mu_0 I}{2\pi r}$

2. Find the maximum magnetic field anywhere

Inside, $B_{\text{in}}\propto r$, so it increases linearly with $r$.

Outside, $B_{\text{out}}\propto \dfrac{1}{r}$, so it decreases with $r$.

Hence the magnetic field is maximum at the surface $r=a$.

So, $B_{\max}=B(a)=\frac{\mu_0 I}{2\pi a}$

We need the positions where magnetic field is half of this maximum value: $B=\frac{1}{2}B_{\max}=\frac{1}{2}\cdot \frac{\mu_0 I}{2\pi a}=\frac{\mu_0 I}{4\pi a}$

3. Inside the wire: find $r$ such that $B_{\text{in}}=\dfrac{1}{2}B_{\max}$

$\frac{\mu_0 I r}{2\pi a^2}=\frac{\mu_0 I}{4\pi a}$

Cancel common factors $\mu_0 I$ and $\pi$: $\frac{r}{2a^2}=\frac{1}{4a}$

Multiply both sides by $2a^2$: $r=\frac{a}{2}$

So, inside the wire, the distance is $\boxed{\frac{a}{2}}$

4. Outside the wire: find $r$ such that $B_{\text{out}}=\dfrac{1}{2}B_{\max}$

$\frac{\mu_0 I}{2\pi r}=\frac{\mu_0 I}{4\pi a}$

Cancel common factors: $\frac{1}{2r}=\frac{1}{4a}$

Cross-multiplying: $4a=2r$ $r=2a$

So, outside the wire, the distance is $\boxed{2a}$

5. Match with options

Thus the required pair $[\text{inside},\text{outside}]$ is $\left[\frac{a}{2},\,2a\right]$

This matches:

Option C

6. Comparison with stored answer

Stored correct answer: C

My derived answer: C

They agree.