### Current density and current

Most electrical devices are not electrostatic devices.  Most electrical devices require the flow of a current. A current requires moving charges.
Let ρ+ = n+q+ be the density of the positive charges in some region, i.e. the amount of positive charge per unit volume, and let ρ- = n-q- be the density of the negative charges.  Here n+ and n- are the number of positively and negatively charged particles per unit volume and q+ and q- are the charge of each positively and negatively charged particle, respectively.
(n+ and n- are positive numbers q+ is a positive number with units and q- is a negative number with units.  Therefore ρ+ is a positive number with units and ρ- is a negative number with units.)
In neutral ordinary matter ρ+ + ρ- = 0, i.e. the net charge per unit volume is zero.

The current density is defined as j = ρ+<v+> + ρ-<v->.
In this expression <v+> and <v-> are the average velocities of the positive and negative charges, respectively.
If <v-> = 0 and only positive charges are moving, then the direction of the current density is the direction of the velocity of the positive charges.  In ordinary conductors the positive charges have zero average velocity and only the negative charges are moving.  Then the direction of the current density is opposite to the direction of the velocity of the negative charges, as shown in the diagram on the right.

The SI unit of current density is (C/m3)(m/s) = C/(m2s).  The current density is a vector.  Its magnitude is the net amount of charge that crosses a unit area perpendicular to its direction per second in the direction of j minus the net amount of charge that crosses the same unit area per second opposite to the direction of j.  If just as many negative as positive charges move across a unit area in the same direction per second, then the current density is zero.

The electric current I is the net amount of charge which flows through a surface, i.e. the flux of the current density through a surface area.

I = ∫j∙dA= dQ/dt.

If j is uniform across the surface and constant in time, then
I = jA = j*Aperpendicular = ΔQ/Δt.
The SI unit of current is Ampere = Coulomb/second (A = C/s).  The current I is not a vector.  It is a scalar, but it can be a negative or positive number.  When obtaining the current from the current density using I = j∙A, we have to choose a direction for the unit vector n normal to the surface area through which the current flows.  If the net flow of positive charge is in the direction of n, (or the net flow of negative charge is in a direction opposite to the direction of n), then the current I is positive.  If the reverse is true, then the current I is negative.

Assume that a steady current flows in a wire with cross-sectional area A.  The electrons move with constant average velocity to the left as shown in the diagram.  The current density is a vector pointing towards the right.  Let the normal to the cross-sectional area point towards the right.  Then I = j*A= ΔQ/Δt, and the current is a positive number.  (If the electrons were moving towards the right, then j would point towards the left and I would be a negative number.)
The magnitude of the current I moving through a wire is given by

I = |ρ-|<v->A = n-|q-|<v->A = neqe<v->A,

where ne is the number of free electrons per unit volume.  The current I equals the number of electrons that pass any point along the wire per second times the unit of charge qe.  <v-> is the drift speed of the electrons.  It is their average speed, as they move along the wire.

### What causes the electrons to move in a wire?

Particles accelerate when they acted on by forces.  When a particle with charge q is placed in an electric field E, a force F = qE is acting on the particle and it accelerates.  To produce an electric field, something has to do work and separate charges.  Such a device is called an emf.  Examples of an emf are a battery or a power supply.  Such an emf separates charges by moving electrons internally from its positive to its negative terminal.  This produces an electric field, which points from the positive to the negative terminal outside and inside the emf.  Electrons in a wire placed in this electric field close to the battery will accelerate toward the end of the wire closest to the positive terminal.  But if the wire is not connected to the terminals of the emf, they will pile up at one end of the wire, leaving net positive charge at the other end.  The separated charges in the wire then produce their own electric field opposing the field produced by the emf.  The interior of the wire will be field free and charges will no longer accelerate.

If, however, the wire is connected to the emf, then the emf can pump the electrons through its interior back from the positive to the negative terminal.  Then field inside the wire will not be zero, and electrons will continue to accelerate towards the positive terminal.

Steady currents can only flow in continuous loops.  At any point, just as much charge has to flow out of a small volume surrounding the point as flows into the volume.  If this were not so, charge would accumulate at the point, setting up its own electric field.  This field would exert an additional force on the moving charges, disrupting the steady current.  The electric field in a homogeneous wire with constant cross-sectional area carrying a steady current is the same everywhere.  If it were not, electrons would move with different velocities in different sections, and charges would accumulate in certain regions.  The field produced by these charges would disrupt the steady current.  The diagram on the right shows the field in a wire carrying a steady current.

