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ECE 2317 Applied Electricity and Magnetism. Prof. D. Wilton ECE Dept. Notes 25. Notes prepared by the EM group, University of Houston. Magnetic Field. v.  r. N. S. q. This experimental law defines the magnetic flux density vector B. Lorentz Force Law:.

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Prof d wilton ece dept

ECE 2317

Applied Electricity and Magnetism

Prof. D. Wilton

ECE Dept.

Notes 25

Notes prepared by the EM group, University of Houston.


Prof d wilton ece dept

Magnetic Field

v

r

N

S

q

This experimental law defines the magnetic flux density vector B.

Lorentz Force Law:

In general, (with both E and B present):

The units of B are Webers/m2 or Tesla [T].


Prof d wilton ece dept

Magnetic Gauss Law

z

S (closed surface)

N

N

y

x

S

B

Magnetic pole (not possible) !

S

No net flux out !


Prof d wilton ece dept

Magnetic Gauss Law: Differential Form

Apply the divergence theorem:

Since this applies for any volume,

B field flux lines either close on themselves or at infinity!


Prof d wilton ece dept

Ampere’s Law

I

y

x

Iron filings

(exact value)


Prof d wilton ece dept

Ampere’s Law (cont.)

Note: the definition of one Amp is as follows:

1 [A] current produces:

So


Prof d wilton ece dept

Ampere’s Law (cont.)

Define:

H is called the “magnetic field”

The units of H are [A/m].

(for single infinite wire)


Prof d wilton ece dept

Ampere’s Law (cont.)

y

C

I

x

The current is inside a closed path.


Prof d wilton ece dept

y

C

I

x

Ampere’s Law (cont.)

The current is outside a closed path.


Prof d wilton ece dept

C

Ampere’s Law (cont.)

Hence

Ampere’s Law

Iencl

Although this law was “derived” for an infinite wire of current, the assumption is made that it holds for any shape current.

This is an experimental law.

“Right-Hand Rule”


Prof d wilton ece dept

Amperes’ Law: Differential Form

C

S

(from Stokes’ theorem)

S is a small planar surface

Since the unit normal is arbitrary,


Prof d wilton ece dept

Maxwell’s Equations (Statics)

electric Gauss law

magnetic Gauss law

Faraday’s law

Ampere’s law

In contrast to electric fields, the sources for magnetic fields produce a circulation and are vectors! !


Prof d wilton ece dept

Maxwell’s Equations (Dynamics)

electric Gauss law

magnetic Gauss law

Faraday’s law

Ampere’s law


Prof d wilton ece dept

Displacement Current

Ampere’s law:

A

“displacement current”

insulator

h

z

+ + + + + + + + + + +

(This term was added by Maxwell.)

Q

I

The current density vector that exists inside the lower plate.


Prof d wilton ece dept

Ampere’s Law: Finding H

  • 1) The “Amperian path” C must be a closed path.

  • 2) The sign of Iencl is from the RH rule.

  • 3) Pick C in the direction of H (as much as possible).

  • 4) Wherever there is a component of H along path C , it

  • must be constant. => Symmetry!


Prof d wilton ece dept

z

I

y

C

r

x

Example

Calculate H

First solve for H .

“Amperian path”

An infinite line current along the z axis.


Prof d wilton ece dept

z

I

y

C

r

x

Example (cont.)


Prof d wilton ece dept

Example (cont.)

Hd cancels

2)Hz = 0

Hzdz=0

I

h

C

 = 

3)H = 0

I

Magnetic Gauss law:

h

S


Prof d wilton ece dept

z

I

y

r

x

Example (cont.)


Prof d wilton ece dept

Example

y

coaxial cable

b

I

a

c

a

I

r

x

C

b

z

c

The outer jacket of the coax has a thickness of t = c-b

 < a

Note: the permittivity of the material inside the coax does not matter here.

b <  < c


Prof d wilton ece dept

y

b

a

r

x

C

c

Example (cont.)

The other components are zero, as in the wire example.

This formula holds for any radius, as long as we get Ienclcorrect.


Prof d wilton ece dept

y

b

a

r

x

C

c

Example (cont.)

 < a

a < < b

b < < c

 > c

Note: There is no magnetic field outside of the coax (“shielding property”).


Prof d wilton ece dept

y

b

a

r

x

C

c

Example (cont.)

 < a

a < < b

b < < c

 > c

Note: There is no magnetic field outside of the coax (“shielding property”).


Prof d wilton ece dept

Example

Solenoid

n = N/L= # turns/meter

a

Calculate H

z

L

I

Find Hz

 < a

h

C


Prof d wilton ece dept

 > a

Hz

C

h

Example (cont.)


Prof d wilton ece dept

Example (cont.)

The other components of the magnetic field are zero:

1)H = 0 since

C

2)H = 0from

S


Prof d wilton ece dept

n = N/L = # turns/meter

a

z

L

I

Example (cont.)

Summary:


Prof d wilton ece dept

Example

z

Calculate H

y

-

x

x

+

Infinite sheet of current

top view

y


Prof d wilton ece dept

Hx-

x

Hx+

y

Example (cont.)

(superposition of line currents)

(magnetic Gauss Law)

By symmetry:


Prof d wilton ece dept

Example (cont.)

w

-

2y

x

+

C

y

Note: No contribution from the left and right edges (the edges are perpendicular to the field).


Prof d wilton ece dept

Example (cont.)

Note: The magnetic field does not depend on y.


Prof d wilton ece dept

y

I

h

I

x

w

z

w >> h

Example

Find H everywhere


Prof d wilton ece dept

Example (cont.)

y

h

x

Two parallel sheets of opposite surface current


Prof d wilton ece dept

“bot”

“top”

y

I

h

-I

x

magnetic field due to top sheet

magnetic field due due to bottom sheet

Example (cont.)

Infinite current sheet approximation of finite width current sheets


Prof d wilton ece dept

Example (cont.)

Hence


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