Transcript Magnetic Field of a Moving Charge

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 currentcarrying 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
Magnetic Field of a Moving Charge)
Magnitude of B
Permeability of free space
0 q v sin 
B
4
r2
 0  4x10 7
(28-1)
 
Direction of B determined by v xr
0
 10  7
4
The vector form:
3
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
0 q v
FB 
4 r 2
FB
v2
 2
FE
c
The ratio of the two
forces. Where c=speed of
light. Therefore:
FE>>FB
Magnetic Field of a Current Element
Total magnetic field of several moving charges =
vector sum of fields caused by individual charges
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
I = nqvdA
Figure 28-3
 dQv d sin   0 nqvd Adl sin 
dB  0

2
4
r
4
r2
 0 Idl sin 
dB 
(28-5) Biot-Savart Law
2
4
r
 
Direction of dB determined by Idl xr
Vector form of Biot-Savart Law
5
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
Magnetic Field of a Current-Carrying Conductor
 0 Idl sin 
dB 
4
r2
sin   sin(    ) 
x
x2  y2
dl  dy
0 Idy
0 Ix
x
dy

4 ( x 2  y 2 ) x 2  y 2
4 ( x 2  y 2 )3 / 2
0 Ix a
dy
B
4 a x 2  y 2 3 / 2
dB 
Figure 28-5
Based upon symmetry around
the y-axis the field will be a
circle
0 Ix  1
B
 2
4  x
If a  x
a

 0 I  2a 
y
 
 2

2
2
2
4

x
x  y  a
 x a 
 0 I  2a   0 I
B
 2
4 x  a  2 x
7
Magnetic Field of a Current-Carrying Conductor
Figure 28-6
where r = perpendicular distance from the current-carrying wire.
8
Force between Parallel Conductors
Only field due
to I shown
Each conductor lies in the field set up
by the other conductor

 
'
F  I LB
Note: If I and I’ are in the same
direction, the wires attract. If I
and I’ are in opposite directions,
the wires repel.
 0 I 
F  I LB sin   I LB  I L 

2

r


0 II '
Substitute for B
FL
2 r
See Example 28.5 Page 966
F 0 II '

L 2 r
'
'
'
9
Magnetic Field of a Circular Current Loop
 0 Idl sin 
dB 
4
r2
sin   sin 90  1
r 2  x2  a2
0 I
dl
cos
4 ( x 2  a 2 )


0 I
dl
a
dBx  dB cos  


4 ( x 2  a 2 )  ( x 2  a 2 )1/2 
dB 
Figure 28-12
0 I
0 Ia
a
Bx 
dl 
(2a)
2
2 3/ 2 
2
2 3/ 2
4 ( x  a )
4 ( x  a )
By= 0
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Magnetic Field of a Circular Current Loop
Bx 
0 NIa 2
2( x 2  a 2 )3 / 2
(on the axis of N circular loops)
(x=0)
x
Figure 28-13
Figure 28-14
11
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
Ampere’s Law (Chapter 28, Sec 6)
Figure 28-15
For Figure 28-15a
 B  dl   B cos dl   Bdl  B  dl
 I
B 0
2 r
 dl  2r
0 I
 B  dl  B(2r )  2r (2r )  0 I
For Figure 28-15b
 B  dl   0 I
For Figure 28-15c
b
c
d
a
a
b
c
d
 B  dl   B  dl   B  dl   B  dl   B  dl
b
c
d
a
c
d
 B  dl  B  dl  0 dl  B  dl  0 dl
1
2
a
b
0 I
0 I
 B  dl  2r1 (r1 )  0  2r2 (r2 )  0
 B  dl  0
14
Ampere’s Law
 B  dl   B cos dl
cos dl  rd
0 I
B
2 r
0 I
0 I
 B  dl   2r (rd )  2  d
Figure 28-16
 d  2
I
B

dl

(2 )   I

2
0
0
 d  0
 B  dl 0
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Applications of Ampere’s Law Example 28-9
Field of a Solenoid (magnetic field is concentrated in side the coil)
n = turns/meter
Figure 28-20
 B  dl 0 I encl  0 nLI
b
Figure 28-21
c
d
a
 B  dl   B  dl   B  dl   B  dl   B  dl
a
b
b
c
c
d
N
n
l
d
a
 B  dl  B dl  0 dl  0 dl  0 dl  BL
a
BL  0 nLI
b
c
where N = total coil turns
l = total coil length
d
N
B   0 nI   0 I
l
turns/meter
(28-23)
16
Applications of Ampere’s Law
Example 28-9
Field of a Solenoid
N
n
l
turns/meter
where N = total coil turns
l = 4a = total coil length
Figure 28-22
17
Applications of Ampere’s Law Example 28-10
Field of a Toroidal Solenoid – (field is inside the toroid)
N turns
Path 1
 B  dl  I
0 encl
0
B0
No current enclosed
Path 3
 B  dl  I
0 encl
0
B0
Figure 28-23
Current cancels
Path 2
 B  dl  I   NI
 B  dl B  dl  B(2r )
0 encl
B(2r )  0 NI
0
NI
B  0
2r
(28-24)
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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
Bohr Magneton- In
atoms electron spin
creates current a loop,
which produce magnetic
their own field
19
Ferromagnetism and Hysteresis loops
•
•
•
The larger the loops the more energy that is lost magnetizing and demagnetizing.
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
M 
B
0