A coil is formed by winding $$250$$ turns of insulated $$16$$-gauge copper wire (diameter $$1.3 \,mm$$) in a single layer
on a cylindrical form of radius $$12$$ cm. What is the resistance
of the coil? Neglect the thickness of the insulation.
The resistance of the coil is given by $$R = ρL/A$$, where L is the length of the wire, ρ is
the resistivity of copper, and A is the cross-sectional area of the wire. Since each turn of
wire has length $$2πr$$, where r is the radius of the coil, then
$$L = (250)2\pi r = (250)(2\pi)(0.12 m) = 188.5 m.$$If $$r_w$$ is the radius of the wire itself, then its cross-sectional area is $$A = \pi {r_w}^2$$ = $$\pi (0.65 \times 10^{–3})m^2$$= $$1.33 \times 10^{–6} m^2$$
According to Table , the resistivity of copper is $$ρ=1.69 \times10^{-8}\, Ω⋅m$$ .Thus,
$$R=\dfrac{\rho L}{A}=\dfrac{(1.69\times 10^{-9}\,\Omega.m)(188.5\,m)}{1.33 \times 10^{-6}\,m^2}=2.4\Omega$$