Magnetostatic Fields

Dr. Talal Skaik

Islamic University of Gaza – Palestine

2012

2
 Introduction
 In chapters 4 to 6, static electric fields characterized by E or D
(D=εE) were discussed.
 This chapter considers static magnetic fields, characterised by H
or B (B=μH).
 As we have noticed, a distribution of static or stationary charges
produces static electric field.
 If the charges move at a constant rate (direct current- DC), a
static magnetic field is produced (magnetostatic field).
 Static magnetic field are also produced by stationary permanent
magnets.
3
Magnet and Magnetic Field
The iron filings form circles around
the wire along the magnetic field

Magnetic Field Around A compass needle is deflected by the

Current Carrying Wires direct current flowing in a conductor
Electrostatic Fields have dual equations for magnetostatic fields

6
 Biot-Savart’s Law
Biot-Savart’s law states that the magnetic field intensity dH
produced at a point P by a differential current element Idl is
proportional to the product Idl and the sine of the angle α 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 and its direction can be obtained by right handed screw rule.

I dl sin 
dH 
R2
kI dl sin 
or dH = 2
, k  1/ 4
R
I dl sin 
dH =
4 R 2

7
 I dl sin 
dH =
4 R 2
I dl  aR I dl  R
dH  
4 R 2
4 R 3
where R =|R| and a R =R/R

The direction of dH can be

determined by the right-
hand rule or right-handed
screw rule.

8
 Biot-Savart’s Law
The direction of magnetic field intensity H (or current I) can be
represented by a small circle with a dot or cross sign depending on
whether H (or I) is out of, or into the page.

9
 Biot-Savart’s Law
For different current distributions:

Current distributions:
(a) line current,
(b) surface current,
(c) volume current.

I dl  aR
H  (line current)
L
4 R 2

K dS  aR
H  (surface current) (K: surface current density)
S
4  R 2

J dv  aR
H  (volume current) (J: volume current density)
v
4 R 2 10
 Magnetic Field of straight Conductor
Consider conductor of finite length AB, carrying current from point A
to point B. Consider the contribution dH at P
I dl  R
due to dl at (0,0,z): dH  ,
4 R 3

But dl  dz az and R= a   za z , so,

I  dz
dl  R   dz a  , Hence H=  a
2 3/2
4    z 
2

Letting z= cot , dz    cosec 2 d ,

2 2
I  2 cosec2 d I
H
4 1  3cosec3 a   4 a 1 sin  d
I
or H=  cos  2  cos 1  a
tan    / z 4
 z   cot 1+cot x=cosec x
2 2 11
 I
H=  cos  2  cos 1  a
4
When the conductor is semi-infinite, so that point A is now at O(0,0,0)
while B is at (0,0,), 1 =90o ,  2 =0o :
I
H= a
4
When the conductor is of infinite length, point A is at (0,0,  ) while
B is at (0,0,) ,1 =180o ,  2 =0o :
I
H= a
2
A simple approach to determine a :
a  al  a
where al is is a unit vector along the line current, and a is a unit vector
along the perpendicular line from the line current to the field point. 12
 I
H=  cos  2  cos 1  a
4

α1:outside, α2: inside 13

 Example 7.1
The conducting triangular loop in the figure carries a current of 10 A.
Find H at (0,0,5) due to side 1 of the loop.

(a) conducting triangular loop,

(b) side 1 of the loop.

