Download Chapter 28: Sources of Magnetic Field

Chapter 28
1
Gravitational and Electric Fields

m
g  G 2 rˆ
r



q
E  k 2 rˆ
r
We expect that B will
1.
have a 1/r2 dependence
2.
Directed along the radius
Recall F= iLxB and F=qE and F=mg

Based on the pattern

i
B   2 rˆ
r
2
Broken Symmetry

But those other fields
diverge and B does
not
 Based on our
experience with bar
magnets, B must
return to the magnet
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Biot-Savart Law
B into page
X
X
X
q
i ds
Point P
X
X
X
1. B goes into the page
2. i ds and r are both
perpendicular to B
3. B proportional to i ds x r
4. But needs to be 1/r2 so
  o i ds  rˆ
dB 
4 r 2
or
  o i ds  r
dB 
4 r 3
4
A new constant, 0
Called the permeability of free space
 Value = 4 x 10-7 T*m/A
 And yes, 0/ 4 10-7 T*m/A

5
Modified Biot-Savart

If a charged particle has a constant velocity, v,
then I can modify Biot-Savart:

  o q v  rˆ
B
2
4 r
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Solving Biot-Savart Problems





Biot-Savart problems are
typically one of geometry
You must integrate about the
limits of the current loop while
evaluating the cross-product
Sometimes, ds and r are
parallel, which eliminates the
contribution. You should
examine the problem to see
where this is true.
Sometimes, ds and r are
perpendicular, which forces the
cross-product to its maximum
value.
Mostly, the ds and r are related
only through sin q. This means
that you may have to create an
integrand over the angle, not
necessarily the length of the
current loop.

 o i ds  rˆ
dB 
2
4 r
In conclusion, Biot-Savart
problems are strongly dependent
on creating the appropriate
geometry over which to integrate.
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Magnetic Field Due to a Long,
Straight Wire
ds
q
sin q  sin(   q ) 
R
s2  R2
Direction of current
s
r
X
X
X
X
R
Point P
i
Direction of B
X
r  s2  R2
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Integrating


 oi  sin q ds
B   dB  2 dB 
2

2

r

0
0
oi 
oi 
R
ds
Rds
B

2
2

2 0 s 2  R 2 s  R
2 0 s 2  R 2 3 2


oi 
oi
s

 
B
1
2R  s 2  R 2  2 
2R

0
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Magnetic Field due to Current of a
Circular Arc of Wire
oi sin q ds
B   dB 
4  r 2
oi ds
B
4  R 2

oi
B
dq

4R 0
 o i
B
4R
r always perpendicular to
ds so cross-product is ds
R
where ds  Rd q
ds
r
B out of page
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A Complete Loop

We can take the
previous result and
evaluate it at =2
 The result is the
same as your text on
pg. 1077 Eq. 28-17
 oi 2  oi
B

4R
2R
B
oi
2a
where R  a
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Force Between Two Long Wires with
Parallel Currents
Force on b caused
by the magnetic
field of a
Fa on b =ibL x Ba
d
X
X
X
X
ia

oia
 zˆ 
Ba 
2d
L
ib
Fa on b
oia oia ib L
 ib L

2d
2d
By RH rule, Fa on b is towards a
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By Symmetry
Fb on a is equal and opposite to Fa on b
 So the two wires with parallel currents
are attracted to one another
 If I reverse the current on b (antiparallel), then the forces generated by
the wires will repel one another.
 Parallel currents attract; anti-parallel
currents repel.

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Derivation of Ampere’s Law
Consider a distance R from a wire carrying current I.
Now consider all of the points which are a distance R
from the wire. They form a circle of circumference 2R.
Now we evaluate the closed loop integral at this point.
 
o I
 B  ds  B  ds  B(2R)  2R (2R)  o I
So the circle that we made is called an “Amperian Loop”.
 
 B  ds  o I enclosed
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Ampere’s Law


Ampere’s Law says that
the magnitude of B is
proportional to the net
current enclosed within
the Amperian Loop.
Amperian loops do not
have to be circles.
They could be
rectangles but a circle is
usually more convenient
to integrate. The only
rule is that they have to
be closed!
 
B

d
s


I
o
enclosed

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What is the direction of the magnetic
field?

