E-Field Through Copper Wire given drift velocity

In summary, the conversation discusses finding the electric field in a conductor using the drift velocity of free electrons and the resistivity of copper. The equations used are vd=(qEτ)/me and ρ=1/σ=me/(nq2τ). The conversation also mentions using the current density formula J = (1/ρ) E and the relation between current density and drift velocity, J = n e vd. The solution involves finding the number of electrons per m3, n, which cannot be calculated without looking up information such as the density of copper.
  • #1
stephen8686
42
5

Homework Statement


If the magnitude of the drift velocity of free electrons in a copper wire is 7.84×10-4 m/s, what is the electric field in the conductor? (Also gives chart that states that the resistivity of Cu is 1.7×10-8 Ωm)

Homework Equations


[/B]
vd=(qEτ)/me (where τ is avg. time between collisions)

ρ=1/σ=me/(nq2τ)

The Attempt at a Solution



first I solved for τ in the second equation:
τ=me/(nq2ρ)

Then I put this value in for τ in the first equation:
vd=(qE(me/(nq2ρ))/me
Simplify:
vd=E/(qnp)

When I put in values:
7.84×10-4=E/(1.6×10-19)(1.7×10-8)(n)
I didn't know how to get n (number of e- per m3), so I googled the density of copper (8.96 g/ml) and did some conversions to get n=8.475×1028, which gave me a correct answer of E=.18V/m, but I don't think I was supposed to use google, and no density chart is given in the book. Is there another way to do this that I overlooked?
 
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  • #2
stephen8686 said:
Is there another way to do this that I overlooked?

There is a more direct way that does not involve τ. The current density is related to the electric field by J = (1/ρ) E and to the drift velocity by J = n e vd. You still need n which you cannot get from what is given unless you look things up.
 
  • #3
kuruman said:
There is a more direct way that does not involve τ. The current density is related to the electric field by J = (1/ρ) E and to the drift velocity by J = n e vd. You still need n which you cannot get from what is given unless you look things up.

Thanks kuruman. Just wanted to make sure I wasn't missing anything important.
 

1. What is an E-Field and how does it relate to copper wire?

An E-Field, or electric field, is a region in space where an electrically charged particle experiences a force. In the context of copper wire, an E-Field is created when a voltage is applied to the wire, causing an electric current to flow. The strength of the E-Field is directly related to the voltage applied and the properties of the wire, such as its conductivity and diameter.

2. How does the drift velocity of electrons affect the E-Field in a copper wire?

The drift velocity of electrons refers to the average speed at which electrons move through a conductor, such as copper wire. In a copper wire, electrons are constantly moving in a random fashion, colliding with other electrons and the atoms of the wire. The E-Field helps to guide these electrons in a specific direction, resulting in a net flow of current. The higher the drift velocity, the larger the current that can flow through the wire.

3. What factors can affect the E-Field through a copper wire?

The strength of the E-Field through a copper wire is affected by several factors, including the voltage applied, the length and diameter of the wire, and the material and temperature of the wire. The conductivity of the wire also plays a role, as materials with higher conductivity allow for a stronger E-Field and larger current flow.

4. How is the E-Field through a copper wire measured?

The E-Field through a copper wire can be measured using a device called a voltmeter, which measures the voltage across the wire. The current flowing through the wire can also be measured using an ammeter. By using these two measurements and Ohm's Law (V=IR), the strength of the E-Field can be calculated.

5. What are some practical applications of understanding the E-Field through copper wire?

Understanding the E-Field through copper wire is important in many practical applications, such as in the design and operation of electrical circuits and devices. It is also crucial in industries such as telecommunications, where the transmission of information through wires relies on the strength and stability of the E-Field. Additionally, understanding the E-Field can help in troubleshooting issues with electrical systems and ensuring the safety of workers and the general public.

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