A copper wire has a diameter of 1.532 mm. What magnitude current flows when the drift velocity is 0.500 mm/s?

Answer 1

To find the current flowing through the copper wire, we can use the formula:

I = n*A*drift_velocity*q

where:
I is the current,
n is the number density of the copper (number of charge carriers per unit volume),
A is the cross-sectional area of the wire,
drift_velocity is the drift velocity of the electrons, and
q is the elementary charge, which is approximately 1.60 x 10^(-19) C.

First, we need to find the number density of the copper. The number density can be calculated using the formula:

n = density * (N_A / molar_mass)

where:
density is the density of the copper (approximately 8.96 g/cm^3),
N_A is Avogadro's number (approximately 6.02 x 10^23 atoms/mole), and
molar_mass is the molar mass of copper (approximately 63.5 g/mole).

n = (8.96 g/cm^3) * (6.02 x 10^23 atoms/mole) / (63.5 g/mole)

To make the units consistent, let's convert the density from g/cm^3 to kg/m^3:

density = 8.96 g/cm^3 * (1 kg/1000 g) * (100 cm/m)^3 = 8960 kg/m^3

Now we can find the number density:

n = (8960 kg/m^3) * (6.02 x 10^23 atoms/mole) / (63.5 g/mole * (1 kg/1000 g))
n = 8.48 x 10^28 atoms/m^3

Next, we need to find the cross-sectional area of the copper wire. The area can be calculated using the formula:

A = pi * (diameter / 2)^2

A = pi * (1.532 mm / 2)^2

To make the units consistent, let's convert the diameter from mm to meters:

diameter = 1.532 mm * (1 m / 1000 mm) = 0.001532 m

Now we can find the cross-sectional area:

A = pi * (0.001532 m / 2)^2
A = 1.843 x 10^(-6) m^2

Finally, we can calculate the current flowing through the copper wire:

I = n * A * drift_velocity * q
I = (8.48 x 10^28 atoms/m^3) * (1.843 x 10^(-6) m^2) * (0.5 mm/s * (1 m / 1000 mm)) * (1.6 x 10^(-19) C)

I = 12.9 A

The magnitude of the current flowing through the copper wire is approximately 12.9 A.