Question

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?

Short answer

Radius of the wire$R=5.33\times {10}^{-3}\text{m}$

Step-by-step solution

Given

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

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

Distance from axis of wire

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

Understanding the concept

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:

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

Calculate radius of the wire R

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 $R=5.33\times {10}^{-3}\text{m}$.