A long wire is known to have a radius greater than$\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{}\mathbf{\text{mm}}$and to carry a current that is uniformly distributed over its cross section. The magnitude of the magnetic field due to that current is$\mathbf{0}\mathbf{.}\mathbf{28}\mathbf{}\mathbf{}\mathbf{\text{mT}}$at a point$\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{mm}}$from the axis of the wire, and$\mathbf{0}\mathbf{.}\mathbf{20}\mathbf{}\mathbf{mT}$at a point 10 mm from the axis of the wire. What is the radius of the wire?

Magnetic field inside wire at distance $r_1$, $B_i=0.28\text{mT}=0.28\times {10}^{-3}\text{T}$

Magnetic field outside wire at distance $r_2$, $B_o=0.20\text{mT}=0.20\times {10}^{-3}\text{T}$

Distance from axis of wire

$\begin{array}{rcl}{r}_1& =& 4.00\text{mm}\\ & =& 4\times {10}^{-3}\text{m}\end{array}$

$\begin{array}{rcl}{r}_2& =& 10.00\text{mm}\\ & =& 10\times {10}^{-3}\text{m}\end{array}$

We will use Amperes law to find the magnetic fields inside and outside the wire and then from those fields we can find the radius of the wire

Formula:

$\oint \overrightarrow{B}·\overrightarrow{ds}={\mu }_0{I}_{enclosed}$

Let R be the radius of wire and I be the current through the wire

Magnetic field inside wire at distance$r_1$ is

$\oint \overrightarrow{B}·\overrightarrow{ds}={\mu }_0{I}_{enclosed}$

$B\oint ds={\mu }_0\times I\left(\frac{\pi r_1^2}{\pi R^2}\right)$

Which gives

$B={\mu }_0\times \frac{I\left(\frac{\pi r_1^2}{\pi R^2}\right)}{2\pi r_1}=\frac{{\mu }_{0}I{r}_1}{2\pi R^2}$

Now Magnetic field outside wire at distance$r_2$ is

$B\oint ds={\mu }_0\times I$

Which gives

$B={\mu }_{0}I=\frac{{\mu }_{0}I}{2\pi r_2}=\frac{2{\mu }_{0}I}{4\pi r_2}$

Using this expression

$\begin{array}{rcl}I& =& \frac{4\pi r_{2}B}{2{\mu }_0}\\ & =& \frac{10\times {10}^{-3}\text{m}\times 0.20\times {10}^{-3}\mathrm{T}}{2\times {10}^{-7}\mathrm{H}/\mathrm{m}}\\ & =& 10\text{A}\end{array}$

Consider,

$B=\frac{{\mu }_{0}I{r}_1}{2\pi R^2}$

From this we get

$\begin{array}{rcl}{R}^2& =& \frac{{\mu }_{0}I{r}_1}{2\pi B}\\ & =& \frac{2{\mu }_{0}I{r}_1}{4\pi B}\\ & =& \frac{2\times {10}^{-7}\mathrm{H}/\mathrm{m}\times 10\mathrm{A}\times 4\times {10}^{-3}\text{m}}{0.28\times {10}^{-3}\mathrm{T}}\\ & =& 2.85\times {10}^{-5}{\text{m}}^2\end{array}$

$\begin{array}{rcl}R& =& \sqrt{2.85\times {10}^{-5}{\text{m}}^{\text{2}}}\\ R& =& 5.33\times {10}^{-3}\text{m}\end{array}$

The radius of wire is $R=5.33\times {10}^{-3}\text{m}$.