91Ó°ÊÓ

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

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.

$\begin{array}{r}\mathrm{n}=\frac{\mathrm{B}}{{\mathrm{μ}}_0\mathrm{l}}=\frac{0.0270\mathrm{T}}{4\mathrm{π}×{10}^{−7}\mathrm{Tm}/\mathrm{A}×12.0\mathrm{A}}\\ =1790\text{turns}/\mathrm{m}\end{array}$

Therefore, the number of turns is 1790turns/m

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

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

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

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

Therefore, thetotal length of wire is 63.0m