Electric conduction current density and examples

4402
David Holt

It is called current density to the amount of current per unit area through a conductor. It is a vector quantity, and its modulus is given by the quotient between the instantaneous current I that passes through the cross section of the conductor and its area S, so that:

Stated like this, the units in the International System for the current density vector are amps per square meter: A / mtwo. In vector form the current density is:

The current density vector. Source: Wikimedia Commons.

Current density and current intensity are related, although the former is a vector and the latter is not. The current is not a vector despite having magnitude and meaning, since having a preferential direction in space is not necessary to establish the concept.

However, the electric field that is established inside the conductor is a vector, and it is related to the current. Intuitively, it is understood that the field is more intense when the current is also more intense, but the cross-sectional area of ​​the conductor also plays a determining role in this regard..

Article index

  • 1 Electric conduction model
    • 1.1 What happens when the conductor is connected to a battery?
    • 1.2 The entrainment speed vd and the current density
  • 2 Conductivity of a material
  • 3 Ohm's Law
  • 4 Application examples
    • 4.1 -Resolved example 1
    • 4.2 -Resolved example 2
  • 5 References

Electric conduction model

In a piece of neutral conductive wire like the one shown in Figure 3, cylindrical in shape, the charge carriers move randomly in any direction. Inside the conductor, according to the type of substance with which it is made, there will be n charge carriers per unit volume. This n should not be confused with the normal vector perpendicular to the conducting surface.

A piece of cylindrical conductor shows current carriers moving in different directions. Source: self made.

The proposed conductive material model consists of a fixed ionic lattice and a gas of electrons, which are current carriers, although they are represented here with the + sign, since this is the convention for current.

What happens when the conductor is connected to a battery?

A potential difference is then established between the ends of the conductor, thanks to a source that is responsible for doing the work: the battery..

A simple circuit shows a battery that by means of conductive wires lights a light bulb. Source: self made.

Thanks to this potential difference, the current carriers accelerate and march in a more orderly way than when the material was neutral. In this way it is able to light the bulb of the circuit shown.

In this case, an electric field has been created inside the conductor that accelerates the electrons. Of course, their path is not free: although the electrons have acceleration, as they collide with the crystalline lattice they give up some of their energy and are dispersed all the time. The overall result is that they move a little more orderly within the material, but their progress is certainly very little..

As they collide with the crystalline lattice they set it to vibrate, resulting in heating of the conductor. This is an effect that is easily noticed: the conductive cables heat up when they are crossed by an electrical current.

Crawling speed vand the current density

Current carriers now have a global motion in the same direction as the electric field. That global speed they have is called crawl speed or drift speed and is symbolized as vd.

Once a potential difference is established, the current carriers have a more orderly movement. Source: self made.

It can be calculated by some simple considerations: the distance traveled inside the conductor by each particle, in a time interval dt it is vd . dt. As said before, there is n particles per unit volume, the volume being the product of the cross-sectional area A and the distance traveled:

V = A.vd dt

If each particle has charge q, what amount of charge dQ passes through the area TO in a time interval dt?:

dQ = q.n. Avd dt

The instantaneous current is just dQ / dt, therefore:

J = q.n.vd

When the charge is positive, vd is in the same direction as AND  Y J. If the charge were negative, vd  it is opposite to the field AND, but J Y AND they still have the same address. On the other hand, although the current is the same throughout the circuit, the current density does not necessarily remain unchanged. For example it is smaller in the battery, whose cross-sectional area is greater than in the conducting wires, thinner.

Conductivity of a material

It can be thought that the charge carriers moving inside the conductor and continuously colliding with the crystalline lattice, face a force that opposes their advance, a kind of friction or dissipative force Fd which is proportional to the average speed that they carry, that is, the drag speed:

Fd v

Fd = α. vd

It is the Drude-Lorentz model, created at the beginning of the 20th century to explain the movement of current carriers inside a conductor. It does not take quantum effects into account. α is the constant of proportionality, whose value is in accordance with the characteristics of the material.

If the drag speed is constant, the sum of forces acting on a current carrier is zero. The other force is that exerted by the electric field, whose magnitude is Fe = q.E:

what - α. vd = 0

The entrainment speed can be expressed in terms of the current density, if it is conveniently solved:

From where:

J = nqtwoE / α

The constants n, q and α are grouped in a single call σ, so that finally we obtain:

 J = σAND

Ohm's law

The current density is directly proportional to the electric field established inside the conductor. This result is known as Ohm's law in microscopic form or local Ohm's law.

The value of σ = n.qtwo / α is a constant that depends on the material. It's about the electric conductivity or just conductivity. Its values ​​are tabulated for many materials and its units in the International System are amps / volt x meter (A / V.m), although there are other units, for example S / m (siemens per meter).

Not all materials comply with this law. Those that do are known as ohmic materials.

In a substance with high conductivity it is easy to establish an electric field, while in another with low conductivity it takes more work. Examples of materials with high conductivity are: graphene, silver, copper and gold.

Application examples

-Worked example 1

Find the entrainment velocity of the free electrons in a copper wire of cross-sectional area 2 mmtwo when a current of 3 A passes through it. Copper has 1 conduction electron for each atom.

Fact: Avogadro's number = 6.023 102. 3 particles per mole; electron charge -1.6 x 10-19 C; copper density 8960 kg / m3; molecular weight of copper: 63.55 g / mol.

Solution

From J = q.n.vd The magnitude of the drag velocity is solved:

How come the lights come on instantly?

This speed is surprisingly small, but it must be remembered that cargo carriers are continually colliding and bouncing inside the driver, so they are not expected to go too fast. It may take an electron almost an hour to go from the car battery to the headlight bulb for example.

Fortunately, you don't have to wait that long to turn on the lights. An electron in the battery quickly pushes the others inside the conductor, and thus the electric field establishes itself very quickly as it is an electromagnetic wave. It is the disturbance that propagates within the wire.

The electrons manage to jump at the speed of light from one atom to the adjacent one and the current begins to flow in the same way that water does through a hose. The drops at the beginning of the hose are not the same as at the outlet, but it is still water.

-Worked example 2

The figure shows two connected wires, made of the same material. The current that enters from the left to the thinnest portion is 2 A. There the entrainment velocity of the electrons is 8.2 x 10-4 m / s. Assuming that the value of the current remains constant, find the entrainment velocity of the electrons in the portion to the right, in m / s.

Solution

In the thinnest section: J1 = n.q. vd1 = I / A1

And in the thickest section: Jtwo = n.q. vd2 = I / Atwo

The current is the same for both sections, as well as n Y what, Thus:

References

  1. Resnick, R. 1992. Physics. Third expanded edition in Spanish. Volume 2. Compañía Editorial Continental S.A. by C.V.
  2. Sears, Zemansky. 2016. University Physics with Modern Physics. 14th. Ed. Volume 2. 817-820.
  3. Serway, R., Jewett, J. 2009. Physics for Science and Engineering with Modern Physics. 7th Edition. Volume 2. Cengage Learning. 752-775.
  4. Sevilla University. Department of Applied Physics III. Density and intensity of current. Recovered from: us.es
  5. Walker, J. 2008. Physics. 4th Ed. Pearson. 725-728.

Yet No Comments