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ELEC 3105 BASIC EM AND POWER ENGINEERING
MAGNETIC FIELD OF A LONG STRAIGHT WIRE (FINITE
LENGTH/ INFINITE LENGTH)
AMPERE’S LAW
MAGNETIC FIELD OF A LONG SOLENOID
B FIELD FOR A LONG STRAIGHT WIRE (FINITE LENGTH)
LP
X
dL 𝑑𝜃
𝜃2
𝜃1
��𝐻��
r ��𝐵
��1
��2
��
Biot-Savard Law
��= ∫𝐿 𝑖𝑛𝑒
❑
��𝐵 ∫𝐿𝑖𝑛𝑒
❑ 𝜇𝑜
4𝜋��𝑑𝐿× ��𝑅2
��= ∫𝐿 𝑖𝑛𝑒
❑
��𝐻 ∫𝐿𝑖𝑛𝑒
❑1
4𝜋��𝑑𝐿× ��𝑅2
LP
X
dL 𝑑𝜃
𝜃2
𝜃1
��𝐻��
r ��𝐵
��1
��2
��
Biot-Savard Law
��= ∫𝐿 𝑖𝑛𝑒
❑
��𝐵 ∫𝐿𝑖𝑛𝑒
❑ 𝜇𝑜
4𝜋��𝑑𝐿× ��𝑅2
��𝑑𝐿=𝑑𝑧 ��
��𝑑𝐿× ��=Isin (𝜃 )dz ��
��=��𝐼 𝜇𝑜
4 𝜋 ∫𝐿𝑖𝑛𝑒
❑ 𝑠𝑖𝑛 (𝜃 )𝑅2 𝑑𝑧
B FIELD FOR A LONG STRAIGHT WIRE (FINITE LENGTH)
LP
X
dL 𝑑𝜃
𝜃2
𝜃1
��𝐻��
r ��𝐵
��1
��2
��
��=��𝐼 𝜇𝑜
4 𝜋 ∫𝐿𝑖𝑛𝑒
❑ 𝑠𝑖𝑛 (𝜃 )𝑅2 𝑑𝑧
R=r csc (𝜃 )
𝑧=−𝑟 𝑐𝑜𝑡 (𝜃 )
𝑑𝑧=𝑟 𝑐𝑠𝑐2 (𝜃 )𝑑𝜃
��=��𝐼 𝜇𝑜
4 𝜋 ∫𝜃 1
𝜃 2 𝑠𝑖𝑛 (𝜃 )𝑟 𝑐𝑠𝑐2 (𝜃 )𝑟 2𝑐𝑠𝑐2 (𝜃 )
𝑑𝜃
B FIELD FOR A LONG STRAIGHT WIRE (FINITE LENGTH)
LP
X
dL 𝑑𝜃
𝜃2
𝜃1
��𝐻��
r ��𝐵
��1
��2
��
R=r csc (𝜃 )
𝑧=−𝑟 𝑐𝑜𝑡 (𝜃 )
𝑑𝑧=𝑟 𝑐𝑠𝑐2 (𝜃 )𝑑𝜃
��=��𝐼 𝜇𝑜
4 𝜋 ∫𝜃 1
𝜃 2 𝑠𝑖𝑛 (𝜃 )𝑟 𝑐𝑠𝑐2 (𝜃 )𝑟 2𝑐𝑠𝑐2 (𝜃 )
𝑑𝜃
��=��𝐼 𝜇𝑜
4𝜋𝑟∫𝜃1
𝜃2
𝑠𝑖𝑛 (𝜃 )𝑑𝜃
��=��𝐼 𝜇𝑜
4𝜋𝑟 [𝑐𝑜𝑠 (𝜃1 )−𝑐𝑜𝑠 (𝜃2 ) ]
B FIELD FOR A LONG STRAIGHT WIRE (FINITE LENGTH)
LP
X
dL 𝑑𝜃
𝜃2
𝜃1
��𝐻��
r ��𝐵
��1
��2
��
��=��𝐼 𝜇𝑜
4𝜋𝑟 [𝑐𝑜𝑠 (𝜃1 )−𝑐𝑜𝑠 (𝜃2 ) ]
B FIELD FOR A LONG STRAIGHT WIRE (INFINITE LENGTH)
𝜃1→0
𝜃2→𝜋
��=��𝐼 𝜇𝑜
2𝜋𝑟infinite length
AMPERE’S LAW
r
�� 𝐼��=��𝐼 𝜇𝑜
2𝜋𝑟
infinite length
𝑑𝑙
∮ �� ∙𝑑𝑙=𝐼 𝜇𝑜RHR gives ��
AMPERE AND THE LONG STRAIGHT WIRE
This is how it is done
RHR for field direction
��=��𝐼 𝜇𝑜
2𝜋𝑟
𝐵=𝐼 𝜇𝑜
2𝜋𝑟
9
DIFFERENTIAL FORM OF AMPERE’S LAW
Lecture 16 slide 3
Area enclosed A
J
Current density flowing through loop.