Even though an electric force is continuously acting on each electrons in a wire when a steady current is flowing, the electrons do not continue to accelerate indefinitely, but reach an average terminal velocity, called the drift velocity <v->.  When the electrons move with the drift velocity, then the resistive forces (frictional forces) acting on them are equal in magnitude and opposite in direction to the electric forces.  Typical drift speeds are on the order of 1 mm/s.

The resistive forces depend on the type of material the wire is made of.  The terminal velocity depends on the electric force and the resistive force.  The higher the terminal velocity, the larger is the current.  The current will increase if the voltage across the wire increases, (this increases the strength of the electric field), and decrease if the resistive forces increase in strength.  We write
I = ΔV/R, or R = ΔV/I.

This expression defines the resistance R of the wire.  The resistance is a measure of the resistive forces.  The unit of resistance is Ohm = Volt/Ampere (Ω = V/A).

#### Problem:

An emf source of 6.0 V is connected to a purely resistive lamp and a current of 2.0 A flows.  All the wires are resistance-free.  What is the resistance of the lamp?

Solution:
Ohm's law:  R = ΔV/I = (6 V)/(2 A) = 3 Ω.

The resistivity ρ of a material is the resistance of a piece of material with unit cross sectional area and unit length.  The resistance of a wire is proportional to its length l and inverse proportional to its cross-sectional area A.  We have

R = ρl/A.
The unit of resistivity is Ωm.  The conductivity σ of a material is σ = 1/ρ.

The resistivity is a property of the material a conductor is made of.  For many conductors it only weakly depends on the voltage ΔV across the conductor or the current I flowing through the conductor over a wide range of voltages and currents as long as the temperature of the material is held constant.  This is called Ohm's law.  R = ΔV/I is approximately constant for many conductors at constant temperature.

Ohm's law breaks down under extreme conditions for all conductors.  Many modern materials do not obey Ohms law under any circumstances.  (Such materials are called non-ohmic materials.)

The resistivity and therefore the resistance of most materials depends on the temperature.  Tables often list the resistivity at some reference temperature T0, (for example at 20oC), and then list a temperature coefficient α = (1/ρ0)Δρ/ΔT.  To find the resistance at some temperature T use ρ = ρ0[1 +α (T - T0)].  A thermistors is a semiconductor crystal whose resistance strongly depends on temperature.  Thermistor thermometers use thermistors to measure temperature.  The resistance is measured to obtain the temperature.  Thermistors are used as clinical temperature sensors in stethoscopes, as probes during surgery, and in other medical devices where temperature detection and control is vital.

#### Problem:

A copper wire has a length of 160 m and a diameter of 1.00 mm.  If the wire is connected to a 1.5 V battery, how much current flows through the wire?

Solution:
The current can be found from Ohm's Law, V = IR.  V is the battery voltage, so if R can be determined, the current can be calculated.
The resistance of the wire is R = ρl/A.
For copper ρ = 1.72*10-8 Ωm.
The cross-sectional area of the wire is A = πr2 = π(0.0005)2 = 7.85*0-7 m2.
The resistance of the wire then is ((1.72*10-8)*160/(7.85*10-7))Ω = 3.5 Ω.
The current is I = V/R = (1.5/3.5)A = 0.428 A.

#### Problem:

An aluminum wire has a resistance of 0.10 Ω.  If you draw this wire through a die, making it thinner and twice as long, what will be its new resistance?

Solution:
The initial resistance Ri of the aluminum wire with length L and cross-sectional area A is equal to is Ri = ρl/A.  The initial volume of the wire is AL. After passing the wire through the die, its length has changed to L' and its cross-sectional area is equal A'.  Its final volume is therefore equal to A'L'.  Since the density of the aluminum does not change, the volume of the wire does not change, and therefore the initial and final dimensions of the wire are related by AL = A'L'.
We have A' = L/L'A = A/2.
The final resistance Rf of the wire is given by Rf = ρL'/A' = 4ρL/A = 4Ri = 0.40 Ω.

#### Problem

A high voltage transmission line has an aluminum cable of diameter 3.0 cm, 200 km long.  What is the resistance of this cable?

Solution:
The resistivity of aluminum is 2.8*10-8 Ωm.  The length of the cable is 2*105 m.  The diameter of the cable is 3 cm and its cross-sectional area is equal to π(d/2)2 or 7.1*10-4 m2.  Substituting these values into R = ρL/A the resistance of the cable can be determined.
R = (2.8*10-8*2*105) Ω/( 7.1*10-4) = 7.9 Ω.