Consider a long straight wire of a circular cross-section (radius a) carrying a steady current I. The current is uniformly distributed across this cross-section. The distances from the centre of the wire’s cross-section at which the magnetic field [inside the wire, outside the wire] is half of the maximum possible magnetic field, any where due to the wire, will be :

<p>Maximum possible magnetic field is at the surface</p> <p>$$\begin{aligned} & \mathrm{B}_{\max }=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{a}} \\ & \frac{\mathrm{~B}_{\max }}{2}=\frac{\mu_0 \mathrm{I}}{4 \pi \mathrm{a}} \end{aligned}$$</p> <p>It can be obtained inside as well as outside the wire</p> <p>For inside,</p> <p>$$\begin{aligned} & \frac{\mu_0 \mathrm{I}}{4 \pi \mathrm{a}}=\frac{\mu_0 \mathrm{Ir}}{2 \pi \mathrm{a}^2} \\ & \Rightarrow \mathrm{r}=\frac{\mathrm{a}}{2} \end{aligned}$$</p> <p>For outside</p> <p>$$\begin{aligned} & \frac{\mu_0 \mathrm{I}}{4 \pi \mathrm{a}}=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}} \\ & \Rightarrow \mathrm{r}=2 \mathrm{a} \end{aligned}$$</p> <p>Correct answer $\left[\frac{\mathrm{a}}{2}, 2 \mathrm{a}\right]$</p>