enclosedoIdB ∫
10
Ampere’s Law:
enclosedoIdB ∫
Area enclosed A
J
Current density flowing through loop.
∫ ∫ S
oadJdB
∫ ∫ S
odanJdanB ˆˆ
JBo
∫ ∫ S
adFdF Using Stoke’s theorem
Differential equation for B
Can we solve this equation?
DIFFERENTIAL FORM OF AMPERE’S LAW
MAGNETIC FIELD OF A INFINITELY LONG SOLENOID
Current out of page
Current into page
Infinite coil of wire carrying a current I
Axis of solenoid
P
Evaluate B field here
1
In the vicinity of the point P
P
2 3 4 5
3
12
45Axis of solenoid
resultant
Expect B to lie along axis of the solenoid
Current out of page
0 B
Implies that B field has no radial component. I.e. no component pointing towards or away from the solenoid axis.
B
MAGNETIC FIELD OF A INFINITELY LONG SOLENOID
Current out of page
Current into page
P
Claim: The magnetic field outside of the solenoid is zero.
1 Closed pathaB
bB
MAGNETIC FIELD OF A INFINITELY LONG SOLENOID
Current out of page
Current into page
P
Must have since there is no net enclosed current.1 Closed pathaB
bB
01
∫
dB
Conceivably there might be some non-zero field components outside of the solenoid, and as shown.aB
bB
MAGNETIC FIELD OF A INFINITELY LONG SOLENOID
Current out of page
Current into page
P
We would need to have to give .
1 Closed pathaB
bB
01
∫
dB
Conceivably there might be some non-zero field components outside of the solenoid, and as shown.aB
bB
ba BB
MAGNETIC FIELD OF A INFINITELY LONG SOLENOID
MAGNETIC FIELD OF A LONG SOLENOID
Current out of page
Current into page
P
Now we distort the path by moving one side away from the solenoid.
2Closed path
aB
bB
02
∫
dBStill valid
MAGNETIC FIELD OF A INFINITELY LONG SOLENOID
Current out of page
Current into page
P
To have ,B would have to have the same magnitude no matter how far we moved away from the solenoid. This is possible only if B = 0 outside of the solenoid. 2
Closed pathaB
bB
02
∫
dBStill valid
02
∫
dB
0 ba BB
MAGNETIC FIELD OF A INFINITELY LONG SOLENOID
Current out of page
Current into page
Using the closed path shown we can now obtain an expression for the magnetic field inside the long solenoid.
3 Closed path
0bB
P
L
MAGNETIC FIELD OF A INFINITELY LONG SOLENOID
Current out of page
Current into page
3 Closed path
0bB
P
L
NIBLdB o∫3
enclosedo IdB ∫
Ampere’s Law
L
NIB o
N : number of turns enclosed by length L
MAGNETIC FIELD OF A INFINITELY LONG SOLENOID
Current out of page
Current into page
0bB
P
L
NIB o
N : number of turns enclosed by length L
• B is independent of distance from the axis of the long solenoid as we are inside the solenoid!• B is uniform inside the long solenoid.
