Question

A coil contains N turns of insulated copper wire of diameter d and resistivity ρ wound on a cylinder of diameter D. What is the total resistance between the two ends of the coil of copper wire?

(Given D >> d)

Options

• `(4 rho ND)/d^2`

• `(8 rho ND)/d^2`

• `(2 rho ND)/d^2`

• `(12 rho ND)/d^2`

Answer 1

`bb((4 rho ND)/d^2)`

Explanation:

Diameter of copper wire = d

Diameter of cylindrical core = D

Number of turns = N (given D >> d)

The length of the copper wire wound on the cylinder is

L = NπD

The cross-sectional area of the wire is

A = `(pid^2)/4`

Using the resistance formula:

R = `L(rho)/A`

= `rho(N pi D)/((pid^2)/4)`

= `rho(ND)/(d^2/4)`

= `(4 rho ND)/d^2`