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

Resistance (R) of a copper wire length / and cross-section A is given by the expression.
`R=rho (l)/(A)`
Where, p is resistivity of copper = `1.6 xx 10^(-8) omega m R= 10 omega` , Radius of wire r = 0.5`//`2 mm = 0.25 mm = 0.00025 m.
A = `pir^(2) = 3.14 xx (0.00025)^(2) = 0.000000019625 m^(2)`
`l=(RA)/(rho)=(10xx0.000000019625)/(1.6xx10^(-8))=122.72 m` If the diameter (radius) is doubled, the new radius r = 0.5 mm = 0.0005 m
so the new resistance will be
`(R )/(R )=1.25/10=1/4`
` rarr R=1/4 R` Hence the new resistance will be become `1/4` times of the original resistance