MAGNETIC FIELD OF A INFINITELY LONG SOLENOID
21
ELEC 3105 BASIC EM AND POWER ENGINEERING
MAGNETIC VECTOR POTENTIAL
22
MAGNETIC VECTOR POTENTIAL
To find for general problems, we need more sophisticated techniques than Ampere’s law and a few postulates.
B
7
MagnetostaticsPOSTULATE POSTULATE 11FOR THE MAGNETIC FIELDFOR THE MAGNETIC FIELD
A current element immersed in a magnetic field will experience a force given by:
dBIFd
dI
B
Fd
Units of Newtons {N}
26
MagnetostaticsPOSTULATE POSTULATE 22FOR THE MAGNETIC FIELDFOR THE MAGNETIC FIELD
A current element produces a magnetic field which at a distance Ris given by:
d
R
RIBd o
2
ˆ
4
dI
B
Bd
Units of {T,G,Wb/m2}
3
A m p e r e ’s L a w
T h e s t a r t i n g p o in t i s a m o d i f i c a t i o n o f p o s tu la t e 2 .
W e n e e d a m e a n s o f c o m p u t in g t h e m a g n e t i c f i e ld f o r a k n o w n c u r r e n t d i s t r ib u t i o n .
enclosedo IdB ∫
L in e in te g r a l a r o u n d c lo s e d p a th
C u r r e n t e n c lo s e d b y p a th
∫ S
adJI
Suppose we can find a function such that: AB
zyxA ,,
zyxA ,, Is called the Magnetic Vector Potential.
It is a function of the coordinates.It has direction.
It is possible to show that for any zyxA ,,
: 0 A
0 B
MAGNETIC VECTOR POTENTIAL
24
Recall: JBo
Differential form of Ampere’s Law
Ampere’s Law:enclosedoIdB ∫
Area enclosed A
J
Current density flowing through loop.
∫ ∫ S
oadJdB
∫ ∫ S
odanJdanB ˆˆ
JBo
∫ ∫ S
adFdF
Using Stoke’s theorem
Differential equation for B
Can we solve this equation?And: AB
Then:
JAo
MAGNETIC VECTOR POTENTIAL
Add in mathematical manipulations
25
But for any vector function :
JAo
A
AAA
2
And if 0 A
AA
2
use here
MAGNETIC VECTOR POTENTIAL
26
JAo
2
zozJA 2
GIVES
In component form:
yoyJA 2
xoxJA 2
We have managed to generate independent equations governing the x, y and z components of
zyxA ,,
MAGNETIC VECTOR POTENTIAL
27
zyxA ,,
Is called the Magnetic Vector Potential.It is a function of the coordinates.It has direction.Is not unique.Three scalar equations involving derivatives
MAGNETIC VECTOR POTENTIAL
zozJA 2
yoyJA 2
xoxJA 2
Natural progression of the student’s perspective of the magnetic vector potential
Today Next week Before exam After exam
I’m doing the deferred exam
28
Similar to Poisson’s equation:
zozJA 2
yoyJA 2
xoxJA 2
zyxAAA ,,
3 4
P o i s s o n ’s / L a p l a c e ’s E q u a t i o n
I n m a n y r e g i o n s o f s p a c e = 0 , n o n e t c h a r g e d e n s i t y . I n t h i s c a s e :
o
V
2
L a p l a c e ’ s E q u a t i o nL a p l a c e ’ s E q u a t i o n02 V
P o i s s o n ’ s E q u a t i o nP o i s s o n ’ s E q u a t i o n
∫∫∫
volo zzyyxx
dzdydxzyxzyxV
2
21
2
21
2
21
222222111
,,
4
1,,
Solution of this form in (x, y, z)
Integration over volume containing charge .
