Transcript Slide 1

Magnetostatics
1
EEL 3472
Magnetostatics
If charges are moving with constant velocity, a static
magnetic (or magnetostatic) field is produced. Thus,
magnetostatic fields originate from currents (for instance,
direct currents in current-carrying wires).
Most of the equations we have derived for the electric fields
may be readily used to obtain corresponding equations for
magnetic fields if the equivalent analogous quantities are
substituted.
2
EEL 3472
Magnetostatics
Biot – Savart’s Law
dH  k
The magnetic field intensity
dH produced at a point P by
the differential current
element Idl is proportional
to the product of Idl and the
sine of the angle  between
the element and the line
joining P to the element and
is inversely proportional to
the square of the distance R
between P and the element.
Idlsin 
R2
1
IN SI units, k  4
, so
dH  k
3
Idlsin 
4 R 2
EEL 3472
Magnetostatics
Using the definition of cross product (A  B  ABsinAB en)
we can represent the previous equation in vector form as
Idl  er Idl  R
dH 

er  R
RR
2
3
R
4R
4R
The direction of d H can be determined by the right-hand rule
or by the right-handed screw rule.
If the position of the field point is specified by r and the
position of the source point by r 
dH 
I(dl  er r ) I[dl  (r  r)]
2 
3
4 r  r
4  r  r
where err is the unit vector directed from the source point to
the field point, and r  r  is the distance between these two

points.
4
EEL 3472
Magnetostatics
Example. A current element is located at x=2 cm,y=0, and
z=0. The current has a magnitude of 150 mA and flows in
the +y direction. The length of the current element is 1mm.
Find the contribution of this element to the magnetic field
at x=0, y=3cm, z=0
dl  103 ey
r  3 102 ey r   2 * 102 ex
(r  r)  (2ex  3ey ) 102
r  r  13 102



dl  (r  r )  103 ey  (2ex  3ey )  102
 10-5 (2ez )
ey  ex  ez
ey  ey  0

dH 
I[dl  (r  r )]
3
4  r  r 
0.15(2 10-5 )ez

4  ( 13  102 ) 3
 5.09 10-3 ez A m
5
EEL 3472
Magnetostatics
The field produced by
a wire can be found
adding up the fields
produced by a large
number of current
elements placed head
to tail along the wire
(the principle of
superposition).
6
EEL 3472
Magnetostatics
Consider an infinitely long straight wire located on the z axis
and carrying a current I in the +z direction. Let a field point
be located at z=0 at a radial distance R from the wire.
e
z



R
(R  Z2 )1 / 2
z 0
r  r   (R2  Z2)1 / 2

H
7

dH 
Idz
R
2
2
2
2 1/2

4

(
R

Z
)
(
R

Z
)
z  
2
I(dl  err )
4 r  r 
2

Idz(ez  err )
4 r  r 
2


Idz sin  e
4(R 2  Z2 )
1
IR 
Z
  2
dz

2 3 /2
(
R

Z
)
4  R2 R2  Z2

I
I

(1  1) 
4R
2R




 
EEL 3472
Magnetostatics
8
EEL 3472
Magnetostatics
dH 
r  r  
I(dl  er'r )
2
4 r  r
a2  b2  const
dl  ad
(perpendicular to both dl
and er 'r )
The horizontal components of H
(x-components and ycomponents add to zero
c os  sin(900  ) 
9
a
a2  b2
H 

2
0
Iad
a
 2
2
2
4 (a  b ) (a  b2 )1/ 2
Ia2
Ia2

 2 
4 (a 2  b 2 ) 3 / 2
2(a 2  b 2 ) 3 / 2
H  H ez
EEL 3472
Magnetostatics
Magnetic Flux Density
The magnetic flux density vector is related to the magnetic
field intensity H by the following equation
B  H,
T (tesla) or Wb/m2
where  is the permeability of the medium. Except for
ferromagnetic materials ( such as cobalt, nickel, and iron),
most materials have values of  very nearly equal to that for
vacuum,
o  4  107
H/m
(henry per meter)
The magnetic flux through a given surface S is given by
   B  d s,
S
10
Wb (weber)
EEL 3472
Magnetostatics
Magnetic Force
One can tell if a magnetostatic field is present because it
exerts forces on moving charges and on currents. If a current
element Idl is located in a magnetic field B , it experiences a
force
dF  Idl  B

This force, acting on current element Idl , is in a direction
perpendicular to the current
element and also perpendicular

to B . It is largest when B and the
 wire are perpendicular.
The force acting on a charge q moving through a magnetic
field with velocity v is given by
F  q(v  B )

The force is perpendicular to the direction of charge motion,
and is also perpendicular
to B .

11
EEL 3472
Magnetostatics
The Curl Operator
This operator acts on a vector field to produce another vector
field. Let B(x, y, z) be a vector field. Then the expression
for the curl of B in rectangular coordinates is
 Bz By 
 By Bx 
 Bx Bz 
  B  



 ex  
 ez
 ey  

y

z

z

x

x

y






where Bx , B y , and Bz are the rectangular components of
The curl operator can also be written in the form of a
determinant:
12
ex
ey
ez

B 
x

y

z
Bx
By
Bz
EEL 3472
Magnetostatics
The physical significance of the curl operator is that it
describes the “rotation” or “vorticity” of the field at the point
in question. It may be regarded as a measure of how much
the field curls around that point.
The curl of B is defined as an axial ( or rotational) vector
whose magnitude is the maximum circulation of B per unit
area as the area tends to zero and whose direction is the
normal direction of the area when the area is oriented so as
to make the circulation maximum.