14
 Example 7.1 Solution
I
Applying H=  cos  2  cos 1  a , that is valid for any straight,
4
thin, current carrying conductor. 1 ,  2 , and a are found as follows:
2
cos 1  cos 90o  0, cos  2  ,  =5
29
a  al  a , but al  ax , and a  az
so a  ax  az  a y , Hence
I
H1   cos  2  cos 1  a
4
10  2 
H1    0  (a y )
4 (5)  29 
H1  59.1a y mA/m

15
 Example 7.2
Find H at (-3,4,0) due to the current filament shown in the figure.

(a) current filament along semi-infinite x- and z-axes, aℓ and a for H2 only;
(b) determining a for H2. 16
 Example 7.2 Solution
Let H=H1 +H 2 at P(-3,4,0), where H1 is due
to current filament along x-axis, and H 2 is due to current filament
along z-axis.
I
H2   cos  2  cos 1  a
4
At P(-3,4,0),  =(9+16)1/2  5,
1 =90o ,  2 =0o
a  al  a ,
but al  az ,
3 4
and a  ax  a y
5 5
 3 4  4 3
a  az   ax  a y   ax  a y
 5 5  5 5 17
or alternatively, from figure b,
4 3
a  sin  ax  cos  a y  ax  a y
5 5
3  4ax  3a y 
Thus, H 2  1  0 
4 (5) 5
H 2  38.2ax  28.65a y mA/m
a in this case is the negative of the regular a of
3
cylindrical coordinates: H 2  1  0  (a )  47.75a mA/m
4 (5)
To find H1 at P:
 =4,  2  0o , cos 1  3 / 5, and a  al  a  ax  a y  az . Hence,
3  3
H1  1   az  23.88az mA/m
4 (4)  5 
Thus H=H1 +H 2  38.2ax  28.65a y  23.88az mA/m
or H=  47.75a  23.88az mA/m 18
 Example 7.3
A circular loop located at X2+y2=9, z=0 carries a direct current of 10 A
along aФ. Determine H at (0,0,4) and (0,0,-4).

(a) circular current loop, (b) flux lines due to the current loop. 19
 Example 7.3 - solution
The magnetic field intensity dH at point P(0,0,h)
contributed by current element Idl is :
I dl  R
dH 
4 R 3
where dl   d a , R=(0,0, h )  ( x, y,0)
R    a  ha z , and
a a az
dl  R  0  d 0   h d a   2d a z
 0 h
Hence,

3/2 
d a z   dH  a  dH z a z
I
dH=  h d  a    2

4   2  h 2 
By symmetry, the contributions along a add up to zero  H   0 20
 Example 7.3 - solution
2
I  2 d a z I  2 2 az
Thus, H=  dH z az   4   
2 3/2 2 3/2
0

2
 h  4    h 
2

I2
or H= az
2 3/2
2   2  h 
(a) Substituting I=10 A,  =3, h=4 gives
10(3)2 az
H  0, 0, 4  =  0.36 az A/m
2 9  16
3/2

(b) replacing h by -h,

dl  R    h d a   2 d az
(z-component remains the same)
Hence, H  0, 0, 4   H  0, 0, 4   0.36 az A/m 21
 AMPERE'S CIRCUIT LAW - MAXWELL'S EQUATION
Ampere's circuit law states that the line integral of the tangential
component of H around a closed path is the same as the net
current Ienc enclosed by the path.

In other words, the circulation of H equals I enc ; that is,

 H  dl=I enc (integral form of 3rd Maxwell's equation)

Apply Stoke's theorem,
I enc   H  dl      H   dS
L S

But, I enc   J  dS
S

   H  J (differential form of third Maxwell's equation)

22
 AMPERE'S CIRCUIT LAW - MAXWELL'S EQUATION

 H  dl=I enc

  H  dl  apply to Amperian path (L).

L

 I enc  current enclosed by amperian path.

I enc is found if J(A/m 2 ) is known: I enc   J  dS
S

or if K(A/m) is known for surface current density.

23
 APPLICATIONS OF AMPER'S LAW
A. Infinite Line Current
Consider infinite filamentary current I along the z-axis. To determine H
at point P, allow Amperian path passing through P, such that H is
constant provided ρ is constant.

