Question

A copper wire has a diameter of `0.5 mm` and a resistivity of `1.6 xx 10^-6 Omegacm`. How much of this wire would be required to make a `10 Omega` coil ? How much does the resistance change if the diameter is doubled ?

Answer 1

(a) First of all we will calculate the area of cross-section of the copper wire. Here the diameter of copper wire is 0.5mm, so its radius (r) will be `(0.5)/(2)mm` or 0.25 mm. This radius of 0.25mm will be equal to `(0.25)/(1000)m` or `0.25 xx 10^(-3)m`. Thus, the radius r of this copper wire is `0.25 xx 10^(-3)m`. We will now find out the area of cross-section of the copper wire by using this value of the radius. So,
Area of cross-section of wire, `A = pir^(2)`
`=(22)/(7) xx (0.25 xx 10^(-3))^(2)`
`= 0.1964 xx 10^(-6)m^(2)`
Restistivity, `rho = 1.6 xx 10^(-8) Omega m`
Resistance, `R = 10 Omega`
And, Length, `l = ?` (To be calculated)
Now, putting these values in the formula:
`rho = (R xx A)/(l)`
We get: `1.6 xx 10^(-8) = (10 xx 0.1964 xx 10^(-6))/(l)`
`l =(10 xx 0.1964 xx 10^(-6))/(1.6 xx 10^(-8))`
`l = (1964)/(16)`
`l = 122.7 m`
Thus, the length of copper wire required to make `10 Omega` resistance will be 122.7 metres. (b) The resistance of a wire is inversely proportional to the square of its diameter. So, when the diameter of the wire is doubled (that is, made 2 times), then its resistance will become `((1)/(2))^(2)` or `(1)/(4)` (one-fourth).