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1
ELEC 3105 Basic EM and Power Engineering
Biot-Savard LawApplications of Biot-Savard Law
2
Biot-Savard Law
Usually we deal with current localized to wires rather than spread over space.
I
I
d
1r
2r
P
Origin
When the current is contained by thin wires, “we can use the magnetic vector potential to derive an integral from which the magnetic field can be found directly
from the wire location and currents”.
3
Biot-Savard Law KISS Rule
I
I
d
1r
2r
P
Origin
We will find the magnetic field at some point P located at vector from the origin due to a short segment of wire located at from the origin, carrying
current I, where the current is in the direction of the vector , as shown below.
B
1111
,, zyxr
d 2222
,, zyxr
d
small segment of wire at ),,(
222zyx
find hereB
4
Biot-Savard Law
Consider a small segment of wire of overall length
I
d
P
21r
This expression isThe Biot-Savard Law
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}
d
Bd
Same result as postulate 2 for the magnetic field
2
21
211
ˆ
4 r
rdIrBd o
5
Biot-Savard Law
The magnetic field produced by all the short current elements along the line is the Biot-Savard law.
Consider a small segment of wire of overall length
I
d
P
R
This expression isThe Biot-Savard Law
d
Bd
Line
o
Line R
RdIrBdrB 21
ˆ
4
An integral is another way of saying “principle of superposition” applies to magnetic fields.
6
Biot-Savard Law & Current Density J (A/m2)
I
I
d
1r
2r
P
Origin
find hereB
1rBd
dIdxdydzJ The current in the differential volume element dxdydz is:
volume
o
volume R
RJdvolrBdrB 21
ˆ
4
𝑅
7
Biot-Savard Law & Surface Current Density K (A/m)
I
I
d
1r
2r
P
Origin
find hereB
1rBd
dIKdA The current in the differential surface element dA is:
surface
o
surface R
RKdArBdrB 21
ˆ
4
𝐾
𝑅dA
8
ELEC 3105 Basic EM and Power EngineeringSTART Applications of Biot-Savard Law
BIOT from
RAMA
9
Magnetic field produced by a circular current ring
y
10
Magnetic field produced by a circular current ring
We will first consider the magnetic field present at the
center of a current carrying ring.
I
x
y
a
We can use the Biot-Savard Law
11
Magnetic field produced by a circular current ring
All contributions from each line segment point in the same direction: in the direction of the axis of the ring.
I
x
y
ad
add
Top viewSide view
I out of page
I into page
a aBdBd
Bd
12
Magnetic field produced by a circular current ring
I
x
y
ad
add Top viewSide view
a aStarting from the Biot-Savard Law for a single segment.
We get: ar 21
ˆadd rr ˆ
21
r
12r
2
21
211
ˆ
4 r
rdIrBd o
24 a
adIdB o
13
Magnetic field produced by a circular current ring
I
x
y
ad
add Top view
a a
ar 21
ˆadd rr ˆ
21
r
12r
Direction:
zr ˆˆˆ
Side view
As shown by arrows in side view
24 a
adIdB o
2
21
211
ˆ
4 r
rdIrBd o
14
Magnetic field produced by a circular current ring
I
x
y
ad
add Top view
Summation over all segments around the ringd
Magnetic field at the center of the circular current ring
2
0 4 a
dIdBB o
a
IB o
2
za
IB o ˆ
2
15
Magnetic field produced by a circular current ring
We will now consider the magnetic field present on
the central axis of a current carrying ring.
Biot-Savard Law
The Biot-Savard law applied to the small segment gives an element of magnetif field at the point P.
Consider a small segment of wire of overall length
I
d
P
21r
This expression isThe Biot-Savard Law
26
MagnetostaticsPOSTULATE POSTULATE 22 FOR THE MAGNETIC FIELDFOR THE MAGNETIC FIELD
A current element produces a magnetic field which at a distance R is given by:
d
R
RIBd o
2
ˆ
4
dI
B
Bd
Units of {T,G,Wb/m2}
d
Bd
Same result as postulate 2 for the magnetic field
Bd
2
21
211
ˆ
4 r
rdIrBd o
We can use the Biot-Savard Law
I
x
y
a
3-D viewz
z
16
Magnetic field produced by a circular current ring
d add
I out of page
I into page
a a
Bd
Bd
I
x
y
a
3-D viewz
z
All contributions from each line segment point in different directions: We must consider components of .
Bd
Bd
21r
17
Magnetic field produced by a circular current ring
I out of page
I into page
a a
Bd
Bd
z21
r Bd Bd
zdB
zdB
hdB
hdB
hdB
hdB
Horizontal components will cancel in pairsaround the segments of the current ring
zdB
zdB
Z components will add around the segments of the current ring
18
Magnetic field produced by a circular current ring
hdB
hdB
Horizontal components will cancel in pairsaround the segments of the current ring
zdB
zdB
Z components will add around the segments of the current ring
19
Magnetic field produced by a circular current ring
I out of page
I into page
a a
Bd
Bd
z21
r Bd z
dB
cosdBdBz
22
2121zarr
21
cosr
a
3-D view
Can use Biot-Savard Law
We need only concern ourselves with the magnitude of since we know the final direction of is along the z direction.
Bd
B
2
21
211
ˆ
4 r
rdIrBd o
20
Magnetic field produced by a circular current ring
I out of page
I into page
a a
Bd
Bd
z21
r Bd z
dB
cosdBdBz
22
2121zarr
21
cosr
a
3-D view
Expression for magnetic field produced by one current segment of the ring.
