Chapter 29: Sources of Magnetic Field
Sources are moving electric charges
single charged particle:
  o qv  r
B
field point
4 r 2

qv  r
B
^
 k' 2
^
r
r
r
r
compare
v

1 qr
E
4 o r 2
qr
k 2
field lines circulate around
r
“straight line trajectory”
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from the definition of a Coulomb and other standards,
o= 4x10-7 N s2/C2 = 4x10-7 T m/A
with o , the speed of light as a fundamental constant can be
determined (in later chapters) from
c2 = 1/(o o)
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Consider: Two protons move in the x direction at a speed v.
Calculate all the forces each exerts on the other.
magnetic:
F1 on 2 = q2v2xBat2due to 1
= q2v2x (k’ (q1v1xr^12)/(r12)2)
electric
F1 on 2 = (k q1q2)/(r12)2 ^r12
Simple geometry (all 90 degree angles)
q1
Fmagnetic = k’|q2v2 q1v1|/ (r12)2
= k’ e2v2/r2
F1 on 2
v1
^
r12
r12
B
v2
Felectric = k e2/r2
Fmagnetic /Felectric = k’ v2 /k = (o/4v2/ (1/4o
= v2 /c2
but... depends upon frame?
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Gauss’s Law for magnetism:
Magnetic field lines encircle currents/moving
charges. Field lines do not end or begin on any “charges”.
 
 B  dA  0
for any closed surface.
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Current elements as magnetic field sources
superposition of contributions of all charge carriers
+ large number of carriers + small current element


 o qi vi  rˆi
B
field point
4 ri 2

 o qvd  rˆ
B
^
^

N
r
r
r
4
r2

o
qvd  rˆ
I dl

(nAdl )
4
r2

  o Idl  rˆ
dB 
4 r 2

  o Idl  rˆ
B
4  r 2
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Field of a long straight wire
I

dl
y
r  x2  y2
sin   y / r
r^
dB
x

  o Idl  rˆ
dB 
4 r 2
 o Idy sin 

( zˆ ) (into page)
2
4
r
o
Ix

dy ( zˆ )
32
2
2
4 x  y 
 o 
Ix
B
dy ( zˆ )
32

2
2
4  x  y 
 o I
B
( zˆ )
2x
o I
B
2r
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Example: A long straight wire carrries a current of 100 A. At what distance will
the magnetic field due to the wire be approximately as strong as the earth’s field
(10-4 T)?
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Force between two long parallel current carrying wires
(consider, for this example, currents in the same direction)
I2
I1
F = I2 Bdue to I1 L
= I2 (oI1/(2r))L
F/L = oI1 I2 /(2r)
force on I1 is towards I2
force is attractive (force is
repulsive for currents in
opposite directions!)
Fdue to I1
Bdue to I1
Example: Two 1m wires seperated by 1cm each carry 10 A in the same direction.
What is the force one wire exerts on the other
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Magnetic Field of a Circular Current Loop


r^
I | dl  rˆ | I dl
dB
| dB | k '
r2
r
a

Idl

 k' 2
dl rˆ

r
x
dBx
r  x2  a2
cos( 90   )  sin 
a

Bx   dBx
x2  a2

2 a
dBx | dB | cos( 90   )
aIds
2a 2 I
  k' 2
 k' 2
2
3
2
2 32
Idl
a
(
x

a
)
(
x

a
)
0
 k' 2
x  a2 x2  a2
2M
Bx  k ' 2
dl  ds (segment of arc)
( x  a 2 )3 2
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Bx 
oa 2 I
2( x 2  a 2 )3 2
o I
Bx 
at center
2a
oa 2 I
Bx 
2 x3
x  a
2M
 k' 3
x
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Example: A coil consisting of 100 circular loops .2 m in radius carries a current of
5 A. What is the magnetic field strength at the center?
N loops => N x magnetic field of 1 loop.
At what distance will the field strength be half that at the center?
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Ampere’s Law
•Equivalent to Biot-Savart Law
•Useful in areas of high symetry

  o Idl  r
dB 
4 r 2
•Analogous to Gauss’s Law for Electric Fields
Formulated in terms of:
 
