1 Expression for curl by applying Ampere’s Circuital Law might be too lengthy to derive, but it...
25
1 Expression for curl by applying Ampere’s Circuital Law might be too lengthy to derive, but it can be described as: J H The expression is also called the point form of Ampere’s Circuital Law, since it occurs at some particular point. AMPERE’S CIRCUITAL LAW (Cont’d)
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Transcript of 1 Expression for curl by applying Ampere’s Circuital Law might be too lengthy to derive, but it...
1
Expression for curl by applying Ampere’s Circuital Law might be too lengthy to derive, but it can be described as:
JH The expression is also called the point form of Ampere’s Circuital Law, since it occurs at some particular point.
AMPERE’S CIRCUITAL LAW (Cont’d)
2
The Ampere’s Circuital Law can be rewritten in terms of a current density, as:
SJLH dd
Use the point form of Ampere’s Circuital Law to replace J, yielding:
SHLH dd
This is known as Stoke’s Theorem.
AMPERE’S CIRCUITAL LAW (Cont’d)
3
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:r 0 HB
Where μ is the permeability with:
mH 70 104
4
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:
SB d
5
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
MAGNETIC FLUX DENSITY (Cont’d)
6
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.
7
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:
0 SB d
By applying divergence theorem, the point
form of Gauss’s Law for static magnetic fields:
0 B
8
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!
9
SOLUTION TO EXAMPLE 6
The relation between B and H is:
aHB200I
To find flux crossing the portion, we need to use:
SB d
where dS is in the aφ direction.
10
So, aS dzdd Therefore,
WbI
dzdI
d
z
aa
SB
60
2
0
05.0
01.0
0
1061.101.0
05.0ln
2
2
2
SOLUTION TO EXAMPLE 6 (Cont’d)
11
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.
12
MAGNETIC FORCES (Cont’d)
The force is actually acting on the individual charges moving in the conductor, given by:
BuF qm
By the definition of electric field intensity, the
electric force Fe acting on a charge q within an
electric field is:
EF qe
13
A total force on a charge is given by Lorentz force equation:
BuEF q
MAGNETIC FORCES (Cont’d)
The force is related to acceleration by the equation from introductory physics,
aF m
14
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:
BuF dQd
dt
dLu
But,
15
BLF Idd
So,BLF d
dt
dQd
Since corresponds to the current I in the line,
dtdQ
MAGNETIC FORCES (Cont’d)
We can find the force from a collection of current elements
12212 BLF dI
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Consider a line of current in +az direction on the z
axis. For current element a,
zaa IdzId aL
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.
MAGNETIC FORCES (Cont’d)
17
EXAMPLE 7
If there is a field from a
second line of current
parallel to the first, what
will be the total force?
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The force from the magnetic field of line 1
acting on a differential section of line 2 is:
12212 BLF dId
Where,
aB
210
1I
By inspection from figure,
xy aa , Why?!?!
SOLUTION TO EXAMPLE 7
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0210
12
21010212
2
22
Ly
yxz
dzy
II
dzy
II
y
IdzId
aF
aaaF
zdzd aL 2Consider , then:
yy
LIIaF
2
21012
SOLUTION TO EXAMPLE 7
20
Generally,
212
121212
012 4 R
ddII
aLLF
• 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 to find F12 . 12212 BLF dI
MAGNETIC FORCES (Cont’d)
21
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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The figure is as shown.
SOLUTION TO EXAMPLE 8
23
SOLUTION TO EXAMPLE 8 (Cont’d)
B L F xIdloop
AI 30
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:
Which in our case becomes with,
zyx zyx aaaB 253 and
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02.0
05.001.0
01.0
03.005.0
05.0
02.003.0
03.0
01.002.0
5330
5330
5330
5330
yxxy
yyxx
yxxy
yyxx
yxxdy
yxxdx
yxdy
yxdx
aa a
aa a
aa a
aa aFSo,
SOLUTION TO EXAMPLE 8 (Cont’d)
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Simplifying these becomes:
N
dydx
dydx
z
zz
zz
a
aa
aaF
027.0150.0081.006.0
)01.0)(3(30)05.0)(5(30
)03.0)(3(30)02.0)(5(30
02.0
05.0
01.0
03.0
05.0
02.0
03.0
01.0
mNz aF 36
SOLUTION TO EXAMPLE 8 (Cont’d)
