Preparatory_Notes_Exam2.ppt

Preparatory Notes
and Examples for
Exam #2
EEE241: Fundamentals of
Electromagnetic Theory
Instructor: Dragica Vasileska

Electromagnetic Force
The first term in the Lorentz Force Equation represents the electric force Fe
acting on a charge q within an electric field is given by.
e
q

F E
The electromagnetic force is given by Lorentz Force Equation (After Dutch
physicist Hendrik Antoon Lorentz (1853 – 1928))
 
q
  
F E u B
The electric force is in the direction of the electric field.
The Lorentz force equation is quite useful in determining the paths charged
particles will take as they move through electric and magnetic fields. If we also
know the particle mass, m, the force is related to acceleration by the equation
.
m

F a

Since the magnetic force is at right angles to the magnetic field, the work done
by the magnetic field is given by
cos90 0
W d FdL
  
 
F L
Magnetic Force
The magnetic force is at right angles to the magnetic field.
The magnetic force requires that the charged particle be in motion.
It should be noted that since the magnetic force acts in a direction normal to the
particle velocity, the acceleration is normal to the velocity and the magnitude of
the velocity vector is unaffected.
The second term in the Lorentz Force Equation represents magnetic force
Fm(N) on a moving charge q(C) is given by
m
q
 
F u B
where the velocity of the charge is u (m/sec) within a field of magnetic flux
density B (Wb/m2). The units are confirmed by using the equivalences
Wb=(V)(sec) and J=(N)(m)=(C)(V).

Magnetostatics – Biot-Savart’s Law
Shortly following Oersted’s discovery that currents produce magnetic
fields, Jean Baptiste Biot (1774-1862) and Felix Savart (1791-1841)
arrived at a mathematical relation between the field and current.
The Law of Biot-Savart is
1 1 12
2
12
2
4
I d
d
R



L a
H
Note: The Biot-Savart law is analogous to the Coulomb’s law
equation for the electric field resulting from a differential charge
1 12
2 2
12
.
4
dQ
d
R


a
E
To get the total field resulting from a current,
you can sum the contributions from each
segment by integrating
2
4
.
R
Id
R


 
L a
H
(A/m)
(A/m)

Magnetostatics – Ampere’s Circuital Law
In electrostatics problems that featured a lot of symmetry we were able to
apply Gauss’s Law to solve for the electric field intensity much more easily
than applying Coulomb’s Law.
Likewise, in magnetostatic problems with sufficient symmetry we can employ
Ampere’s Circuital Law more easily than the Law of Biot-Savart.
Ampere’s Circuital Law says that the integration of H around any closed
path is equal to the net current enclosed by that path.
The line integral of H around a closed path is termed
the circulation of H.
enc
d I

 H L
H
enc
I

Magnetostatics – Magnetic Flux Density
The magnetic flux density, B, related to the magnetic field intensity in free
space by
o


B H
where o is the free space permeability, given in units of henrys per meter, or
7
4 x10 /
o
H m
  

The units of B are therefore (H)(A)/m2, but it is more instructive to write
webers per meter squared, or Wb/m2, where Wb=(H)(A).
But for brevity, and perhaps to honor a deserving scientist, a tesla , T, equivalent
to a Wb/m2, is the standard unit adopted by the International System of Units.
The amount of magnetic flux, , in webers, from magnetic field passing
through a surface is found in a manner analogous to finding electric flux:
d
  B S

Therefore, the algebraic sum of the currents
entering any closed surface is zero.
1
0
n
i
i
I



Kirchhoff’s Current Law: The algebraic sum
of the currents entering any node is zero.
Gauss’s Law and Kirchhoff’s Current Law
Gauss’s Law: The net magnetic flux passing through
a closed surface (Gaussian surface) must be zero
cos 0
cos
d HdS I

  
  
  
B S
Node
Closed
Surface
I1
I2
I3
I4
I5
0
I 

1 2 3 4 5
1
0
n
i
i
I I I I I I

     

This is analogous to Kirchhoff’s Current Law (KCL)!