Same form of equation, same form of solutionFigure next page
FINDING THE MAGNETIC VECTOR POTENTIAL
Need to obtain all three components
General solution for scalar potential
29
RECAL FOR POTENTIAL V
3 4
P o i s s o n ’s / L a p l a c e ’s E q u a t i o n
I n m a n y r e g i o n s o f s p a c e = 0 , n o n e t c h a r g e d e n s i t y . I n t h i s c a s e :
o
V
2
L a p l a c e ’ s E q u a t i o nL a p l a c e ’ s E q u a t i o n02 V
P o i s s o n ’ s E q u a t i o nP o i s s o n ’ s E q u a t i o n
∫∫∫
volo zzyyxx
dzdydxzyxzyxV
2
21
2
21
2
21
222222111
,,
4
1,,
x y
z
222
,, zyx
),,(111zyxV
2r
1r
Charge distribution
∫∫∫
volo rr
dzdydxrrV
21
22221 4
1
Could write solution in short hand form as:
21rr
30zyx AAA ,,
In the short hand vector notation
FINDING THE MAGNETIC VECTOR POTENTIAL
∫∫∫
vol
xox
zzyyxx
dzdydxzyxJzyxA
2
21
2
21
2
21
222222111
,,
4,,
∫∫∫
vol
yoy
zzyyxx
dzdydxzyxJzyxA
2
21
2
21
2
21
222222
111
,,
4,,
∫∫∫
vol
zoz
zzyyxx
dzdydxzyxJzyxA
2
21
2
21
2
21
222222111
,,
4,,
∫∫∫
vol
o
rr
dzdydxrJrA
21
22221 4
31
EXAMPLE B USING AFind and for a long wire aligned along the z-axis.
A
B
y
z
I
wire
111
,, zyxP
r
Evaluate hereA
By geometry:
0yx
JJ
0yx
AASimplifies solving for A
∫∫∫
vol
xox
zzyyxx
dzdydxzyxJzyxA
2
21
2
21
2
21
222222111
,,
4,,
∫∫∫
vol
yoy
zzyyxx
dzdydxzyxJzyxA
2
21
2
21
2
21
222222
111
,,
4,,
∫∫∫
vol
zoz
zzyyxx
dzdydxzyxJzyxA
2
21
2
21
2
21
222222111
,,
4,,
32
y
z
I
wire
111
,, zyxP
r
Becomes
A
Where
Current in the wire
∫∫∫
vol
zoz
zzyyxx
dzdydxzyxJzyxA
2
21
2
21
2
21
222222111
,,
4,,
EXAMPLE B USING A
∫
wire
oz
zzyx
dzIzyxA
2
21
2
1
2
1
2111 4
,,
IdydxJ z ∫ sectioncross
wire22
33
y
z
I
wire
111
,, zyxP
r
Az should have the same value for any z1.(Once again by a symmetry argument)
A
Chose to evaluate for z1 = 0 for simplicity.
Wire extends for - to +
EXAMPLE B USING A
∫
wire
oz
zzyx
dzIzyxA
2
21
2
1
2
1
2111 4
,,
∫
wire
oz
zyx
dzIzyxA
2
2
2
1
2
1
2111 4
,,
2
1
2
1111 ln2
,, yxI
zyxA oz
34
y
z
I
wire
111
,, zyxP
rA
Since is the distance from the point P to the wire
EXAMPLE B USING A
2
1
2
1111 ln2
,, yxI
zyxA oz
rIzyxA o
z ln2
,, 111
2
1
2
1 yxr
35
y
z
I
wire
111
,, zyxP
rA
Now for B
We have
and
AB
Then we can evaluate the curl of A to get B.Note that A has only a z component.
EXAMPLE B USING A
rIzyxA o
z ln2
,, 111
22 r
IyB o
x
22 r
IxB o
y
0z
B
38
y
z
I
wire
111
,, zyxP
rA
Now for B
We have
and
AB
Then we can evaluate the curl of A to get B.Note that A has only a z component.
EXAMPLE B USING A
rIzyxA o
z ln2
,, 111
22 r
IyB o
x
22 r
IxB o
y
0z
B
39
y
z
I
wire
111
,, zyxP
rB
22 r
IyB o
x
22 r
IxB o
y
Gives
22
yxBBB
r
IB o
2
As would be obtained from Ampere’s Law
circles wire
EXAMPLE B USING A
THE FORCE AND TORQUE ON A DIPOLE IS TAKEN UP AT THE START OF THE NEXT LECTURE
ELEC 3105 BASIC EM AND POWER ENGINEERING
FORCE AND TORQUE ON MAGNETIC DIPOLE
Magnetic dipole = product of current in
loop with surface area of loop
FORCE ON A MAGNETIC DIPOLE B
Consider a circular ring of current I placed at the end of
a solenoid as shown in the figure. The current in the
solenoid produces a magnetic field in which the current loop is placed into.
