Question

A copper wire has a diameter of 0.5 mm and a resistivity of $1.6\times {10}^{-6}ohmcm$. How much of this wire would be required to make a 10 ohm coil? How much does the resistance change if the diameter is doubled?

Answer 1

Solution:
Given:
Diameter of wire (d) = 0.5 mm = $5\times {10}^{-4}m$
Resistivity (ρ) = $1.6\times {10}^{-6}\Omega cm=1.6\times {10}^{-8}\Omega m$
Required resistance (R) = 10 Ω
Calculating length of wire:
R = ρ l/A
⇒ l = RA/ρ
⇒ $l=10\times \pi \times \frac{{(2.5\times {10}^{-4})}^2}{1.6\times {10}^{-8}}$
⇒ l = 122.72 m
Let’s just leave this here only and move ahead to the next step.

Calculating new resistance:
R’ = ρ l/A’
⇒ A' = 4A (diameter is doubled)
⇒ R’ = (ρ l)/(4A)
⇒R' = R/4
⇒R' = 10/4 = 2.5 Ω
Change in resistance is: R' - R = 10 - 2.5 = 7.5 Ω