 B  dl   B||dl
For simplicity, consider single long straight wire (source) and
paths for the integral confined to a plane perpendicular to the
wire.
dl
B
I
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dl
r

B
d
dl

B
r
d
 
B  dl  Bdl cos
 Brd
oI

rd
2r
oI

d
2
 
oI
 B  dl   2 d
oI

d

2
  o I path encloses current
0
path does not
enclose current
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 
 B  dl   o I enclosed
Right Hand Rule

relates conventions for directions of I and dl .
In general
 
 
 B  dl   o  J  dA
•Amperian Loop analogous to Gaussian surface
•Use paths with B parallel/perpendicular to path
•Use paths which reflect symetry
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Aplication of Ampere’s Law: field of a long straight wire
Cylindrical Symetry, field lines circulate around
wire.
r
I


 B (r )  dl  B (r )  dl
 B (r )2 r
 oI
oI
B (r ) 
2 r
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Field inside a long conductor
I
r


 B (r )  dl  B (r )  dl
 B (r )2 r
  o I encl
I encl  JA
I
2


r
 R2
R
J
I 2
B (r )2 r   o 2 r
R
o I r
B(r ) 
2 R 2
o I
[ B(r ) 
outside]
2 r
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B/Bmax
r=R
r
p202c29: 17
Homework: Coaxial cable
I
I
p202c29: 18
Magnetic Field in a Solenoid
Close packed stacks of coils form cylinder
I
B
I
B
Fields tend to cancel in region right between wires.
Field Lines continue down center of cylinder
Field is negligible directly ouside of the cylinder
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B
L
I
 
 B  dl  BL
I enc  LnI (n turns per unit length)
BL   o LnI
B   o nI
B  o
N
I (for finite solenoid)
L
p202c29: 20
Example: what field is produced in an air core solenoid with 20 turns per cm
carrying a current of 5A?
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Toroidal Solenoid
Field along center of torus
 
 B  dl  B 2 r
I enc  NI
B 2 r   o NI
B
 o NI
2 r
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Magnetic Materials
Microscopic current loops:
L
electron “orbits”
electron “spin”
v
I e
2r
v 2
  IA  e r
2r
mvr
e
e

L
2m 2m
h
Ln
 n
2


e
  Bohr Magneton
2m
 9.274  10  24 J / T
 
U    B
  
  B
B 
Quantum Effects: quantized L, Pauli Exclusion Principle important in macroscopic
magnetic behaviour.
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Magnetic Materials: Microscopic magnetic moments interact
with an external (applied) magnetic field Bo and each other,
producing additional contributions to the net magnetic field B.
Magnetization M = tot/V
B= Bo + o M
linear approximation: M proportional to Bo
o => = Km o = permeability
m = Km-1 magnetic Susceptibility
Types of Materials
Diamagnetic: Magnetic field decreases in strength.
Paramagnetic: Magnetic field increases in strength.
Ferromagnetic: Magnetic field increases in strength!
Diamagnetic and Paramagnetic are often approximately
linear with  m
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Ferromagnetism:
Greatly increases field
M
Saturation
Highly nonlinear, with Hysteresis:
Bo
Hysteresis= magnetic record
Permanent
Magnetization
Magnetization forms in Magnetic Domains
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Displacement Current
“Generalizing” displacement current for Ampere’s Law
conduction current creates magnetic field
Amperian loop with surface
Ienc = iC
iC
Parallel Plate
Capacitor
Amperian loop with surface
Ienc = 0???
iC
Parallel Plate
Capacitor
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Define Displacement Current between plates so that iD = iC
A
q  Cv   Ed  EA   E
d
dq
d E
iC 

 iD
dt
dt
 
 B  dl   o (iC  iD )
d  
  oiC   o  E  dA
dt


dE
jD  
dt
changing Electric Fields can create Magnetic Fields!
p202c29: 27