A copper wire has diameter 0.5 mm and resistivity of `1.6 xx 10^(-8) Omega m`. What will be the length of this wire to make its resistance `10 Omega` ? How much does the resistance change if the diameter is doubled?

Explanation

Diameter of the wire, `D = 0.5 mm `
`= 0.5 xx 10^(-3) m `
Resistivity of the wire, `rho = 1.6 xx 10^(-8) Omega m`
Resistance of the coil, `R = 10 Omega`
Let required length of the wire = l
Using
`R = rho . l/A ,` we have `R = rho. l/(pi D^2//4)`
`[because A = pi r^2 = pi (D/2)^2 = (pi D^2)/(4)]`
or `R = (4rho l)/(pi D^2) " " .....(i)`
or `l = (R xx pi D^2)/(4 rho)`
`= (10 xx 3.142 xx (0.5 xx 10^(-3))^2)/(4 xx 1.6 xx 10^(-8)) m`
`= (3.142 xx 0.5 xx 0.5)/(4 xx 1.6) m`
`= 0.1227 xx 10^3 m = 122.7 m`
i.e., Required length of the wire = 122.7 m
From equation (i), we have `R prop 1/(D^2)`
`therefore` If the diameter of the wire is doubled its resistance will become four times its initial resistance.
i.e. It will become `4 xx 10 Omega = 40 Omega`