Question

500 turns of insulated 18-gauge copper wire (diameter 1.024 mm) are wound in a single layer on a cylindrical form of diameter 10.00 cm.

What is the resistance of the coil?

(Hint: Take into account the thickness of the wire when calculating the radius of a single turn.)

Answer 1

resistance is R = rho*l/A

rho is the resistivity of the copper wire = 1.69*10^-8 ohm-m

A is the area of cross section = pi*r^2 = 3.142*(1.024*10^-3/2)^2 = 8.23*10^-7 m^2

No.of turns are N = 500

length of the wire is l = 500*2*pi*R = 500*2*3.142*0.05 = 157.1 m

then R = rho*l/A = (1.69*10^-8*157.1/(8.23*10^-7)) =3.23 ohm