91Ó°ÊÓ

The total length of the copper wire is$1.0\text{m}$, the current through the wire is$1.0\text{A}$and the magnetic field at the center of the coil is$1.0\text{mT}$.

The formula to calculate the total length of the wire can be written as

$l=\mathrm{Ï€}Nd........................\left(1\right)$

Here, $l$is the total length of the wire, $N$i the number of turns when the wire is bent into a circular loop and $d$is the diameter of the loop.

Simplify equation (1) to obtain the number of turns in the loop.

role="math" $N=\frac{l}{\mathrm{Ï€}d}......................\left(2\right)$

The formula to calculate the magnetic field at the center of a circular loop is given by

role="math" $B=\frac{{\mathrm{μ}}_{0}NI}{d}.........................\left(3\right)$

Here, $B$is the magnetic field, ${\mathrm{μ}}_0$is the permeability of free space and $I$is the current through the loop.

Substitute the expression for the number of turns from equation (2) into equation (3) and simplify to obtain the diameter of the loop.

$\begin{array}{rcl}B& =& \frac{{\mathrm{μ}}_{0}I}{d}\left(\frac{l}{\mathrm{π}d}\right)\\ & =& \frac{{\mathrm{μ}}_{0}Il}{\mathrm{π}{d}^2}\\ d& =& \sqrt{\frac{{\mathrm{μ}}_{0}Il}{\mathrm{π}B}}..........................\left(4\right)\end{array}$

Substitute $4\mathrm{π}×{10}^{-7}{\text{N/A}}^2$for ${\mathrm{μ}}_0$, $1.0\text{A}$for $I$, $1.0\text{m}$for $l$and $1.0\text{mT}$for $B$into equation (4) to calculate the required diameter.

$\begin{array}{rcl}d& =& \sqrt{\frac{4\mathrm{π}×{10}^{-7}{\text{N/A}}^2×1.0\text{A}×1.0\text{m}}{\mathrm{π}×1.0\text{mT}×{\displaystyle \frac{1\text{T}}{1000\text{mT}}}}}\\ & =& 0.02\text{m}×\frac{100\text{cm}}{1\text{m}}\\ & =& 2.0\text{cm}\end{array}$

The required diameter of the loop is$2.0\text{cm}$.