Prof. David R. JacksonECE Dept.

Fall 2014

Notes 3

ECE 2317 Applied Electricity and Magnetism

Notes prepared by the EM Group University of Houston

1

Current is the flow of charge: the unit is the Ampere (Amp)

Current

1 Amp 1 [C/s]

Convention (Ben Franklin): Current flows in the direction that positive charges move in.

+ + + +

Current flows from left to right.

Velocity

Ampere

2

Current (cont.)

72 2 10 N/m when 1 mxF d

Definition of I =1 Amp:

I

I

d

Fx2 x

# 1

# 2

Two infinite wires carrying DC currents

The Amp is actually defined first, and from this the coulomb is defined.

3

20

2 2x

IF

d

70 4 10 [A/m] From later in the semester:

Note: 0 = permeability of free space.

Postulate: Positive charges moving one way is equivalent to negative charges moving the other way (in terms of measurable physical effects).

Current (cont.)

1 [A]

- - - -

Flow rate is 1 [C] per second

Velocity

This is equivalent to:

+ + + +

Flow rate is 1 [C] per second

Velocity

4

In both cases:

Current flows from right to left.

Sign convention: A positive current flowing one way is equivalent to a negative current flowing the other way.

Current (cont.)

+ + + +

1 [A]

Flow rate is 1 [C] per second

Velocity

-1 [A]or

Note: The blue arrow is called the reference direction arrow.

5

It is very useful in circuit theory to assume reference directionsand allow for negative current.

Mathematical definition of current

Current (cont.)

Reference direction arrow

Q = amount of charge (positive or negative) that crosses the plane in the

direction of the reference arrow in time t.

I

QI

t

Q

dQi t

dtMore generally,

6Uniform current Non-uniform current

Example

i(t)Q

dQi t

dt

10 2 cos 2 60 [A]i t t

Find the charge Q(t) that crosses the dashed line going from left to right in the time interval (0, t) [S].

1

2

0

0

( )

( ) 0 0

( )

t

t

Q t i t dt C

i t dt C Q

i t dt

7

Example (cont.)

0

0

0

( )

10 2 cos 2 60

110 2 sin 2 60

2 60

t

t

t

Q t i t dt

t dt

t

C10 2

sin 2 60 [ ]120

Q t t

8

Current Density Vector J

The current-density vector points in the direction of current flow (the direction of positive charge motion).

v

++++

+ +

+

+ +

++

+

Consider a general “cloud” of charge density, where the charge density as well as the velocity of the charges may be different at each point.

9

ˆ ˆJ v

The magnitude of the current-density vector tells us the current density (current per square meter) that is crossing a small surface that is perpendicular to the current-density vector.

2A/m[ ]I

J JS

I = the current crossing the surface S in the direction of the velocity vector.

Current Density Vector (cont.)

2A/mˆ [ ]I

J vS

10

v+++

++ +

+

+ +

++

+

S

v

Hence

Current Density Vector (cont.)

11

Consider a small tube of moving charges inside the cloud:

v+++

++ +

+

+ +

++

+

S

v++

++ +

+

+ +

++

+

+S

Tube

Current Density Vector (cont.)

// Q t LI Q t Q LJ

S S S L V t

L = distance traveled by charges in time t.

vJ vor ˆ ˆvJ J v v v

vJ vHence

so

12

+ + + +S I

L

Q v

Reference plane

Tube

Current Density Vector (cont.)

vJ v

Summary

13

2A/mˆ [ ]I

J vS

v+++

++ +

+

+ +

++

+

S

v

Current Crossing a Surface

ˆcos cosI J S J S J n S

This is the current crossing the surface S in the direction of the unit normal.

n̂

S

J

cosS

14

The surface S does not have to be perpendicular to the current density vector.

ˆ ˆ cos cos cosJ n J n J J

Current Crossing Surface

ˆI J n S

Integrating over a surface,

ˆS

I J n dS Note: The direction of the unit normal vector determines whether the current is measured going in or out.

S

Jn̂

15

Example

Cloud of electrons

Ne = 8.47 1028 [electrons / m3]

4 m/sˆ 1.0 10 [ ]v z

(a) Find: current density vector

vJ v

28 3 19 10 38.47 10 [1/ m ] 1.602 10 [C] 1.36 10 [C/m ]v e eN q

26 A/mˆ 1.36 10 [ ]J z

The cloud of electrons is inside of a copper wire.

Hence

There are 8.47 1028 atoms / m3

There is one electron/atom in the conduction band.

Notes:

16

- - - - --- - - - --

z

10 4ˆ1.36 10 1.0 10J z

Example (cont.)

(b) Find: current in wire for the given reference direction

In̂

6

6

6 2

ˆ

ˆ ˆ1.36 10

1.36 10

1.36 10

S

S

S

I J n dS

z z dS

dS

a

6 2

26

1.36 10

1.36 10 0.001

I a

A4.26 [ ]I

Radius a = 1 [mm]

I z- - - - --

6 2ˆ 1.36 10 [A/m ]J z

Note: The wire is neutral, but the positive nuclei do not move.

17

Example (cont.)

Density of Cu: 8.94 [g/cm3]

Avogadro’s constant: 6.02214179301023 atoms/mol

1 mol = amount of material in grams equal to the atomic weight of atom

Atomic weight of Cu: 63.546 (atomic number is 29, but this is not needed)

1 electron per atom in the conduction band (Ne = atoms/m3)

Properties of copper (Cu)

3 3 233

28 3

28 3

18.94 10 [kg / m ]  [mol/kg] 6.0221417930 10 [atoms/mol]

63.546 10  

8.47 10 [atoms/m ]

8.47 10 [electrons/m ]

eN

18

28 38.47 10 [electrons/m ]eN

Using knowledge of chemistry, calculate the value of Ne (that is used in this example).

3 63.546 [g/mol] 63.546 10  [k g/mol] Atomic weight

Example

ˆ ˆ 3S S S

I J n dS J z dS xy dS

2 3 2 2ˆ ˆ ˆ3 3 3 [A/m ]J x x y y z y z xy

x

y

z

(1,0,0)

(0,1,0)

Find the current I crossing the surface S in the upward direction.

19

(1,1,0)

S

Example (cont.)

1 1

0 0

1 1

0 0

3

3

3

1 13

2 2

3

4

S

I xy dS

xy dx dy

x dx y dy

20

x

y

1

1

A0.75 [ ]I

Note: The integrand is separable, and the limits of integration are fixed numbers. Hence, we can split this into a product of two one-dimensional integrals.

Example

ˆ ˆ 3S S S

I J n dS J z dS xy dS

2 3 2 2ˆ ˆ ˆ3 3 3 [A/m ]J x x y y z y z xy

x

y

z

(1,0,0)(0,1,0)

S

Find the current I crossing the surface S in the upward direction.

21

Example (cont.)

( )1

0 0

12

0 0

12

0

14 3 2

0

3

3

13

2

31

2

3 1 2 1

2 4 3 2

S

y x

y x

I xy dS

xy dy dx

x y dx

x x dx

x x x

A0.125 [ ]I

x

y

1

1y(x) = 1 - x

22

Note: The integrand is separable, but the limits of integration are not fixed numbers. Hence, we cannot split this into a product of two one-dimensional integrals.