Question

A current of 5 A exists in a copper wire of length 50 m and diameter of 2.5 mm when applying a

Answer 1

Answer:

a) 1273.23 A/m^2

b) 7.19*10^-5 m/s

c) 236881.7 Ohms

Explanation:

(a) To find the current density you use the following formula:

$J=\frac{I}{A}=\frac{I}{\pi r^2}$

I: current in the wire

A: cross area of the wire

r: radius of the wire

$J=\frac{5A}{\pi(1.25*10^{-3}m)^2}=1273.23\frac{A}{m^2}$

(b) The electron drift speed is given by:

$v_d=\frac{I}{nqA}=\frac{I}{nq\pi r^2}$

n: number of conduction electrons per m^3

q: charge of the electron = 1.6*10^-19C

The number of free electrons is calculated by using:

$n=\frac{\rho N_A}{M}\\\\n=\frac{(9*10^{3}kg/m^3)(6.22*10^{23})}{63.54*10^{-3}}=8.5*10^{28}$

Next, you replace the values of the parameters in the equation for vd:

$v_d=\frac{5A}{(8.85*10^{28}m^{-3})(1.6*10^{-19}C)\pi (1.25*10^{-3}m)^2}\\\\v_d=7.19*10^{-5}\frac{m}{s}$

(c) The conductivity is given by:

$\sigma=\frac{L}{RA}$

You first calculate R:

$R=VI=(0.86V)(5A)=4.3\Omega$

Next, replace for sigma:

$\sigma=\frac{50m}{(4.3\Omega)(\pi (1.25*10^{-3}m)^2)}=236881.7\Omega^{-1}m^{-1}$