Download magnetostatic (cont`d)

AMPERE’S CIRCUITAL LAW (Cont’d)
Expression for curl by applying Ampere’s
Circuital Law might be too lengthy to derive, but
it can be described as:
H  J
The expression is also called the point form of
Ampere’s Circuital Law, since it occurs at
some particular point.
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AMPERE’S CIRCUITAL LAW (Cont’d)
The Ampere’s Circuital Law can be rewritten in
terms of a current density, as:
 H  dL   J  dS
Use the point form of Ampere’s Circuital Law to
replace J, yielding:
 H  dL     H  dS
This is known as Stoke’s Theorem.
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3.3 MAGNETIC FLUX DENSITY
In electrostatics, it is convenient to think in terms
of electric flux intensity and electric flux density.
So too in magnetostatics, where magnetic flux
density, B is related to magnetic field intensity by:
B  H
  0  r
Where μ is the permeability with:
0  4  10 7 H m
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MAGNETIC FLUX DENSITY (Cont’d)
The amount of magnetic flux, φ in webers
from
magnetic
field
passing
through
a
surface is found in a manner analogous to
finding electric flux:
   B  dS
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MAGNETIC FLUX DENSITY (Cont’d)
Fundamental features of magnetic fields:
• The field lines form a
closed loops. It’s different
from electric field lines,
where it starts on positive
charge and terminates on
negative charge
Figure 3-26 (p. 125)
Magnetic field lines form closed loops, sot he netflux through a Gaussian surface is
zero.
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
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MAGNETIC FLUX DENSITY (Cont’d)
• The magnet cannot be
divided in two parts, but it
results in two magnets.
The magnetic pole cannot
be isolated.
Figure 3-27 (p. 125)
Dividing a magnet in two parts results in two magnets. You cannot isolate a
magnetic pole.
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
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MAGNETIC FLUX DENSITY (Cont’d)
The net magnetic flux passing through a
gaussian surface must be zero, to get Gauss’s
Law for magnetic fields:
 B  dS  0
By applying divergence theorem, the point form
of Gauss’s Law for static magnetic fields:
B  0
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EXAMPLE 6
Find the flux crossing the portion of the
plane φ=π/4 defined by 0.01m < r < 0.05m
and 0 < z < 2m in free space. A current
filament of 2.5A is along the z axis in the az
direction.
Try to sketch this!
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SOLUTION TO EXAMPLE 6
The relation between B and H is:
B  0 H  0
I
2
a
To find flux crossing the portion, we need to use:
   B  dS
where dS is in the aφ direction.
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SOLUTION TO EXAMPLE 6 (Cont’d)
So,
dS  ddza
Therefore,
   B  dS
0 I
  
a  ddza
z  0   0.01 2
2
0.05
20 I 0.05
6

ln
 1.61  10 Wb
2
0.01
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3.4 MAGNETIC FORCES
Upon application of a magnetic field, the wire is
deflected in a direction normal to both the field and the
direction of current.
(a)
(b)
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MAGNETIC FORCES (Cont’d)
The force is actually acting on the individual
charges moving in the conductor, given by:
Fm  qu  B
By the definition of electric field intensity, the
electric force Fe acting on a charge q within an
electric field is:
Fe  qE
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MAGNETIC FORCES (Cont’d)
A total force on a charge is given by Lorentz force
equation:
F  qE  u  B 
The force is related to acceleration by the
equation from introductory physics,
F  ma
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MAGNETIC FORCES (Cont’d)
To find a force on a current element, consider a
line conducting current in the presence of
magnetic field with differential segment dQ of
charge moving with velocity u:
dF  dQu  B
But,
dL
u
dt
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MAGNETIC FORCES (Cont’d)
So,
dQ
dF 
dL  B
dt
Since dQ
the line,
dt corresponds to the current I in
 dF  IdL  B
We can find the force from a collection of
current elements
F12   I 2dL2  B1
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MAGNETIC FORCES (Cont’d)
Consider a line of current in +az direction on the z
axis. For current element a,
IdL a  Idz aa z
But, the field cannot exert magnetic force
on the element producing it. From field of
second element b, the cross product will
be zero since IdL and aR in same
direction.
Figure 3-28 (p. 128)
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(a) Differential current elements on a line. (b) A pair
EXAMPLE 7
If there is a field from a
second line of current
parallel to the first, what
will be the total force?
Figure 3-28 (p. 128)
(a) Differential current elements on a line. (b) A pair of current-carrying lines will
exert magnetic force on each other.
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SOLUTION TO EXAMPLE 7
The force from the magnetic field of line 1 acting
on a differential section of line 2 is:
dF12  I 2 dL 2  B1
Where,
0 I1
B1 
a
2
By inspection from figure,
  y, a  a x
Why?!?!
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SOLUTION TO EXAMPLE 7
Consider
dL 2  dza z
, then:
0 I1
0 I1I 2
 a x  
dF12  I 2dza z 
dz a y 
2y
2y
0
0 I1I 2

F12 
 a y  dz
2y
L
0 I1I 2 L
 F12 
ay
2y
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MAGNETIC FORCES (Cont’d)
Generally,
0
dL 2  dL1  a12 
F12 
I 2 I1  
4
R12 2
• Ampere’s law of force between a pair of current-
carrying circuits.
• General case is applicable for two lines that are not
parallel, or not straight.
• It is easier to find magnetic field B1 by Biot-Savart’s
law, then use
F12   I 2dL2  B1 to find F12
.
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EXAMPLE 8
The magnetic flux density in a region of free space
is given by B = −3x ax + 5y ay − 2z az T. Find the
total force on the rectangular loop shown which
lies in the plane z = 0 and is bounded by x = 1, x =
3, y = 2, and y = 5, all dimensions in cm.
Try to sketch this!
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SOLUTION TO EXAMPLE 8
The figure is as shown.
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SOLUTION TO EXAMPLE 8 (Cont’d)
First, note that in the plane z = 0, the z component
of the given field is zero, so will not contribute to the
force. We use:
F  loop IdL x B
Which in our case becomes with,
I  30 A and B  3xa x  5 ya y  2 za z
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SOLUTION TO EXAMPLE 8 (Cont’d)
So,
F
0.03



30 dxa x   3 xa x  5 y y  0.02 a y 
0.01
0.05
 30dya y   3x x 0.03 a x  5 ya y  
0.02
0.01



30 dxa x x  3 xa x  5 y y  0.05 a y 
0.03
0.02
 30dya y x  3x x 0.01 a x  5 ya y 
0.05
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SOLUTION TO EXAMPLE 8 (Cont’d)
Simplifying these becomes:
F
0.03
0.05
0.01
0.01
0.02
0.02
0.03
0.05
 30(5)(0.02)a z dx    30(3)(0.03) a z dy

 30(5)(0.05)a z dx    30(3)(0.01) a z dy
 0.06  0.081  0.150  0.027  a z N
 F  36a z mN
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