ene 325 electromagnetic fields and waves lecture 4 magnetostatics
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ENE 325ENE 325Electromagnetic Electromagnetic Fields and WavesFields and Waves
Lecture 4 Lecture 4 MagnetostaticsMagnetostatics
IntroductionIntroduction (1)(1) source of the steady magnetic field may be a per
manent magnet, and electric field changing linear ly with time or a direct current.
a schematic view of a bar magnet showing the m agnetic field. Magnetic flux lines begin and termi
nate at the same location, more like circulation.
IntroductionIntroduction (2)(2) Magnetic north and south poles are alwa
ys together.
N
S
N
S
N
S
N
S
N
S
N
S
IntroductionIntroduction (3)(3) Oersted’s experiment shows th at current produ
ces magnetic fields that loop around the conducto r. The field grows weaker as one compass moves a
way from the source of the current.
- Bi o Savart l aw- Bi o Savart l aw(1)(1) - The law of Bio Savart states that at any point P the magni
tude of the magnetic field intensity produced by the differ ential element is proportional to the product of the current
, the magnitude of the differential length, and the sine of t he angle lying between the filament and a line connecting the filament to the point P at which the filed is desired.
The magnitude of the magnetic field intensity is inversely proportional to the square of the distance from the differe
ntial element to the point P.
- Bi o Savart l aw- Bi o Savart l aw(2)(2) The direction of the magnetic field intensity is n
ormal to the plane containing the differential fila ment and the line drawn from the filament to the point P.
- Bio Savart law is a method to determine the mag netic field intensity. It is an analogy to Coulomb’s
law of Electrostatics.
- Bi o Savart l aw- Bi o Savart l aw(3)(3)
2 34 4rId L a Id L R
dHr r
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from this picture:
1 121
2 2124
I d L adH
R
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Total field A/m
24rId L a
Hr
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Magnetic field intensity resulting fr Magnetic field intensity resulting fr om om an infinite length line of current an infinite length line of current
(1)(1)
Pick an observation point P located on axis.
The current
The vector from the source to the test point is
zId L Idza
��������������
R zRa za a
a unit vector
2 2
zR
za aa
z
Magnetic field intensity resulting fr Magnetic field intensity resulting fr om om an infinite length line of current an infinite length line of current
(2)(2)
then
From a table of Integral,
3/ 22 24
z zIdza za aH
z
��������������
3/ 22 2.
4
�������������� I a dzH
z
3/ 2 2 2 22 2
dx x
a x ax a
then
2 2 24
I a zH
z
�������������� /
2
I a
A m
Magnetic field intensity resulti Magnetic field intensity resulti ng from ng from a ring of current a ring of current (1)(1)
A ring is located on z = 0 plane with the radius a. The observation point is at z = h.
Magnetic field intensity resulti Magnetic field intensity resulti ng from ng from a ring of current a ring of current (2)(2)
Id L Iad a��������������
R zRa ha aa
2 2
zR
ha aaa
h a
A unit vector
Magnetic field intensity resulti Magnetic field intensity resulti ng from ng from a ring of current a ring of current (3)(3)
Consider a symmetry 2zId L R had a a d a
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components are cancelled out due to symmetry of two segments on the opposite sides of the ring.
Therefore from
24rId L a
Hr
����������������������������
we have
2
3/ 22 20 4
zIad a ha aaH
h a
��������������
Magnetic field intensity resulti Magnetic field intensity resulti ng from ng from a ring of current a ring of current (4)(4)
then
2 2
3/ 22 2 0.
4
zIa aH d
h a
��������������
We finally get
2
3/ 22 22
��������������z
IaH a
h a
Magnetic field intensity resulti Magnetic field intensity resulti ng from ng from a rectangular loop of c a rectangular loop of c
urrenturrent (1)(1)
Find the magnetic field intensity at the origin. By symmetry, there will be equal magnetic field intensity at each half width (w/2).