The tangential magnetic field intensity is
continuous across the boundary when the
surface current density is zero.
o r
 

B H
We know that
Important Note:
Special Case: If the surface current density K = 0, we get
Magnetostatic Boundary Conditions
T1 T2
H H

If K = 0
T1 T2
H H K
 
1 2
T1 T2
o o
B B
   

Using the above relation, we obtain
T1 T2
H H

The tangential component of the magnetic flux density B is not continuous
across the boundary.
Therefore, we can say that T1 T2
B B

o r
 

B
H
(or)

Gauss’s Law for Magnetostatic fields:
= 0
d
 B S
To find the second boundary condition, we center a Gaussian pillbox
across the interface as shown in Figure.
We can shrink h such that the flux out of the side of the pillbox is
negligible. Then we have
 
N1 N N N2 N N
N1 N2
( )
0.
d B dS B dS
B B S
  
   
  
B S a a a a
N1 N2 .
B B

Normal BC:

Magnetic Force
Example: At a particular instant in time, in a region of space where E = 0 and B
= 3ay Wb/m2, a 2 kg particle of charge 1 C moves with velocity 2ax m/sec.
What is the particle’s acceleration due to the magnetic field?
2
sec
1
3 3
2
2
m
q
m
    
x y z
a u B a a a
To calculate the units: 2 2 2
sec
sec sec sec
C m Wb kg m N m J V m
kg m N J C V Wb
 
    
    
    
Example: A 10. nC charge with velocity 100. m/sec in the z direction enters a
region where the electric field intensity is 800. V/m ax and the magnetic flux
density 12.0 Wb/m2 ay. Determine the force vector acting on the charge.
  9
2
10 10 800 100 12 2
x z y x
V m Wb
q x C N
m s m

  
       
 
 
F E u B a a a a
Given: q= 10 nC, u = 100 az (m/sec), E = 800 ax V/m, B = 12.0 ay Wb/m2.
Given: q= 1 nC, m = 2 kg, u = 2 ax (m/sec), E = 0, B = 3 ay Wb/m2.
m

F a
Newtons’ Second Law Lorentz Force Equation
   
q q
   

F E u B u B
Equating

Magnetic Force
Example: A pair of parallel infinite length lines each carry current I = 2A in the
same direction. Determine the magnitude of the force per unit length between
the two lines if their separation distance is (a) 10 cm, (b)100 cm. Is the force
repulsive or attractive?
1 2
12
2
o
I I L
y


  y
F a
1 2
12
2
o
I I
y
L


  y
F
a
Case (a) y = 10 cm
Magnetic force between two current elements
when current flow is in the same direction
Magnetic force per unit length
12
7
2
(4 10 )(2)(2)
2 (10 10 )
8 N/m
L






 


y
F
a
Case (a) y = 10 cm
12
7
2
(4 10 )(2)(2)
2 (100 10 )
0.8 N/m
L






 


y
F
a

Magnetostatics – An Infinite Line current
Example: Consider an infinite length line along the z-axis conducting current I in
the +az direction. We want to find the magnetic field everywhere.
An infinite length line of current
We first inspect the symmetry and see that
the field will be independent of z and  and
only dependent on ρ.
So we consider a point a distance r from the line
along the ρ axis.
2
4
R
Id
R


 
L a
H
The Biot-Savart Law
IdL is simply Idzaz, and the vector from the source to the test point is
R z 

  
R z
a a a
 
 
3
2 2 2
.
4
Idz z
z


 


  



z z
a a a
H
The Biot-Savart Law becomes

Pulling the constants to the left of the integral and realizing
that az x az = 0 and az x aρ = a, we have
 
3
2 2 2
4
I dz
z


 





a
H
The integral can be evaluated using the formula given in Appendix D
 
3 2 2 2 2
2 2
dx x
a x a
x a




2 2 2 2 2 2 2 2 2
2
t t t
I
t t t
     
   
   
  
   
  
   
 
3
2 2 2
2
2 2
2 2
2 2 2
1 1
dz
z
I
t


 
 

 
   
   
 
   

   



t
    


































 2
2
2
2
2
2
2
2
2
2
2
2
2
3
2
2
t
t
t
t
z
z
z
z
z
dz
I
t
t
t
t
t
t 








When the limit

2
I 


a
H
We find the magnetic field intensity resulting
from an infinite length line of current is
An infinite length line of current
 
3
2 2 2
2
2
dz
z
I



 
 

 
3
2 2 2
4
I dz
z


 





a
H
Using
Magnitude: The magnitude of the
magnetic field is inversely
proportional to radial distance.
1



H a
Direction: The direction of the
magnetic field can be found using
the right hand rule.