By postulates 1 and 2 of magnetic fields, the current ring will be subjected to a
magnetic force.out of page into page
I
z
FORCE ON A MAGNETIC DIPOLE B
I
z
outF
outF
downF
downF
Circular ring
Cancel in pairs around the ringout
F
outF
downF
downF
Will add in same direction on ring giving a net force.
FORCE ON A MAGNETIC DIPOLE B
I
z
outF
outF
downF
downF
Circular ring
downF
downF
Will add in same direction on ring giving a net force.
Using postulate 1: dBIFd
r2
B
zB
rB
downF
We need for find Br
Gives:
rdown rIBF 2
FORCE ON A MAGNETIC DIPOLE
z
z
Gaussian cylinder
We will relate Br to z
Bz
Total magnetic flux through Gaussian cylindrical surface must be zero. As many magnetic field lines that enter the surface, leave the surface. No magnetic charges or monopoles.0 B
11
Another important property of B
0 B
Recall everywhere
No net magnetic flux through any closed surface.
Closed surface S
∫ S
adB 0
∫ vol
voldvB 0
Using divergence theorem
0 3-D view
FORCE ON A MAGNETIC DIPOLE
z
z
Flux through side:
3-D view
siderzBr 2
Flux through top:
topz
zzBr 2
Flux through bottom:
bottomz
zBr 2
0bottomtopside
FORCE ON A MAGNETIC DIPOLE
z
z
3-D view
02 22 zBrzzBrzBrzzr
0bottomtopside
z
zBzzBrB zz
r
2
z
BrB z
r
2
We can now use this in our force on current ring expression
FORCE ON A MAGNETIC DIPOLE B
I
z
outF
outF
downF
downF
Circular ring
rdownrIBF 2
r2
B
zB
rB
downF
We have found Br
z
BrB z
r
2
z
BrrIF z
down
2
2
FORCE ON A MAGNETIC DIPOLE B
I
z
outF
outF
downF
downF
Circular ring
r2
B
zB
rB
downF
z
BrrIF z
down
2
2
z
BIrF z
down
2
z
BIAF z
down
z
BmF z
down
z
BmF z
z
Force pulls dipole into region of stronger magnetic field
FORCE ON A MAGNETIC DIPOLE
3-D view
z
In general
xxBmF
yyBmF
zzBmF
BmF
TORQUE ON A MAGNETIC DIPOLE
We will consider a dipole in a uniform magnetic field. We can use any shape we want for the dipole. Here we will select a square loop of wire.
I out of page
I into page
m
B
a
a
I
Side view
Topview
Wire loop
a
a
I
Topview
Wire loop
a
2
a
TORQUE ON A MAGNETIC DIPOLE
TORQUE ON A MAGNETIC DIPOLEm
B
a
a
I
Side view
Topview
Wire loop
F
F
Torque attempts to align dipole
moment with .m
B
Pivot point
Pivot line
TORQUE ON A MAGNETIC DIPOLE
m
B
Side view
F
F
Torque attempts to align dipole
moment with .m
B
Pivot point
sin2
2a
F
Fr
2
a
Total torque
F => Magnetic force on wire of length a
TORQUE ON A MAGNETIC DIPOLE
m
B
Side view
F
F
Pivot point
sin2
2a
F
2
a
F => Magnetic force on wire of length a
IBaF Through postulate 1 for magnetic fields
sin2IBaThen
TORQUE ON A MAGNETIC DIPOLE
TORQUE ON A MAGNETIC DIPOLEm
B
Side view
F
F
Pivot point
2
a
sin2IBa
a
a
I
Wire loopIam 2
sinBm
Bm
TORQUE ON A MAGNETIC DIPOLE
Recommended