An aluminum wire with a diameter of 0.100 mm has a uniform...

An aluminum wire with a diameter of $0.100 \mathrm{~mm}$ has a uniform electric field of $0.200 \mathrm{~V} / \mathrm{m}$ imposed along its entire length. The temperature of the wire is $50.0^{\circ} \mathrm{C}$. Assume one free electron per atom. (a) Use the information in Table 26.2 to determine the resistivity of aluminum at this temperature. (b) What is the current density in the wire? (c) What is the total current in the wire? (d) What is the drift speed of the conduction electrons? (e) What potential difference must exist between the ends of a 2.00 -m length of the wire to produce the stated electric field?

An aluminum wire with a diameter of $0.100 \mathrm{~mm}$ has a uniform electric field of $0.200 \mathrm{~V} / \mathrm{m}$ imposed along its entire length. The temperature of the wire is $50.0^{\circ} \mathrm{C}$. Assume one free electron per atom. (a) Use the information in Table 26.2 to determine the resistivity of aluminum at this temperature. (b) What is the current density in the wire?
(c) What is the total current in the wire? (d) What is the drift speed of the conduction electrons? (e) What potential difference must exist between the ends of a 2.00 -m length of the wire to produce the stated electric field?

Key Concepts

Current Density

Current density is defined as the amount of electric current flowing per unit cross-sectional area of a material. It is a vector quantity that helps in analyzing how current is distributed within a conductor, and it is essential for applications in circuit design and analysis.

Drift Velocity of Charge Carriers

Drift velocity is the average velocity acquired by charge carriers, such as electrons, in a conductor under the influence of an electric field. It results from the balance between the accelerating force of the field and collisions that impede motion. This concept is vital in linking microscopic carrier dynamics to the overall current in a material.

Ohm’s Law (Microscopic Form)

In its microscopic form, Ohm’s law relates the current density to the electric field by the equation J = ?E, where ? is the electrical conductivity. This relationship is fundamental in linking the macroscopic behavior of circuits to the microscopic motion of charge carriers in materials.

Potential Difference and Its Relation to Electric Field

The potential difference (voltage) between two points in a conductor is the work done per unit charge and is directly related to the electric field through the relation V = E·L, where L is the distance between the points. This concept helps in understanding how energy is transferred in electrical systems.

Temperature Dependence of Resistivity

Many materials, particularly metals, exhibit changes in resistivity with temperature. As temperature increases, lattice vibrations typically increase electron scattering, leading to higher resistivity. This concept is crucial in applications where temperature variations can affect circuit performance.

Resistivity

Resistivity is an intrinsic property of a material that quantifies its opposition to the flow of electric current. It is measured in ohm?meters and depends on the material's structure and temperature. Understanding resistivity is fundamental in assessing how materials perform under electrical stress.

Electric Field

An electric field represents the force per unit charge acting on charged particles and is measured in volts per meter. It is the driving force that moves charge carriers through a conductor, making it a central concept in understanding electrical conduction.

Charge Carrier Density in Conductors

Charge carrier density is the number of free charge carriers per unit volume within a material. It is a key parameter in determining electrical conductivity, as a higher density of carriers typically leads to more efficient conduction. Estimations often involve assumptions about the number of free electrons contributed by each atom.

Transcript

00:01 So, given in the question the diameter of the aluminum wire d is given to be 0 .100 mm which is 0 .100 into 10 to the power minus 3 meters.

00:17 The electric field imposed along the length of the wire is e that is 0 .200 volts per meter.

00:30 The temperature of the wire is capital t that is 50 .0 degree centigrade and number of free electrons per atom is given to be equal to 1.

00:47 Now from the table 26 .2 the resistivity of aluminum at 20 degree centigrade is given to be 2 .82 into 10 to the power minus 8 ohm meters let it be expressed as rho naught and the temperature coefficient of resistance alpha is given to be 3 .9 into 10 to the power minus 3 per degree centigrade.

01:17 Now it has been asked to find the resistivity of aluminum at 50 .0 degree centigrade, the current density in the wire, the total current in the wire, the drift speed of the conduction electrons and the potential difference that exists between the ends of a 2 .00 meter length of the wire to produce the electric field capital e to produce the electric field e.

02:09 Now solving the first part of the question resistivity resistivity of a metal varies with temperature as per the following relation rho which is a function of temperature t rho which is a function of temperature is rho naught 1 plus alpha delta t so rho t is 2 .82 into 10 to the power minus 8 ohm meter 1 plus alpha is 3 .9 into 10 to the power minus 3 per degree centigrade and temperature difference is 50 minus 20 that is 30 degree centigrade.

02:55 Solving we get this rho at temperature t to be equal to 3 .15 into 10 to the power minus 8 ohm meters.

03:13 Now coming to part b current density j is sigma e which is equal to e by rho where sigma is the conductivity of the wire of aluminum and capital e is the electric field imposed on the wire rho is the resistivity of the wire so j e is 0 .200 volt per meter and rho just now calculated as 3 .15 into 10 to the power minus 8 ohm meter so j comes out to be 0 .0635 into 10 to the power 8 ampere per meter square which can be written as 6 .35 into 10 to the power 6 ampere per meter square.

04:04 Now area of cross -section of the wire let it be represented by capital a is expressed as pi d square by 4 where d is the diameter of the wire which is given to be 0 .100 into 10 to the power minus 3 meters.

04:21 To put the value we get a to be 7 equal to 7 .85 into 10 to the power minus 9 meter square therefore the total current in the wire i with that will be equal to j into a so j is 6 .35 into 10 to the power 6 ampere per meter square and a is 7 .85 into 10 to the power minus 9 meter square which comes out to be 49 .85 into 10 to the power minus 3 amperes or 49 .9 milli amperes...

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