How much wire is required?
Length of wire is L = R*A/p SI units must be used, so convert radius to meters: r = 0.0254 / 64 m using 25.4mm = 1 inch r = 3.969*10^-4 m (0.3969mm) So cross-sectional area is A = pi*(3.969*10^-4)^2 m^2 A = 4.95*10^-7 m^2 The average resistivity of nichrome is p = 1.1*10^-6 ohm-meter So the required length of wire is L = 2.2 ohm * 4.95*10^-7 m^2 / 1.1*10^-6 ohm-meter L = 0.990 m
R = p*L/A Solve for L, length of wire. L = R*A/p p - resistivity of nichrome A - pi*(1/64)^2 R - 2.2 ohm
