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No Slide Title Slide Presentations for ECE 329, Introduction to Electromagnetic Fields, to supplement “Elements of Engineering Electromagnetics, Sixth Edition” by Nannapaneni Narayana Rao Edward C Jor[.]

to supplement “Elements of Engineering

Electromagnetics, Sixth Edition”

by

Nannapaneni Narayana Rao

Edward C Jordan Professor of Electrical and Computer Engineering

University of Illinois at Urbana-Champaign, Urbana, Illinois, USA

Distinguished Amrita Professor of Engineering Amrita Vishwa Vidyapeetham, Coimbatore, Tamil Nadu, India

The Magnetic Field

dF m

B

I dl

The Magnetic Field

acts to exert force on charge when it is in motion

B = Magnetic flux density vector

Alternatively, since charge in motion constitutes current, magnetic field exerts forces on current elements

Fm B

v

q

Fm qv B

dF m I dl B

Units of B:

Sources: Currents;

Time-varying electric field

2

2

=

Wb

m

Ampère’s Law of Force

R

a12a21

dl1

dl2

I1

I2

0 1 1 12

4

4

I d

d I d

R

I d

I d

d I d

R

I d













l ×a

l × B

l ×a

l × B

I dl

aR R P B

(Biot-Savart Law)

B  0

4

I dl a R

R2

Note B  sin 

B  1

R2

B circular to the axis of the current element

0 Permeability of free space

= 4 10–7 H m



Ex.    

I A situated at 1, 2, 2 Find at 2, 1, 3

x y



B

since

R

R

x y z

R



l ×a l × R B

R a

a a a

 

0

12 3 x y

I dx





 a  a

 

0

3

=

I dx





B

w

Current Distributions

(a) Filamentary Current

I (A)

(b) Surface Current

Surface current density, JS (A/m)

wire

JS  I

w







max

area A

J

(c) Volume Current

Density, J (A/m2)

J A I

 max

r

y

r I

z – z

z

a2

dz

z

aR

 1



 2 P(r, , z)

0

2 2

0

2 2

4

sin 4

z R

dz d

I dz













    





    

a ×a B

a

 

2 1

2 1

0

2 0

2 0

0

cot

sin

cosec sin

cos 4

cos cos 4

z d

r

d z

I

r I r I r







 



































   









1

a

a

a a

a

For infinitely long wire,

a1  – , a2  ,

1  0 , 2  

B  0I

2r a

Magnetic Field Due to an Infinite Plane Sheet of

Uniform Surface Current Density

This can be found by dividing the sheet into

infinitely long strips parallel to the current density

and using superposition, as in the case of finding the electric field due to an infinite plane sheet of uniform surface charge density Instead of going through this procedure, let us use analogy To do this, we first

note the following:

P

B ar

I dl

(a) Point Charge Current Element

R

ar

Q

40R2 aR B  0I dl a R

4R2

r = 0

P

ar B

r I

(b) Line Charge Line Current

E  L0

20r ar B  20I

r a

 0I

2r az ar

z

r = 0

P

ar E

r

L0

P B

an

Then,

(c) Sheet Charge Sheet Current

P

an E

S0

E  S0

20 an B 0

2 JS an