The whole current I is enclosed by the path,  H  dl=I enc

according to Amper's law:
I   H  a    d a 
2
I =H   d H (2 )
0

I
or H= a
2
24
 B. Infinite Sheet of Current
Consider an infinite current sheet in the z=0 plane with uniform
current desnity K=K y a y A/m. Applying Ampere's law to the
rectangular closed 1-2-3-4-1 path gives:  H  dl I enc  K yb
 Consider the sheet as comprising of filaments
 Consider dH above and below the sheet due to a pair of
filamentary currents.

25
 The resultant dH has only an x-component.
 H on one side is the negative of that on the other side.
H a z0
H=  0 x
  H 0a x z0
Evaluating the line integral of H along the closed path:
2 3 4 1 
 H.dl   1  2  3  4  H.dl=I enc  K yb
1
0(  a )+(  H 0 )(  b)  0(a )  H 0 (b)=2 H 0b  K y b  H 0  Ky
2
1
 2 K y a x z0
H= 
  1 K yax z0
 2
In general, for an infinite sheet of current density K A/m,
1
H= K  a n where a n is a unit normal vector directed from
2
the current sheet to the point of interest 26
Magnetic field of Infinite Sheet of Current

1
H= K  a n
2

27
Infinitely Long Coaxial Transmission Line

28
 Infinitely Long Coaxial Transmission Line
Consider infinitely long coaxial transmission line of two concentric cylinders
The inner conductor has radius a and carries current I , the outer conductor has
inner radius b and thickness t and carries return current  I .
To determine H everywhere, apply Ampere's law in four possible regions:
0    a, a    b, b    b  t ,   b  t

29
For region 0    a : apply Ampere's law to path L1

 H  dl  I
L1
enc   J  dS

Since the current is uniformly distributed over

I
the cross section, J= 2 az , dS= d d  az
a
2 
I I I2
I enc   J  dS  2    d  d  2   2
2

 a  0  0 a a
I2 I
H  dl  H 2  2 or H 
L1
a 2 a 2
For region a    b : Apply Ampere's law to path L 2

 H  dl  I
L2
enc  I (since the whole current is enclosed by L 2 )

I
H 2  I or H  (same as infinite straight filamentary current)
2 30
For region b    b  t :
Apply Ampere's law to path L3

 H  dl  H 2  I
L3
enc

where
I
I enc  I   J  dS, J=  az
  b  t   b 
2 2
 
2 
I   2  b2 
Thus, I enc I
 2 2   
  b  t   b   0  b
 d  d I 1  2
 t  2bt


 
I   2  b2 
 H  1  2 
2  t  2bt 
For region   b  t : Apply Ampere's law to path L 4

 H.dl  I  I  0
L4
or H  0
31
 Infinitely Long Coaxial Transmission Line
 I
 2 a 2 a 0 a

 I
a   b
Putting all equations  2 a
H= 
together:  I   2  b2 
 H  1  2  a b    bt
 2  t  2bt 
   bt
 0

32
Infinitely Long Coaxial Transmission Line

33
 Example 7.5
Planes z=0 and z=4 carry current K= -10 ax A/m and K=10 ax A/m,
respectively. Determine H at
(a) (1,1,1)
(b) (0,-3,10)

34
 Example 7.5
Let H=H 0 +H 4 , where H 0 +H 4 are
the contributions due to current
sheets z=0 and z=4,
(a) At (1,1,1) , which is between the plates (0  ( z  1)  4),
H 0  (1/ 2)K  a n  (1/ 2)( 10a x )  a z  5a y A/m
H1  (1/ 2)K  a n  (1/ 2)(10a x )  ( a z )  5a y A/m
Hence, H=10a y A/m

(b) At (0,-3,10) , which is above the sheets ( z  10  4  0),

H 0  (1/ 2)K  a n  (1/ 2)( 10a x )  a z  5a y A/m
H1  (1/ 2)K  a n  (1/ 2)(10a x )  a z  5a y A/m
Hence, H= 0 A/m 35
 Example 7.6
A toroid whose dimensions are shown in the figure has N turns and
carries current I. Determine H inside and outside the toroid.

