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ENE 325
Electromagnetic Fields
and Waves
Lecture 7 Ampére’s circuital law
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1
Review (1)




Capacitance
Static magnetic field (time invariant)
Source of magnetic field
 permanent magnet
 electric field changing linearly with time
 direct current
Bio-Savart law is a method to determine the magnetic field
intensity. It is an analogy to Coulomb’s law of
Electrostatics.
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Review (2)
Id L  a r Id L  R
dH 

2
4 r
4 r 3
 Magnetic field intensity from an infinite length line of current
(line is located on z-axis)
I a
H
2
 Magnetic field intensity from a ring of current (a ring is located on
x-y plane and the observation point is on z-axis.
H
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Ia 2

2 h2  a

2 3/ 2
az
3
Review (3)
Magnetic field intensity from a rectangular loop of
current (a wire loop is located on x-y plane and the
observation point is at the origin.

2 2I
H
az
L

Right hand rule
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Outline
 Ampére’s
circuital law
 Curl and point form of Ampére’s circuital law
 Magnetic flux density
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Ampére’s circuital law



Analogy to Gauss’s law
Use for magnetostatic’s problems with sufficient symmetry.
Ampere’s circuital law – the integration of around any closed
path is equal to the net current enclosed by that path.
 H d L  I enc

A
To find , choose the proper Amperian path that is everywhere
either tangential or normal to and over which is constant.
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Use Ampere’s circuital law to determine
H from the infinite line of current.
From
 H d L  I enc
H  H  a
d L   d a
2
then  H d L   H a  d a  I enc .
0
 H
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I
2
a
A/m.
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Magnetic field of the uniform sheet
of current (1)
K  K ax
Create path a-b-c-d and perform the integration along the path.
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Magnetic field of the uniform sheet of
current (2)
From
 H d L  I enc
 H y (w)  H z (h)  H y (w)  H z (h)  K x w,
divide the sheet into small line segments along x-axis,
by symmetry Hz is cancelled.
.
 2H y  K x
Kx
Hy 
.
2
Because of the symmetry, the magnetic field intensity
on one side of the current sheet is the negative of that
on the other.
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Magnetic field of the uniform sheet of
current (3)
Above the sheet,
and
or we can write
1
H y1   K x
2
1
H y2  Kx
2 .
H
(z > 0)
(z < 0)
1
A/m
K
an
2
where a n is a unit vector normal to the current sheet.
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Magnetic field inside the solenoid
a
d
x
h
y
b
z
c
w
From
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 H d L  I enc
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Magnetic field inside the toroid that has
a circular cross section (1)
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Magnetic field inside the toroid that has
a circular cross section (2)
From
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 H d L  I enc
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Ex1 Determine H at point P (0.01, 0, 0) m
from two current filaments as shown.
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Ex2 Determine H for the coaxial cable that
has a inner radius a = 3 mm, b = 9 mm, and c
= 12 mm. Given I = 0.8 A.
a) at  < a
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b) at a <  < b
c) at b <  < c
d) at  > c
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Ex3 Determine H at point (10, 0, 0) mm
resulted from three current sheets: K1 = 1.5a y
ay
A/m at x = 6 mm, K2 = -3A/m
at x = 9
mm, and K3 = 1.5A/m
at x = 12 mm.
ay
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Curl and the point form of Ampére’s
circuital law (1)


‘Curl’ is employed to find the point form Ampère’s
circuital law, analogous to ‘Divergence’ to find the
point form of Gauss’s law.
Curl of H or  H is the maximum circulation of H per
unit area as the area shrinks to zero.
H dL

  H  lim
J
S 0
S
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Curl and the point form of Ampére’s
circuital law (2)

‘Curl´ operator perform a derivative of vector and
returns a vector quantity. For Cartesian coordinates,
can be written as
 H
ax
 H  
x
Hx
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ay
az

 .
y
z
Hy
Hz
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Physical view of curl
a) Field lines indicating divergence
b) Field lines indicating curl
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A simple way to see the
direction of curl using
right hand rule
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Stokes’s Theorem

Stokes’s Theorem relates a closed line integral into a
surface integral


 H d L    H d S
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Magnetic flux density, B

Magnetic flux density B is related to the magnetic field intensity H
in the free space by
B  0 H
Weber/m2 or Tesla (T)
1 Tesla = 10,000 Gauss.
where 0 is the free space permeability, given in units of
henrys per meter, or
0 10-7 H/m.

Magnetic flux  (units of Webers) passing through a surface is
found by
   B dS
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Gauss’s law for magnetic fields
 B dS  0
or
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 B  0.
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EX4 A solid conductor of circular cross section is
made of a homogeneous nonmagnetic material. If
the radius a = 1 mm, the conductor axis lies on the z
axis, and the total current in the direction a z is 20 A,
find
a) H at  = 0.5 mm
b) B at  = 0.8 mm
c) The total magnetic flux per unit length inside the conductor
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Maxwell’s equations for static fields
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Integral form
Differential form
 D d S  Qenc
 D  v
 B dS  0
 B 0
 E dL 0
 E  0
 H d L  I enc
 H  J
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