Electric Current Chapter 27

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# Electric Current Chapter 27 - PowerPoint PPT Presentation

Electric Current Chapter 27. Electric Current Current Density Resistivity – Conductivity - Resistance Ohm’s Law (microscopic and macroscopic) Power Dissipated. E. •. F = q E = m a. Motion of a Point Charge in an Electric Field.

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## Electric Current Chapter 27

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### Electric CurrentChapter 27

Electric Current

Current Density

Resistivity – Conductivity - Resistance

Ohm’s Law (microscopic and macroscopic)

Power Dissipated

E

F = q E = m a

Motion of a Point Charge in an Electric Field

A particle of mass m and charge q, placed in an electric field E,

will experience a force F= qE

q

a

The particle will accelerate with acceleration: a = (q/m) E

In one dimension the motion of the particle is described by:

x = x0 + v0 t + a t2/2

v = v0 + a t

v2 = v02 + 2 a (x – x0)

L

q

a

m

V

Motion of a Point Charge in an Electric Field

Between the plates of a parallel plate capacitor (vacuum)

The charge accelerates with

a = (q/m) E = (q/m) (V/L)

If the particle (with charge q) starts at rest, and the potential difference between the plates is V, then the kinetic energy upon reaching the second plate will equal the change in potential energy: K = m v2 / 2 = qV

(e)(1V)=1 eV = 1.6 x 10-19 J ELECTRON VOLT

Electron motion in a conductor

E=0

You probably think of the conduction electrons as normally sitting still unless pushed by an electric field. That is wrong.

Electrons are in constant motion.

(And it is fast – around 10% of the speed of light.)

But their motion is random – and constantly changing as they bounce off of impurities. Because of the random motion no net flow occurs.

Electron motion in a conductor

E=0

E=0

• An electric field accelerates the electrons (along -E) and so modifies the trajectories of electrons between collisions.
• When E is nonzero, the electrons move almost randomly after each bounce, but gradually they drift in the direction opposite to the electric field.
• This flow of charge is called a current.

E

a wire

I

dq passes through in time dt

Electric current
• We define the electric current as the movement of charge, across a given area, per unit time:

I = dq / dt

• SI unit of current: 1 C/s = 1 Ampere (Amp)
• The direction of the current is the direction in which positive charges would move.
• Electrons move opposite to the direction of the current.

I

Current density

Cross-sectional area A

• If currentIflows through a surface A, the current density J is defined as the current per unit area: J = I / A
• After an electron collides with an impurity, it will accelerate under an E field with a = e E / m.
• Suppose the average time between collisions is t.

Then the average velocity is vd = a t = e E t / m.

This velocity is called theelectron drift velocity (which turns out to be much less than the speed of light)

Current density

A

Density of electrons: n

Number of electrons: N=n(AvdDt)

vd

• Construct the above volume.
• In time Dt all the electrons in it move out

through the right end.

• Hence the charge per unit time (the current) is

I = (N e) / Dt = n e A vdDt / Dt = n e A vd

• The current density is

J = I / A = n e vd = (n e2 t / m) E

vdDt

Example: What is the drift velocity of electrons in a

Cu wire 1.8 mm in diameter carrying a current of 1.3 A?

In Cu there is about one conduction electron per atom.

The density of Cu atoms is

Example: What is the drift velocity of electrons in a

Cu wire 1.8 mm in diameter carrying a current of 1.3 A?

In Cu there is about one conduction electron per atom.

The density of Cu atoms is

Find vd from J=I/A=1.3A/(p(.0009m)2)=5.1x105 A/m2

Example: What is the drift velocity of electrons in a

Cu wire 1.8 mm in diameter carrying a current of 1.3 A?

In Cu there is about one conduction electron per atom.

The density of Cu atoms is

Find vd from J=I/A=1.3A/(p(.0009m)2)=5.1x105 A/m2

Now use

Example: What is the drift velocity of electrons in a

Cu wire 1.8 mm in diameter carrying a current of 1.3 A?

In Cu there is about one conduction electron per atom.