2
214 r
dIdB o
2
214 r
adIdB o
cos4 2
21r
adIdB o
z
321
2
4 r
daIdB o
z
21
Magnetic field produced by a circular current ring
d add
I out of page
I into page
a a
Bd
Bd
Ix
y
a
3-D viewz
z21
r
Summation around ring:
Magnetic field on the central axis of the circular current ring
2
03
21
2
4 r
daIdBB o
zz
2
03
22
2
4d
za
aIB o
z
322
2
2 za
aIB o
z
22
Magnetic field produced by a circular current ringWe only considered the central axis field
The full field is obtained when we will discuss the magnetic dipole
magnetotherapy
23
Magnetic field of a long FINITE solenoid
24
Magnetic field of a FINITE solenoid
Current out of page
Current into page
finite coil of wire carrying a current I
Axis of solenoid
P
Evaluate B field here
a
Radius of solenoid is a. Cross-section cut through solenoid axis
L
25
Magnetic field of a FINITE solenoid
1 2 3 4 5
Axis of solenoid
Current out of page
zdB
dSegment of the solenoid coil
r
d
sin
rdd
arc length rd
a r
asin
26
Magnetic field of a FINITE solenoid
L
The solenoid consists on N turns of wire ”rings of current” each carrying a current I.
The number of turns per unit length of solenoid can be expressed as:
L
N
27
Magnetic field of a FINITE solenoid
LThe current in a segment (dI) can be expressed as the current in one ring (I) multiplied by the number of turns per unit length (N/L) and multiplied by the segment length .
dL
NIdI
d
d
dL
NIrdI
sin
sin
rdd
28
Return to this slide for a moment
d add
I
x
y
a
3-D viewz
Magnetic field on the central axis of the circular current ring
Each ring of the solenoid contributes a magnetic field. Can use this expression for our finite solenoid. Sum over the rings.
322
2
2 za
aIB o
z
29
Magnetic field of a FINITE solenoid
1 2 3 4 5
Axis of solenoid
Current out of page
zdB
dSegment of the solenoid coil
r
d
sin
rdd
arc length rd
a r
asin 3
2
2 r
adIdB o
z
30
Magnetic field of a FINITE solenoid
1 2 3 4 5
zdB
d
r
d
a
dL
NIrdI
sin
r
asin
sub in
3
2
2 r
adIdB o
z
d
L
NI
r
ad
La
NIr
r
adB oo
z 22
2
3
2
d
L
NIdB o
z sin2
31
Magnetic field of a FINITE solenoid
L
d
12
We can now sum (integrate) the expression for over the angular extent of the coil. I.e. sum over all the rings of the finite length solenoid.
zdB
z
2
1
sin2
d
L
NIB o
z
21 coscos2
L
NIB o
z
zL
NIB o ˆcoscos
2 21
32
Magnetic field of a INFINITE solenoid
L
d
01180
2
z
INFINITE SOLENOID RESULT
infinite solenoid result
180cos0cos2
L
NIB o
z
zL
NIB o ˆL
NIB oz
33
Magnetic field at one end of a FINITE solenoid
L
d
901180
2
z
Magnetic field is ½ that of center
180cos90cos2
L
NIB o
z
zL
NIB o ˆ
2
L
NIB oz 2
34
35
Magnetic flux
Φ𝑚= ∬𝑆𝑢𝑟𝑓𝑎𝑐𝑒
❑𝐵 ∙𝑑𝐴
𝐵Goes by the name of magnetic flux density vector
36
Magnetic flux: CLOSED SURFACE
Φ𝑚= ∯𝑆𝑢𝑟𝑓𝑎𝑐𝑒
❑𝐵 ∙𝑑𝐴=0
Magnetic field lines loop back on themselvesNo magnetic sourceNo magnetic sink
No magnetic monopoleNo magnetic charge
37
Divergence of a Magnetic Field
Φ𝑚= ∯𝑆𝑢𝑟𝑓𝑎𝑐𝑒
❑𝐵 ∙𝑑𝐴=0
For a closed surface
No magnetic monopole
No magnetic charge∯
𝑆𝑢𝑟𝑓𝑎𝑐𝑒
❑𝐵 ∙𝑑𝐴= ∰
𝑉 𝑜𝑙𝑢𝑚𝑒
❑𝛻 ∙𝐵𝑑𝑣𝑜𝑙
Divergence theorem
𝛻 ∙𝐵=0𝛻 ∙𝐷=𝜌 𝑣
Electric charge exist𝛻 ∙𝐻=0
38
Magnetic flux
Φ𝑚= ∬𝑆𝑢𝑟𝑓𝑎𝑐𝑒
❑𝐵 ∙𝑑𝐴
We have the expression for the magnetic field AND the limits for the integral Both specified in cylindrical coordinates.
Given that the radial magnetic field is , calculate the magnetic flux through the surface defined by and .
Solution
Remainder of solution presented in class
39
Magnetic flux
Φ𝑚= ∬𝑆𝑢𝑟𝑓𝑎𝑐𝑒
❑𝐵 ∙𝑑𝐴
We have the expression for the magnetic field AND the limits for the integral Both specified in cylindrical coordinates.
In cylindrical coordinates , calculate the magnetic flux through the surface defined by and .
Solution
Remainder of solution presented in class