A copper wire has a square cross 2.3 mm on a side. The wire is 4.0 m long and carries a curren

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

A copper wire has a square cross $2.3 \mathrm{~mm}$ on a side. The wire is $4.0 \mathrm{~m}$ long and carries a current of $3.6 \mathrm{~A}$. The density of free electrons is $8.5 \times 10^{28} / \mathrm{m}^{3} .$ 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

Drift Velocity

Drift velocity is the average velocity that free charge carriers, such as electrons, achieve due to an electric field in a material. It links the microscopic movement of electrons to the macroscopic current via the relation v_d = I/(nqA), where n is the free electron density, q is the charge of an electron, I is the current, and A is the cross-sectional area. This concept highlights how electron mobility contributes to conduction.

Current Density

Current density is defined as the electric current per unit cross-sectional area of a conductor. It provides a measure of how concentrated the current flow is within the conductor. The relationship is given by J = I/A, where I is the current and A is the area, illustrating how both the magnitude of the current and the size of the conductor’s cross?section influence the strength of the current density.

Cross-sectional Area

The cross-sectional area of a conductor is the measure of the size of the cut?section perpendicular to the direction of current flow. It is a key geometric factor in calculating current density, determining how the overall current is distributed across the conductor, and is computed using standard geometric formulas depending on the shape of the cross-section.

Electric Field in a Conductor

The electric field in a conductor represents the force per unit charge that drives the movement of free charge carriers. In conducting materials, it is related to the current density by the material’s conductivity and can be expressed using Ohm’s law in the form E = ?J, where ? is the resistivity of the material. This relationship illustrates how the internal field is established to sustain a given current flow through the conductor.

Transcript

00:01 Hi, in the given problem, side of the square cross section of the copper wire is given as a is equal to 2 .3 millimeter or we can say this is 2 .3 into 10 dash to par minus 3 meter.

00:36 Length of the wire is l is equal to 4 .0 meter and current carried by this wire is i is equal to 3 .6 ampere free electron density in it is n is equal to 8 .5 into 10 dash to the power 28 per meter cube number of three electrons per unit volume.

01:20 In the first part of the problem we have to find the current density of the wire means current passing through unit cross -sectional area that will be given by the ratio of current with the area of cross -section.

01:36 And as here the cross -sectional area is a square shape.

01:40 So its formula will be a square means a square of side.

01:45 So it becomes 3 .6 m4.

01:48 Divided by square of 2 .3 into 10 dash per minus 3 meter.

01:57 So it comes out to be either we can write it like 0 .681 into 10 dash to part 6 amp per meter square or it can be written as 6 .81 into 10 dash to the power 5 amp per meter square and here it becomes the answer for the first part of the problem.

02:29 In the second part of the problem, we have to find the electric field across the ends of this wire, copper wire, for which we use the expression for current density as e by row, electric field, divided by resistivity.

02:52 So it gives the expression for electric field as the product of resistivity of copper with the current density.

03:03 And we know the resistivity of copper is 1 .72 into 10 dash to par minus 8 oom into meter and j we have found in the first part as 6 .81 into 10 dash par 5 ampere per meter square...

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