Answer
$1.7\times10^{-4}m$
Work Step by Step
Use equation 18–3 to find the resistance. The cross-sectional area $A=\pi r^2=\frac{\pi d^2}{4}$.
$$R=\rho\frac{\mathcal{l}}{A}=\rho\frac{4 \mathcal{l}}{\pi d^2}$$
Combine this with equation 18–6b and calculate the diameter of the wire.
$$P=\frac{V^2}{R}=\frac{\pi d^2V^2}{4 \rho \mathcal{l}}$$
$$d=\sqrt{\frac{4 \rho \mathcal{l}P}{\pi V^2}}$$
$$ =\sqrt{\frac{4 (100\times10^{-8}\Omega \cdot m)(3.5m)(95W)}{\pi (120V)^2}}$$
$$ =1.7\times10^{-4}m$$