Magnetic field intensity resulti Magnetic field intensity resulti ng from ng from a rectangular loop of c a rectangular loop of c
urrenturrent (2)(2)Consider 0 x w/2, y = -w/2
xId L Idxa
��������������
( / 2)R x yRa xa w a
A unit vector
2 2
( / 2)
( / 2)
x yR
xa w aa
x w
We have
3/ 22 2
( / 2)
4 ( / 2)
x x yIdxa xa w adH
x w
��������������
Magnetic field intensity resulti Magnetic field intensity resulti ng from ng from a rectangular loop of c a rectangular loop of c
urrenturrent (3)(3)
3/ 22 2
( / 2)
4 ( / 2)
zIdx w a
x w
Then the total magnetic field at the origin is
/ 2
3 / 22 20
( / 2)8
4 ( / 2)
wzIdx w a
Hx w
��������������
Look up the table of integral, we find 3/ 2 2 2 22 2
dx x
a x ax a
then A/m. 2 2z
IH a
W
��������������
Bio-Savart law in different Bio-Savart law in different formsforms
We can express - Bio Savart law in terms of surface and volume current densities by re placing with and :
Id L��������������
KdS��������������
Jdv��������������
24rKdS a
Hr
����������������������������
where K = surface current density (A/m) I = K x width of the current sheet
and 24rJdv a
Hr
����������������������������
Ri ght handrul e Ri ght handrul e
The method to determine the result of cros s product (x).
F qv B ������������������������������������������
AmpAmpéé re’s circuital law re’s circuital law
Analogy to Gauss’s law Use for magnetostatic’s problems with sufficient symme
try. Ampere’s circuital law – the integration of around any c
losed path is equal to the net current enclosed by that p ath.
To find , choose the proper Amperian path that is every where either tangential or normal to and over which is
constant.
encH dL I����������������������������
A
Use Ampere’s circuital law to d Use Ampere’s circuital law to d etermine etermine
from the infinite line of current from the infinite line of current . .H��������������
From encH dL I����������������������������
H H a��������������
dL d a ��������������
then 2
0.
���������������������������� encH dL H a d a I
�������������� IH a
2A/m.
Magneti c fi el dof Magneti c fi el dof the the uni foruni for msheet of current msheet of current (1)(1)
xK Ka
��������������
Create path a-b-c-d and perform the integration along the path.
Magnetic field of Magnetic field of the the uniform s uniform s heet of current heet of current (2)(2)
From encH dL I����������������������������
( ) ( ) ( ) ( ) ,y z y z xH w H h H w H h K w
divide the sheet into small line segments along x-axis, by symmetry Hz is cancelled.
y x2H K .
Because of the symmetry, the magnetic field intensity on one side of the current sheet is the negative of that on the other.
.xyK
H2
where is a unit vector normal to the current sheet.
Magnetic field of Magnetic field of the the uniform s uniform s heet of current heet of current (3)(3)
.
Above the sheet,
y1 x1
H K2
(z > 0)
and y2 x1
H K2
(z < 0)
or we can write A/m n
1H K a
2
����������������������������
na
Magnetic field inside the sol Magnetic field inside the sol enoid enoid
a
b c
d
x
y z
h
w
From encH dL I����������������������������
Magnetic field inside the Magnetic field inside the toroid that has a circular cross toroid that has a circular cross section (1)section (1)
Magnetic field inside the Magnetic field inside the toroid that has a circular cross toroid that has a circular cross section (2)section (2) From encH dL I
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Ex1Ex1 Determine Determine at point P (0.01, at point P (0.01, 0, 0) m from two current filaments 0, 0) m from two current filaments as shown. as shown.
H��������������
2Ex2Ex Determine for the coaxial c Determine for the coaxial c able that has a inner radius able that has a inner radius a a = 3 m = 3 m
m, m, bb 9= mm, and 9= mm, and cc 12= mm. Giv 12= mm. Giv en en II AA08 AA08
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a) at < a
b) at a < < b
c) at b < < c
d) at > c
33 Determine at point (10, 0, 0) Determine at point (10, 0, 0) mm resulted from three current sh mm resulted from three current sh
eets: eets: KK11
15 15 A/m at A/m at x x A AAA 6 A AAA 6 KK22
A A -3-3 A/m at A/m at xx A A A A AAA 9 A A A A AAA 9 KK
33 A A1 A A1
55 A/m at A/m at xx AAA12 AAA12
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yaya
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