Question

A coil 20 cm long with a radius of 1 cm has a winding of 1000

turns of copper wire with a cross-sectional area of 1 mm² . The coil is

connected to an alternating current circuit with a frequency of 50 Hz.

What part of the impedance of the coil will be formed by (1) the

resistance, and (2) the inductive reactance?

Answer 1

Find length of the wire:

$$l=2\pi rN.$$

Find area of the wire:

$$R=\frac{\rho l}{A}=\frac{2\pi\rho rN}{A}=1.056\Omega.$$

Inductance of the coil:

$$L=\mu_0\frac{N^2A_\text{coil}}{x}=\mu_0\frac{N^2\pi r^2}{x}.$$

Inductive resistance:

$$X=\omega L=2\pi fL=2 f\mu_0\frac{N^2r^2\pi^2}{x}=0.62\Omega$$

Find the fraction:

$$\alpha_R=\frac{R}{\sqrt{R^2+X^2}}=0.86,\\\space\\ \alpha_X=\frac{X}{\sqrt{R^2+X^2}}=0.51.$$