Current through a wire help

[aftershock]

Homework Statement

A current of I=8.0A is flowing in a typical extension cord of length L=3.00m. The cord is made of copper wire with diameter d=1.5mm.

The charge of the electron is e=1.6*10^-19C. The resisitivity of copper is $$\rho$$=1.7*10^-8$$\Omega$$ m. The concentration of free electrons in copper is n=8.5*10^28m^-3

Homework Equations

I honestly have no idea. There's nothing in my book about collisions at all other than saying that's what causes resistance.

The Attempt at a Solution

Honestly have no idea where to begin... there is literally not a thing about this in my book. It's from an online homework assignment.

 

[tiny-tim]

hi aftershock!

erm … what's the question?

 

[aftershock]

hi aftershock!

erm … what's the question?

Wow I can't believe I left that part off, sorry.

Find the total number of collisions that all free electrons undergo in one second.

I was able to see the answer and the equation used was something along the lines of m/($$\rho$$ne²)

I have no idea where that came from though.

 

[tiny-tim]

Wow I can't believe I left that part off, sorry.

he he

Find the total number of collisions that all free electrons undergo in one second.

I was able to see the answer and the equation used was something along the lines of m/($$\rho$$ne²)

I have no idea where that came from though.

not something I'm familiar with , but you might get some help from http://en.wikipedia.org/wiki/Classical_and_quantum_conductivity#Classical_conductivity http://en.wikipedia.org/wiki/Drude_model or http://en.wikipedia.org/wiki/Ohm's_Law

 

[rwisz]

As a scientist, it is important to approach problems with an open and curious mind. While it may not be explicitly stated in your textbook, there are many equations and principles that can help us solve this problem.

First, let's define some terms. Current is the flow of electric charge, typically measured in Amperes (A). In this case, we are given that the current is 8.0A. Length is measured in meters (m), and we are given that the length of the cord is 3.00m. Diameter is measured in millimeters (mm), but we can convert it to meters for our calculations (1.5mm = 0.0015m).

The equation that relates current, length, and resistance is Ohm's Law, which states that V=IR, where V is voltage, I is current, and R is resistance. In this case, we are given the current and length, so we can solve for the resistance of the wire.

R = V/I = (8.0A)(3.00m) = 24.0Ω

Next, we need to consider the properties of the copper wire. The resistivity of a material is a measure of how well it resists the flow of electricity. The higher the resistivity, the more difficult it is for electricity to flow through the material. The resistivity of copper is given as 1.7*10^-8Ωm.

We can use the formula R=ρL/A, where ρ is the resistivity, L is the length, and A is the cross-sectional area, to calculate the resistance of the wire. We know the length and resistivity, but we need to find the cross-sectional area. We can do this by using the formula A=πr^2, where r is the radius of the wire.

r = d/2 = 0.0015m/2 = 0.00075m

A = π(0.00075m)^2 = 1.767*10^-6m^2

Now we can plug in our values to solve for the resistance of the wire.

R = (1.7*10^-8Ωm)(3.00m)/(1.767*10^-6m^2) = 0.0288Ω

This is a very small resistance, which is expected for a copper wire. But why does copper