L B  dl 

curl B  rot B    B   lim
an

s 0
s 


max
where the area  s is bounded by the curve L, and an is the
unit vector normal to the surface  s and is determined using
the right – hand rule.
13
EEL 3472
Magnetostatics
14
EEL 3472
Magnetostatics
By 
 B
 By Bx 
Bz 
 B
  B   z 

 ex   x 
 ez
 ey  
z 
x 
y 
 z
 y
 x
15
EEL 3472
Magnetostatics
Classification of Vector Fields
A vector field is uniquely characterized by its divergence
and curl (Helmholtz’s theorem). The divergence of a vector
field is a measure of the strength of its flow source and the
curl of the field is a measure of the strength of its vortex
source.
16
EEL 3472
Magnetostatics
If   A  0 then A is said to be solenoidal or
divergenceless. Such a field has neither source nor sink of
flux.
Since   (  F)  0 (for any F ), a solenoidal field A
can always be expressed in terms of another vector F :
A  F
If   A  0 then A is said to be irrotational (or potential, or
conservative). The circulation of A around a closed path is
identically zero.
Since   V  0 (for any scalar V), an irrotational field A
can always be expressed in terms of a scalar field V:
A  V
17
EEL 3472
Magnetostatics
Magnetic Vector Potential
Some electrostatic field problems can be simplified by
relating the electric potential V to the electric field intensity
E(E  V). Similarly, we can define a potential associated with
the magnetostatic field B :
B  A
where A is the magnetic vector potential.
Just as we defined in electrostatics
dq(r ' )
V(r )  
4 r  r '
we can define
A(r ) 
18
I(r ' )dl'
L 4 r  r'
(electric scalar potential)
Wb
m
(for line current)
EEL 3472
Magnetostatics
The contribution to A of each differential current element Idl'
points in the same direction as the current element that
produces it.
The use of A provides a powerful, elegant approach to
solving EM problems (it is more convenient to find B by first
finding A in antenna problems).
The Magnetostatic Curl Equation
Basic equation of magnetostatics, that allows one to find the
current when the magnetic field is known is
H  J
where H is the magnetic field intensity, and J is the current
density (current per unit area passing through a plane
perpendicular to the flow).


The
magnetostatic curl equation is analogous
to the electric
field source equation
D  
19
EEL 3472
Magnetostatics
Stokes’ Theorem
This theorem will be used to derive Ampere’s circuital law
which is similar to Gauss’s law in electrostatics.
According to Stokes’ theorem, the circulation of A around a
closed path L is equal to the surface integral of the curl of A
over the open surface S bounded by L.
LA  dl     A  ds
S
The direction of integration around L is related to the
direction of ds by the right-hand rule.
20
EEL 3472
Magnetostatics
Determining the sense of dl and dS involved in Stokes’s theorem
Stokes’ theorem converts a surface integral of the curl of a
vector to a line integral of the vector, and vice versa.
(The divergence theorem relates a volume integral of the
divergence of a vector to a surface integral of the vector, and
vice versa).
   Adv   A  ds
v
21
S
EEL 3472
Magnetostatics
1

Surface S2 (circular)
22
 A  dl     A  ds
L
S
EEL 3472
Magnetostatics
Ampere’s Circuital Law
Choosing any surface S bounded by the border line L and
applying Stokes’ theorem to the magnetic field intensity
vector H , we have
   H   ds   H  dl
S
L
Substituting the magnetostatic curl equation
we obtain
H  J
I


 J  ds   H  dl
enc
S
L
which is Ampere’s Circuital Law. It states that the circulation
of H around a closed path is equal to the current enclosed by
the path.
23
EEL 3472
Magnetostatics
(Amperian Path)
Ampere’s law is very useful in determining H when there is
a closed path L around the current I such that the
magnitude of H is constant over the path.
24
EEL 3472
Magnetostatics
Boundary conditions are
the rules that relate fields
on opposite sides of a
boundary.
T= tangential components
N= normal components
We will make use of
Gauss’s law for magnetic
fields
for

25
B N 2   B  ds  0
and Ampere’s circuital law
BN1 and

for
HT1
and
HT2
  H  dl  Ienc
EEL 3472
Magnetostatics
Boundary condition on the tangential component of H
In the limit as L2  0
HT1
HT 2
L4  0 we have
 H  dl  H
l  HT 2l  Ienc
T1

L


The tangential
component of H
is continuous
across the
boundary, while
that of B is
discontinuous
where Ienc is the
current on the
boundary surface
(since the integration
path in the limit is
infinitely narrow).
When the
conductivities of both
media are finite,
Ienc  0
and
HT1  HT2
BT1 / 1  BT2 / 2
26
EEL 3472
Magnetostatics
Gaussian surface
(cylinder with its
plane faces
parallel to the
boundary)
27
EEL 3472
Magnetostatics
Applying the divergence theorem to B and noting that   B  0
(magnetic field is solenoidal) we have
 B  ds     B
S
0
V
In the limit as h 0 , the surface integral over the curved
surface of the cylinder vanishes. Thus we have

BN 2 A  BN1A  0

BN1  BN 2

28
(the normal
component of B is
continuous at the
boundary)
1H N1  2 H N 2

(the normal component
of H is discontinuous at
the boundary)
EEL 3472