Question

You have a piece of copper wire of length L = 15.0 cm and radius R = 0.5 mm. Given that the resistivity of copper is 1.68 X 10^-8 Ohm*m, what is:

1. a. The resistance of your piece of wire?

2. b. If a current of i = 0.5 mA is set up in the wire, what is the magnitude of the current density of the wire?

3. c. With a current of i = 0.5 mA, what is the change in potential difference from one side of the wire to the other?

4. d. Given the charge carrier density of copper is 8.5 x 1028 electrons per cubic meter, what is the magnitude of the drift velocity of the electrons when a current of i = 0.5 mA is flowing through the wire?

answers are: a. 0.0032 ohm, b. 640A/m^2, c.1.6 uV, d. 4.7x10^-8 m/s

I need some help understanding how to work it out to get these answers! thank you!!

Answer 1

Part A.

Resistance of wire is given by:

R = rho*L/A

rho = resistivity of copper wire

L = length

A = area = pi*r^2

Using given values:

R = 1.68*10^-8*0.15/(pi*(0.5*10^-3)^2)

R = 0.0032 ohm

Part B.

Current density is given by:

J = I/A

I = Current in wire

Using given values:

J = 0.5*10^-3/(pi*(0.5*10^-3)^2)

J = 636.62 A/m^2 = 640 A/m^2

Part C.

Using Ohm's law:

Potential difference, V = i*R

V = (0.5*10^-3 Amp)*(0.0032 ohm)

V = 1.6*10^-6 V = 1.6*10^-6 uV

Part D.

drift speed of electrons in wire is given by:

Vd = I/(n*q*A)

n = number of electrons per m^3 in wire = 8.5*10^28 electrons/m^3

q = charge on electron = 1.6*10^-19

So,

Vd = 0.5*10^-3/(8.5*10^28*1.6*10^-19*pi*(0.5*10^-3)^2)

Vd = 4.7*10^-8 m/s

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