Magnetostatics – A Ring of Current
Example: Let us now consider a ring of current with radius a lying in the x-y
plane with a current I in the +az direction.
The objective is to find an expression for the field at an arbitrary point a
height h on the z-axis.
 
 
 
 
2 2
z z
3 3
2 2 2 2
2 2
0 0
4
4
Iad h a d h a
Ia
h a h a
 
   
 
 


 
   

 

 
a a a a a a
H
2
4
R
Id
R


 
L a
H
The Biot-Savart Law
R z
h a 
 
R
R = a a a
The differential segment
The vector drawn from the source to the test point is
2 2
R a
h
    R
z
h a 
 
R
a a a
Magnitude: Unit Vector:
dL = ada
The biot-Savart Law can be written as

Magnetostatics – A Ring of Current
We can further simplify this expression by considering the symmetry of the problem
 
 
 
 
2 2
z
3 3
2 2 2 2
2 2
0 0
4 4
z
d h a d h a
h a h a
Ia Ia
 
  
 
 
 
 
  
 

 
a a a a a
H =
A particular differential current element will give a field
with an aρ component (from a x az) and an az
component (from a x –aρ).
Taking the field from a differential current element on the
opposite side of the ring, it is apparent that the radial
components cancel while the az components add.
 
2
2
z
3
2 2 2
0
4
Ia
d
h a






a
H
 
2
z
3
2 2 2
2
Ia
h a


H a
z
2
I
a

H a
aρ Components
Cancel
az components add
At h = 0, the center of the loop, this equation reduces to
H

Magnetostatics – A Solenoid
Solenoids are many turns of insulated wire coiled in the shape
of a cylinder.
A solenoid
Suppose the solenoid has a length h, a radius a, and is made
up of N turns of current carrying wire. For tight wrapping, we
can consider the solenoid to be made up of N loops of current.
To find the magnetic field intensity from a single loop at a point
P along the axis of the solenoid, from we have
 
2
z
3
2 2 2
d
2 '
dIa
a
z


P
H a
The differential amount of field resulting from a differential
amount of current is given by
The differential amount of current can be considered a function
of the number of loops and the length of the solenoid as
'
N
dI Idz
h


Magnetostatics – A Solenoid
Fixing the point P where the field is desired, z’ will range
from –z to h-z, or
 
 
2
2 2
2
2 2
'
2 '
'
.
2 '
h z
z
h z
z
NIa dz
h z a
NIa dz
h z a










z
z
H a
a
 
2 2 2
2
2
NI h z z
h z a
h z a

 

 
 
 
 
 
z
H a
This integral is found from Appendix D, leading to the solution
At the very center of the solenoid (z = h/2), with the assumption that the length is
considerably bigger than the loop radius (h >> a), the equation reduces to
NI
h
 z
H a

Magnetostatics – Ampere’s Circuital Law Application to
Current Sheet
Example: Let us now use Ampere’s Circuital Law to find the magnetic
field intensity resulting from an infinite extent sheet of current.
Let us consider a current sheet with uniform current
density K = Kxax in the z = 0 plane along with a
rectangular Amperian Path of height h and width w.
Amperian Path: In accordance with the right hand
rule, where the thumb of the right hand points in
the direction of the current and the fingers curl in
the direction of the field, we’ll perform the
circulation in the order
.
a b c d a
   
.
enc
b c d a
a b c d
d I
d d d d

   

