# Circular Mils

### What is Circular Mils?

The circular mil is a unit of area used especially when denoting the cross-sectional size of a wire or cable. It is the quantity used in most commercial wire tables, and thus it needs to be carefully defined.
The mil is a unit of measurement for length and is related to the inch by
 $$\bbox[5px,border:1px solid blue] {\color{blue}{\text{1 mil }= {1 \over 1000}}}$$ Eq.(1) $$\text{1000 mil } = \text{1 in.}$$
In general, therefore, the mil is a very small unit of measurement for length. By definition,
A wire with a diameter of 1 mil has an area of 1 CM
Fig.no.1: Defining the circular mil (CM).
The area of a circular wire in circular mils can be defined by the following equation and shown in the Fig.no.2: $$\text{A (CM)}= {d^2} mil. \text{............... eq.2}$$
Fig.no.2: Verification of Eq.2
Remember, to compute the area of a wire in circular mils when the diameter is given in inches, first convert the diameter to mils by simply writing the diameter in decimal form and moving the decimal point three places to the right. For example,
$${1 \over 8} in. = 0.125 in. = 125 mils$$ Then the area is determined by $$\text{A (CM) = (d mils)}^2 =\text{ (125 mils)} = \text{15,625 CM}$$
Sometimes when you are working with conductors that are not circular, you will need to convert square mils to circular mils, and vice versa.
Applying the basic equation for the area of a circle and substituting a diameter of 1 mil results in $$\begin{array} {rcl} \text{Area of a circle (A)} & = &{\pi \over 4}d^2\\ & = &{\pi \over 4}(1mil)^2\\ & = &{\pi \over 4} \text{sq. mils}\\ & = &\text{1 CM}\end{array}$$ from which we can conclude the following:
 $$\bbox[5px,border:1px solid blue] {\color{blue}{\text{1 CM} = {\pi \over 4} \text{sq. mils}}}$$ Eq.(3)
 $$\bbox[5px,border:1px solid blue] {\color{blue}{\text{1 sq. mils} = {4 \over \pi} CM}}$$ Eq.(4)
EXAMPLE 1: What is the resistance of a 100 ft length of copper wire with a diameter of 0.020 in. at 20℃?
SOLUTION: Resistivity of copper wire at 20℃ is given here in the table no.1. $$\begin{array} {rcl} \rho & = & \text{10.37 CM-Ω/ft}\\ \text{diameter(d)} & = & \text{0.020 in. = 20 mils }\\ \text{length(l)} & = & \text{100 ft}\\ \text{A (CM)} & = & \text{(d mils)}^2\\ &=& \text{(20 mils)}^2 = \text{400 CM}\\ R & = & \rho {l \over A}\\ & = & \text{(10.37 CM-Ω/ft)} {\text{(100 ft)} \over \text{400 CM}}\\ R & = & 2.59 Ω\end{array}$$