Sources of Magnetic Field Chapter 28
• Study the magnetic field generated by a moving charge
• Consider magnetic field of a current-carrying conductor
• Examine the magnetic field of a long, straight, current-carrying conductor
• Study the magnetic force between current-carrying conductors
• Consider the magnetic field of a current loop• Examine and use Ampere’s Law
1
The magnetic field of a moving charge
• A moving charge will generate a magnetic field relative to the velocity of the charge.
2
3
Magnetic Field of a Moving Charge)
20
sin
4 r
vqB
(28-1)
70 104
70 104 x
Permeability of free space
Direction of B determined by rxv
Magnitude of B
The vector form:
Force between two moving protons• Two protons moving at the same velocity
(much less than speed of light) in opposite directions.
• The electric force FE is repulsive.
• The right-hand rule indicates the magnetic force FM is repulsive. (i x k=-j)
• Find the ratio of the magnitude of the forces.
2 2
02
2
2
4B
B
E
q vF
r
F v
F c
The ratio of the two forces. Where c=speed of light. Therefore:FE>>FB
5
Magnetic Field of a Current Element
Figure 28-3
Let dQ = charge in wire segment dl
Let A = cross section area of wire segment dl
Let n = charge density in wire segment dl
dQ = nqAdl
20
20 sin
4
sin
4 r
Adlnqv
r
dQvdB dd
I = nqvdA
20 sin
4 r
IdldB
(28-
5)Biot-Savart Law
Direction of dB determined by rxlId
Vector form of Biot-Savart Law
Total magnetic field of several moving charges = vector sum of fields caused by individual charges
Magnetic field of a straight current-carrying conductor
• Biot and Savart contributed to finding the magnetic field produced by a single current-carrying conductor.
6
7
Magnetic Field of a Current-Carrying Conductor
Figure 28-5
22)sin(sin
yx
x
dydl
20 sin
4 r
IdldB
2/3220
22220
)(4)(4 yx
dyIx
yx
x
yx
IdydB
a
a yx
dyIxB 2/322
0
4
22
0
2220 2
4
1
4 ax
a
x
I
yx
y
x
IxB
a
a
If a x
x
I
a
a
x
IB
2
2
40
2
0
Based upon symmetry around
the y-axis the field will be a circle
8
Magnetic Field of a Current-Carrying Conductor
Figure 28-6
where r = perpendicular distance from the current-carrying wire.
9
Force between Parallel Conductors
Each conductor lies in the field set up by the other conductor
' ' ' 0
'0
'0
sin2
2
2
IF I LB I LB I L
r
IIF L
r
IIF
L r
BLIF
'Note: If I and I’ are in the samedirection, the wires attract. If I and I’ are in opposite directions, the wires repel.
See Example 28.5 Page 966
Only field due to I shown
Substitute for B
10
Magnetic Field of a Circular Current Loop
Figure 28-12
20 sin
4 r
IdldB
190sinsin 222 axr
)(4 220
ax
dlIdB
)2()(4)(4 2/322
02/322
0 aax
Iadl
ax
aIBx
By= 0
02 2 2 2 1/2
cos4 ( ) ( )x
I dl adB dB
x a x a
cos
11
Magnetic Field of a Circular Current Loop
2/322
20
)(2 ax
NIaBx
(on the axis of N circular loops) (x=0)
Figure 28-13
x
Figure 28-14
Ampere’s Law I—specific then general
Similar to electric fields if symmetry exists it is easier to use Gauss’s law
12
Ampere’s Law II
• The line integral equals the total enclosed current
• The integral is the sum of the tangential B to line path
14
Ampere’s Law (Chapter 28, Sec 6)
Figure 28-15
Irr
IrBdlB 0
0 )2(2
)2(
r
IB
2
0 rdl 2
dlBBdldlBdlB cos
IdlB 00 dlB
For Figure 28-15a
For Figure 28-15b
For Figure 28-15c
d
c
a
d
b
a
c
b
dlBdlBdlBdlBdlB
0)(2
0)(2 2
2
01
1
0
rr
Ir
r
IdlB
d
c
a
d
b
a
c
b
dldlBdldlBdlB 00 21
15
Ampere’s Law
Figure 28-16
dlBdlB cos
rddl cos
d
Ird
r
IdlB
2)(
200
r
IB
2
0
2dI
IdlB 0
0 )2(2
0 d
0 dlB
16
Applications of Ampere’s Law Example 28-9Field of a Solenoid (magnetic field is concentrated in side the coil)
Figure 28-20
Figure 28-21nLIIdlB encl 00
n = turns/meter
d
c
a
d
b
a
c
b
dlBdlBdlBdlBdlB
BLdldldldlBdlBd
c
a
d
b
a
c
b
000
nLIBL 0 Il
NnIB 00
l
Nn
where N = total coil turns l = total coil length
turns/meter
(28-23)
17
Applications of Ampere’s LawExample 28-9 Field of a Solenoid
Figure 28-22
l
Nn
where N = total coil turns l = 4a = total coil length
turns/meter
18
Applications of Ampere’s Law Example 28-10Field of a Toroidal Solenoid – (field is inside the toroid)
Figure 28-23
00 enclIdlB
00 enclIdlB
NIIdlB encl 00
Path 1
Path 3
Path 2
N turns
)2( rBdlBdlB NIrB 0)2(
r
NIB
20 (28-24)
0B
0BCurrent cancels
No current enclosed
Magnetic materials • The Bohr magneton will determine how to
classify material. • Ferromagnetic – can be magnetized and
retain magnetism• Paramagnetic – will have a weak response
to an external magnetic field and will not retain any magnetism
• Diamagnetic – shows a weak repulsion to an external magnetic field
19
Bohr Magneton- In atoms electron spin creates current a loop, which produce magnetic their own field
Ferromagnetism and Hysteresis loops
• The larger the loops the more energy that is lost magnetizing and de-magnetizing.
• Soft iron produce small loops and are used for transformers, electromagnets, motors, and generators
• Material that produces large loops are used for permanent magnet applications
0
BM