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 ?

Explanation

Area of cross-section of the wire, `A=pi (d//2)2`
Diameter = 0.5 mm = 0.0005 m
Resistance, `R= 10 Omega`
We know that
`R=rhoI/A`
`I=(RA)/rho`
`=(10x3.14x((0.0005)/2)^(2))/(1.6x10^(-8))`
`=(10x3.14x25)/(4x1.6)=122.72m`
` therefore` length of the wire = 122.72 m
If the diameter of the wire is doubled, new diameter `= 2xx0.5=1mm=0.001m` Let new resistance be R'
`R'=rhoI/A`
`=(1.6x10^(-8)x122.72)/(pi(1/2x10^(-3))^(2))`
`(1.6x10^(-8)x122.72x4)/(3.14x10^(-6)`
`=250.2xx10^(-2)=2.5 Omega`
Therefore, the length of the wire is d122.7 m and the new resistance is `2.5 Omega`.