Question

You have a 1.0-m-long copper wire. You want to make an N-turn current loop that generates a 1.0 mT magnetic field at the center when the current is 1.0 A. You must use the entire wire. What will be the diameter of your coil?

Short answer

$2.0\text{cm}$.

Step-by-step solution

Step 1. Given information

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$1.0\text{mT}$.

Step 2. Calculation

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

$l=\pi Nd........................\left(1\right)$

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" $N=\frac{l}{\pi d}......................\left(2\right)$

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

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

Here, $B$is the magnetic field, ${\mu }_0$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.

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

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

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

Step 3. Final answer

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