Question

A copper wire has a diameter of 0.5 mm and a resistivity of 1.6Ω cm. How much of this wire would be necessary to make a 10Ω coil?

Answer 1

Given,

• Resistivity (ρ) = 1.6 Ω m =  1.6 × 10⁻² Ω m

• Resistance (R) = 10 Ω

• Diameter (d) = 0.5 mm

• Radius (r) = 0.25 mm = 0.25 × 10⁻³ m = 2.5 × 10⁻⁴ m

• Area (A) = πr² = 3.14 × (2.5 × 10⁻⁴)² = 6.25 π × 10⁻² m

We are asked to find length L.

R = ρL/A

L = RA/ρ

Putting values,

⇒ L = ( 10 × 6.25 π × 10⁻⁸ ) / 1.6 × 10⁻²  m

⇒  L =  62.5 × 3.14 / 1.6 × 10⁻⁶

⇒ L = 122.65 × 10⁶ m