Question

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? (The resistivity of gold is 2.44 × 10-8 Ω·m.) A. 9.0 μm B. 8.5 μm C. 17 μm D. 48 μm

Answer 1

Explanation:

Given that,

Length of gold wire, l = 4 m

Voltage of battery, V = 1.5 V

Current, I = 4 mA

The resistivity of gold, $\rho=2.44\times 10^{-8}\ \Omega-m$

Resistance in terms of resistivity is given by :

$R=\dfrac{\rho l}{A}$

Also, V = IR

So,

$\dfrac{V}{I}=\dfrac{\rho l}{A}$

A is area of wire,

$\dfrac{V}{I}=\dfrac{\rho l}{\pi r^2}$, r is radius, r = d/2 (diameter=d)

$\dfrac{V}{I}=\dfrac{\rho l}{\pi (d/2)^2}\\\\\dfrac{V}{I}=\dfrac{4\rho l}{\pi d^2}\\\\d=\sqrt{\dfrac{4\rho l I}{V\pi}} \\\\d=\sqrt{\dfrac{4\times 2.44\times 10^{-8}\times 4\times 4\times 10^{-3}}{1.5\times \pi}} \\\\d=18.2\ \mu m$

Out of four option, near option is (C) 17 μm.