Question

A solenoid is designed to produce a magnetic field of 0.0270 T at its centre. It has radius 1.40 cm and length 40.0 cm, and the wire can carry a maximum current of 12.0 A. (a) What minimum number of turns per unit length must the solenoid have? (b) What total length of wire is required?

Short answer

A) The minimum number of turns is 1790turns/m b) The total length of wire is 69.0m

Step-by-step solution

Concept of the magnetic field in solenoid

The magnetic field magnitude B at the center of a current-carrying solenoid is$\mathbf{n}\mathbf{=}\frac{\mathbf{B}}{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{I}}\mathbf{\text{where, where}}{\mathit{\mu }}_{\mathbf{0}}\mathbf{=}\mathbf{4}\mathit{\pi }\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{H}\mathbf{/}\mathbf{m}$is permeability of free space, I is the current passing through the coil

Calculate the number of the turns

Given B= 0.0270 T, I = 12.0 A, r = 1.40 cm, and L 40.0 cm. Solving for n and substituting the known values of B and I, we find the number if turns.

$\begin{array}{r}\mathrm{n}=\frac{\mathrm{B}}{{\mu }_0\mathrm{l}}=\frac{0.0270\mathrm{T}}{4\pi \times {10}^{-7}\mathrm{Tm}/\mathrm{A}\times 12.0\mathrm{A}}\\ =1790\text{turns}/\mathrm{m}\end{array}$

Therefore, the number of turns is 1790turns/m

Calculate the total length of wire

Since the length of the solenoid is L is 40.0 cm, the total number of turns required for the solenoid is

$\mathrm{N}=\mathrm{nL}=1790\text{turns}/\mathrm{m}\times 0.400\mathrm{m}=716\text{turns}$

Therefore, the number of turns is 716

Each turn is wound around the core of the solenoid as a circle of radius r 1.40 cm (so each turn amounts tolength of the wire) and we have a total number of 716 turns, so the total length of the wire required is given by the product

$\mathrm{I}=\mathrm{N}\times 2\pi \mathrm{r}=716\text{turns}\times (2\pi \times 0.014)=63.0\mathrm{m}$

Therefore, thetotal length of wire is 63.0m