A copper wire has a square cross section 2.3 mm on a side....

A copper wire has a square cross section 2.3 mm on a side. The wire is 4.0 m long and carries a current of 3.6 A. The density of free electrons is 8.5 $\times$ 10$^{28}$/m${^3}$. Find the magnitudes of (a) the current density in the wire and (b) the electric field in the wire. (c) How much time is required for an electron to travel the length of the wire?

Key Concepts

Current Density

Current density is the amount of electric current flowing per unit cross-sectional area of a conductor. It is defined by the formula J = I/A, where I is the current and A is the cross sectional area through which the current flows. This concept is fundamental in understanding how current is distributed across different parts of a material and affects other properties such as heating and electric fields within the conductor.

Electric Field

The electric field in a conductor is the force per unit charge that drives the movement of charge carriers. In the context of electrical conduction, it establishes a potential difference that causes electrons to drift. Microscopic Ohm's law relates the electric field to the current density by the conductivity of the material (J = ?E), making it an essential concept in connecting the macroscopic measurements (current and voltage) with the microscopic behavior of charge carriers.

Drift Velocity

Drift velocity refers to the average velocity at which charge carriers, such as electrons, move through a conductor under the influence of an electric field. It is given by the formula v_d = I/(n e A), where I is the current, n is the density of free charge carriers, e is the elementary charge, and A is the cross-sectional area. This concept is crucial for understanding the time scales involved in electron transport and is directly linked to how quickly electrical signals propagate through a material.

Transcript

00:01 We're given a copper wire with a square cross -sectional area and the length of a side of the cross -sectional area that corresponds to a cross -sectional area is 2 .3 millimeters.

00:16 And the length of the wire is 4 meters and we're told that a current of 3 .6 ampairs flows through this wire and it has a free electron density.

00:31 Of 8 .5 times 10 to 28 per meter cube.

00:36 I've given a little diagram over here.

00:40 Just to show that we have deviated from the normal cylindrical wire, and here instead we have a square wire.

00:50 And the things we're asked to find, we'll do on the next page.

00:56 So we're asked to find here in part a, the resistivity of a.

01:12 Sorry we're asked to find the current density this wire so the current density is defined as the current over the cross -sectional area and this is given in equation 25 .3 but we already know the current so this is given to us the current is 3 .6 amperors and we can easily find the area of a square because we have a square cross -sectional area and so it's just the length of the side square which is 2 .3 times tens of minus 3 meters squared which gives us a density of 6805 to 9 amperes per meter squared which you can write as 6 .81 so i just rounded up times 10 to the fifth emperes per meter squared and in part b you're asked to find the magnitude of the electric field so you again use a definition that's given in first time we give you the equation so using equation 25 .5 you find that the resistivity is defined as the magnitude of the electric field over the charge density so then the electric field is a resistivity times the charge density now you can find the resistivity of copper in table 25 .1 so row is given in 25 and the charge density you just computed in part a...

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