Question

A solenoid $\mathbf{1}\mathbf{.}\mathbf{30}$ long and$\mathbf{2}\mathbf{.}\mathbf{60}\mathbf{}\mathbf{cm}$ in diameter carries a current of $\mathbf{1}\mathbf{.}\mathbf{80}\mathbf{}\mathbf{A}$. The magnetic field inside the solenoid is $\mathbf{23}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{mT}$ . Find the length of the wire forming the solenoid.

Short answer

1. The length of the wire forming the solenoid is $108\mathrm{m}$.

Step-by-step solution

Listing the given quantities 

$\mathrm{l}=1.30\mathrm{m}$

$\mathrm{d}=2.60\mathrm{cm}=0.0260\mathrm{m}$

$\mathrm{B}=23\mathrm{mT}=0.023\mathrm{T}$

$\mathrm{i}=18\mathrm{A}$

Understanding the concept of magnetic field and solenoid

We find the number of turns of the solenoid. Using the number of turns, we can find the total length of the wire used in making the solenoid.

$\mathrm{B}={\mathrm{\mu }}_0\mathrm{in}={\mathrm{\mu }}_0\mathrm{i}\left(\frac{\mathrm{N}}{\mathrm{l}}\right)$

Calculations of the length of the wire forming the solenoid  

The number of turns of the solenoid is

$\mathrm{B}={\mathrm{\mu }}_0\mathrm{i}\left(\frac{\mathrm{N}}{\mathrm{l}}\right)$

$\frac{\mathrm{N}}{\mathrm{l}}=\frac{\mathrm{B}}{{\mathrm{\mu }}_0\mathrm{i}}$

$\begin{array}{c}\mathrm{N}=\frac{\mathrm{Bl}}{{\mathrm{\mu }}_0\mathrm{i}}\\ =\frac{0.023\times 1.30}{1.26\times {10}^{-6}\times 18}\\ \approx 1322\end{array}$

As we know that the length of the circular loop is the circumference of that loop. So, the total length of the solenoid is

$\mathrm{L}=2\mathrm{\pi rN}$

Since $\mathrm{r}=\frac{\mathrm{d}}{2}$

Thus,

$\begin{array}{c}\mathrm{L}=2\mathrm{\pi }\times \frac{0.0260}{2}\times 1322\\ =\mathrm{\pi }\times 0.0260\times 1322\\ =107.98\\ \approx 108\mathrm{m}\end{array}$

The length of the wire forming the solenoid is$108\mathrm{m}$ .