   
H L
H L H L H L H L
Ampere’s Circuital Law

Magnetostatics – Ampere’s Circuital Law
Application to Current Sheet
From symmetry arguments, we know that there is no
Hx component.
b c d a
a b c d
d d d d d
   
    
H L H L H L H L H L
 
0
y y
0
y
H y H y
2H
w
w
d d d
w




 
 
 y y y y
a
H L -a a a
Above the sheet H = Hy(-ay) and below the sheet H = Hyay.
The current enclosed by the path is just
0
w
x x
I K dy K w

  

Equating the above two terms gives
2
x
y
K
H 
N
1
2
 
H K a
where aN is a normal vector from the sheet current to the test point.
A general equation for an infinite current sheet:

Example: The magnetic field intensity is given as H1 = 6ax + 2ay + 3az (A/m)
in a medium with r1 = 6000 that exists for z < 0. We want to find H2 in a
medium with r2 = 3000 for z >0.
Step (a) and (b): The first step is to break H1 into its normal component (a) and
its tangential component (b).
Step (c): With no current at the interface, the tangential component is the same
on both sides of the boundary.
Step (d): Next, we find BN1 by multiplying HN1 by the permeability in medium 1.
Step (e): This normal component B is the same on both sides of the boundary.
Step (f): Then we can find HN2 by dividing BN2 by the permeability of medium 2.
Step (g): The last step is to sum the fields .

A coils of 200 turns is uniformly wound around a wooden ring
with a mean circumference of 600 mm and area of cross-section
of 500 mm2. If the current flowing into the coil is 4 A, Calculate
(a) the magnetic field strength , (b) flux density dan (c) total flux
N = 200 turns
l = 600 x 10-3 m
A = 500 x 10-6 m2
I = 4 A
(a) H = NI/l = 200 x 4 / 600 x 10-3 = 1333 A
(b) B = oH = 4 x 10-7 x 1333 = 0.001675 T = 1675 T
(c) Total Flux  = BA = 1675 x 10-6 x 500 x 10-6
= 0.8375 Wb
turns
Example on Magnetic Circuits

Ohm‘s law I = V/R [A]
Where I =current; V=voltage and R=resistance
And the resistance can be relate to physical parameters as
R =  l /A ohm
Where =resistivity [ohm-meter], l= length in meter and A=area
of cross-section [meter square]
Analogy to the Ohm‘s law
V=NI=H l I= and R=S
Reluctance ( S )
 
weber
S
Hl

  
weber
ampere
A
S
o
r
/




where

Magnetic circuit with different
materials
1
1
1
A
a
S


l
2
2
2
B
a
S


l
B
A S
S
S 

2
2
2
1
1
1
a
a 



l
l
and
For A: area of cross-section = a1
mean length = l1
absolute permeability = 1
For B: area of cross-section = a2
mean length = l2
absolute permeability = 2

340
mm
340
mm 150
mm
1mm
A magnetic circuit made of silicon steel is arranged as in the
Figure. The center limb has a cross-section area of 800mm2 and
each of the side limbs has a cross-sectional area of 500mm2.
Calculate the m.m.f required to produce a flux of 1mWb in the
center limb, assuming the magnetic leakage to be negligible.
Example on Magnetic Circuits

  
a
B
A
A S
S
S
f
m
m 





 2
1
.
.
A
B

 T
A
B 25
.
1
10
800
10
1
6
3





 

15915
10
500
10
4
34000
10
340
6
7
3
1
1
1 






 




 A
S
o
r

Looking at graph at B=1.25T r =34000
Apply voltage law in loops A and B 340
mm
340
mm 150
mm
1mm
A B
4388
10
800
10
4
34000
10
150
6
7
3
2 





 



S
994718
10
800
10
4
10
1
6
7
3





 



a
S

1007
999
8 


    
994718
4388
10
1
15915
10
5
.
0
.
. 3
3




 

f
m
m
Since the circuit is symmetry A =B
In the center limb , the flux is 1mWb which is equal to 2 
Therefore =0.5mWb
  
a
S
S
S
f
m
m 



 2
1 2
.
.
A

Preparatory_Notes_Exam2.ppt

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