Chapter 23 - Magnetic Flux and Faraday's Law of Induction - Problems and Conceptual Exercises - Page 834: 89

$0.990Km$

We know that $l=\frac{L}{\mu_{\circ}n^2A}$ We plug in the known values to obtain: $l=\frac{50.0\times 10^{-3}}{(4\pi \times 10^{-2})(3012)^2\pi (0.0267)^2}=1.958m$ Now $N=\frac{l}{d}$ $\implies N=\frac{1.958}{0.000333}=5898turns$ Thus the $wire \space length=(N)(2\pi r)$ $\implies wire\space length=(5898)(2\pi)(0.0267)=0.990Km$