We have an agreed convention:

Curl your fingers around the Amperian loop
with your fingers curling in the direction of
the integration. If the current is in the same
direction as your thumb, then current is
assigned a positive value. If your thumb is
opposite the direction of the current, then
the current is assigned a negative value.
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Magnetic Field of Long, Straight
Wire Revisited
Integrate from 0 to 2
R
I
 
 B  ds  o I enclosed
2
 
 B  ds  B  Rd q B(2R)   o I
0
B
o I
2R
In this case, the
enclosed current
is defined to be
in the negative
direction.
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Differential Version of Ampere’s Law

 
 
Recall    v   dA   v  ds
 

 
 
  B  dA   B  ds   o I enclosed
and
I enclosed
so


  J enclosed  dA


  B   o J enclosed
A current density creates a steady
magnetic field which circulates
around the current density
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Solenoid


Solenoid is a number of
coils packed tightly
together.
It solves an incredibly
tough problem.

Consider: It is easy to
make a uniform E-field.
All you need are two
parallel plates. But
how do you make a
uniform field out of a
bunch of circles (i.e. a
B-field).
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Finding the magnetic field of a
solenoid
  b   c   d   a  
 B  ds   B  ds   B  ds   B  ds   B  ds
a
b
c
d
 
 B  ds  BL
b
a
a 
 

B

d
s

0

B

d
s


c
b
d
 

 B  ds  0 because B  0
d
The solenoid is
characterized by the number
of turns per unit length, N
The magnetic field outside of
the solenoid is so weak that
it is consider equal to zero
c
I enclosed  iNL
so
BL   oiNL
B   oiN
20
Toroid– A circular solenoid
Consider a circular solenoid of radius, R
 
 B  ds  B(2R)
Current
into page
I enclosed  iN
so
B(2R)  oiN
Amperian
Loop
Current
out of
page
oiN
B
2R
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Bohr Magneton

In the last chapter, we discussed how atoms
could be thought of as current loops
 I=e/T where T=2r/v



v=velocity of the electron
r= radius of the electron’s orbit
If =IA then =((ev)/2r)*r2

=evr/2

Recall for an orbiting point, the angular
momentum, L, is equal to mvr
 So =eL/(2m)
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Quantization of angular momentum

Each orbital is an integer value of
Planck’s constant, h divided by 2
=h/ 2
 L=n 


=e /2m
Called the “Bohr Magneton”
 Represents smallest amount of dipole
moment possible

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Magnetic Materials
This represents any material.
The arrows indicate the
direction of the individual
magnetic moments of the
atoms. As you can see, their
orientation is random and
the vector sum of them is
zero.
This represents a permanent
magnet. The magnetic moments
or dipoles have a net vector sum
point to the left.
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Magnetic Materials in an External
Magnetic Field
Before
Let’s turn on an external B which points to the left
Possibility 1: The dipoles align
with the external field
Bexternal
Possibility 2: The dipoles align
to oppose the external field
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The degree to which the magnet moments align or oppose the
magnetic field determines their classification
Ferromagnetic—Strongly aligns with the
magnetic field
 Paramagnetic—Weakly aligns with an
external magnetic field
 Diamagnetic—Weakly opposes the
external magnetic field
 For any of these cases, we define the
total magnetic field as B=Bexternal + B

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B represents the magnetic field created by
the alignment/opposition of the dipoles

B cm *Bexternal

cm is called the magnetic “susceptibility” of the material
Paramagnetic materials and their susceptibilities





2.2 x 10-5
1.4 x 10-5
3.6 x 10-7
Diamagnetic materials and their susceptibilities




Al
Cu
Air
Bi
Ca
H2O
-1.7 x 10-5
-2.2 x 10-5
-9.1 x 10-6
Ferromagnetic materials have susceptibilities from 103 to 106
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--The Permeability of a Material
B=Bexternal + cm *Bexternal
 B=Bexternal (1+ cm
 Relative permeability, Km is defined as



Km= 1+ cm
Actually, we could replace 0 with a new
“” called the permeability of the
material.

= Km 0
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Hysteresis






Let’s say you had a ferromagnet and you measured its
magnetic field
You applied an external B-field and the dipoles are
aligned with the field
Now you remove the field but some of the dipoles get
“stuck” in their new position.
Now you measure the magnetic field of the
ferromagnet and find that it is different.
You repeat the process and yet, you never get back to
your original value as more or less of the dipoles stick
or unstick in their new positions.
This behavior is called hysteresis.
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