Question

g A 4.0 m length of gold wire is connected to a 1.5 V battery, and a current of 4.0 mA flows through it. What is the diameter of the wire

Answer 1

The diameter of the wire is 1.596 x 10⁻⁵ m.

Step-by-step explanation:

Given;

length of the wire, L = 4.0 m

emf of the battery, V = 1.5 V

current through the wire, I = 4 mA = 0.004 A

The resistance of the wire is given by;

R = V / I

R = 1.5 / 0.004

R = 375 ohms

The resistance of the wire in terms of resistivity is given by;

`R = (\rho L)/(A)\\\\ A = (\rho L)/(R)\\\\`

Where;

A is area of the wire

ρ is the resistivity of gold = 2.44 x 10⁻⁸ ohm. m

`A = (2.44*10^(-8)* 4)/(375)\\\\A = 2.603*10^(-10) \ m^2`

The diameter of the wire is given by;

`A = (\pi d^2)/(4)\\\\ \pi d^2 = 4A\\\\d = \sqrt{(4A)/(\pi) }\\\\ d = \sqrt{(4(2.602*10^(-10)))/(\pi) }\\\\d = 1.596 *10^(-5) \ m`

Therefore, the diameter of the wire is 1.596 x 10⁻⁵ m.