36
 Example 7.6
The net current enclosed by the Amperian path is NI . Hence,

 H  dl=I enc  H (2 )  NI

NI
or H= , for  0  a     0  a
2
where  0 is is the mean radius of the toroid.
Outside the toroid, the current enclosed by an Amperian path is
NI - NI  0 and Hence H =0

37
Magnetic Field of a Toroid

38
 Magnetic Flux Density
•Electric flux density and Electric field intensity are related by D=ε0 E
in free space.
•Similarly, the magnetic flux density B is related to the magnetic
field intensity H by:
B  0 H
•Where μ0 is known as the permeability of free space.

0  4  107 H/m
•The magnetic flux through a surface is given by

 =  B  dS
S

•Where Ψ is in webers (Wb), B is in (Wb/m2) or teslas (T).

39
 Magnetic Flux Lines
•Magnetic flux line is a path to which B is tangential at every point on
the line.
•Each flux line is closed and has no beginning or end.

40
•In an electrostatic field the flux •Unlike electric flux lines, magnetic flux
crossing a closed surface is the lines always close upon themselves.
same as the charge enclosed. •This is because it is not possible to have
isolated magnetic poles (or magnetic
  D  dS  Q
S
charges).

 B  dS  0
•So it is possible to have an
isolated electric charge and the

flux lines produced by it need S
not be closed. 41
It is not possible to isolate the north and south poles of a magnet.

42
Broken Magnet

43
 Gauss’s Law for magnetostatic fields
The total flux through a closed surface in a magnetic field is zero.

 B  dS  0
Applying divergence theorem,

 B  dS     B dv 0
S v
Or

 B  0
This is the Maxwell’s fourth equation. It states that magnetostatic
fields have no sources or sinks.
44
Maxwell’s Equations for Static Fields

45
 Magnetic Scalar and Vector Potentials
*In electrostatics electric field intensity and potential are related by:
E  V
*Similar to this we can relate magnetic field intensity with two
magnetic potentials:
•Magnetic scalar potential (Vm)
•Magnetic vector potential (A)

*Magnetic scalar potential Vm is related to H by the relation:

H  Vm if J=0
J=  H=  ( Vm )  0 (since for any scalar,   (V )  0)
so the magnetic scalar potential Vm is only defined in the region where J=0.
Vm satisfies Laplace's equation 2Vm  0 (J=0)
46
 Magnetic Vector Potential
Since for a magnetostatic field   B=0
For any vector   (  A)=0
we can define magnetic vector potential A such that
B=  A
In many EM problems it is more convenient to first find A and then
find B from it.
0 Idl
A for line current
L
4 R

0 KdS
A for surface current
S
4 R

0 Jdv
A for volume current
v
4 R 47
 Magnetic flux from vector potential
The magnetic fluc through a given area can be found from
=  B  dS
S

Applying Stoke's theorem, we obtain,

 =  B  dS   (  A)  dS   A  dl
S S L

  A  dl
L

48
 Example 7.7
Given the magnetic vector potential A=-ρ2/4 az Wb/m, calculate the
total magnetic flux crossing the surface Ф=π/2, 1≤ρ≤2 m, 0≤z≤5 m.
Example 7.7 Solution

Method 1
Az 
B    A=  a  a , dS=d  dz a 
 2
Hence,
5 2 2
1 1 2 15
   B  dS=
2 z0 1
 d  dz   (5) 
4 1 4
  3.75 Wb

49
 Example 7.7 Solution Continued A=-ρ2/4 az
Method 2
  A  dl=
L
1  2  3   4

where L is the path bounding surface S .

Since A has only z-component,
1  0   3
1  5 0

   2   4   1  dz  (2)  dz 
2 2

4 0 5 
1 15
  (1  4)(5)   3.75 Wb
4 4
50