The density of Cu atoms is

Find vd from J = I/A = 1.3A/(p(.0009m)2) = 5.1x106 A/m2

Now use

Much less than one millimeter per second!

Ohm’s Law
• We found that J=(ne2t/m)E, that is, that the current J is proportional to the applied electric field E(both are vectors):
Ohm’s Law
• We found that J=(ne2t/m)E, that is, that the current J is proportional to the applied electric field E(both are vectors):
• J = s E
• Ohm’s Law
Ohm’s Law
• We found that J=(ne2t/m)E, that is, that the current J is proportional to the applied electric field E(both are vectors):
• J = s E
• Ohm’s Law
• s is the Conductivity, s = J / E.
• Units are (A/m2) divided by (V/m) = A/(Vm)
Ohm’s Law
• We found that J=(ne2t/m)E, that is, that the current J is proportional to the applied electric field E(both are vectors):
• J = s E
• Ohm’s Law
• s is the Conductivity, s = J / E.
• Units are (A/m2) divided by (V/m) = A/(Vm)
• It is useful to turn this around and define the
• Resistivity asr = 1/s, so E= rJ.
• Units of r are (V/A)m
Ohm’s Law

s and r are dependent only on the material,- not its length or area.

However, consider a metal rod of resistivity r:

+V

r, area A

0 volts

L

Ohm’s Law

s and r are dependent only on the material,- not its length or area.

However, consider a metal rod of resistivity r:

+V

r, area A

0 volts

L

E = r J

(V/L) = r (I/A)

V = (rL/A) I

Ohm’s Law

s and r are dependent only on the material,- not its length or area.

However, consider a metal rod of resistivity r:

+V

r, area A

0 volts

L

E = r J

(V/L) = r (I/A)

V = (rL/A) I

V=IR

with R=rL/A

Ohm’s Law

The macroscopic formV = I R is the most

commonly used form of Ohm’s Law.

R is the Resistance

It depends on the material type and shape:R = r L / AUnits: ohms (W).

As r = R A / L, common units for the

resistivity r are Ohm-meters.

Similarly, common units for the

conductivity s = 1 / r are (Ohm m)-1or Mho/m

Ohm’s Law

A

L

J = s Emicroscopic form

 = conductivity  = 1/ = resistivity

 and r = 1/ are dependent only on the material,

(NOT on its length or area)

V = I Rmacroscopic form

R depends on the material type and shape

R = L / A = resistance

Electrical Power Dissipation

V

• In traveling from a to b , energy decrease of dq is:dU = dq V
• Now, dq = I dt
• Therefore, dU = I dt V
• Rate of energy dissipation is dU / dt = I V
• This is the dissipated power, P. (Watts, or Joules /sec)

b

a

dq

Resistance, R

Electrical Power Dissipation

V

• In traveling from a to b , energy decrease of dq is:dU = dq V
• Now, dq = I dt
• Therefore, dU = I dt V
• Rate of energy dissipation is dU / dt = I V
• This is the dissipated power, P. [Watts, or Joules /sec]
• Hence,

b

a

dq

Resistance R

P = I V

P = I2 R = V2 / R

or

Resistivities of Selected Materials

Material Resistivity [ m]

Aluminum 2.65x10-8

Cooper 1.68x10-8

Iron 9.71x10-8

Water (pure) 2.6x105

Sea Water 0.22

Blood (human) 0.70

Silicon 640

Glass 1010 – 1014

Rubber 1013 – 1016

What is the resistance of a Cu wire, 1.8 mm in diameter, and 1 m long ?.

R =  L / A  R = (1.68x10-8) 1 /  (0.0009)2 

R = 6.6x10-3 

What is the voltage difference between the extremes of a Cu wire, 1.8 mm in diameter, and 1 m long, when the current is 1.3 A ?.

V = I R = (1.3 A) 6.6x10-3 = 8.6x10-3 V

What is the power dissipated in a Cu wire, 1.8 mm in diameter, and 1 m long, when the current is 1.3 A ?.

P = I2 R = (1.3)2 6.6x10-3 W = 1.12x10-2 W