Download Chapter 30 - Sources of Magnetic Fields

Chapter 28 - Sources of Magnetic Fields
I.
Introduction - we've already looked at how moving charges are affected by magnetic fields. Now
let's take a look at the magnetic fields produced by the moving charges (current)
Two methods:
II.
1) Law of Biot-Savart
2) Ampere's Law
Magnetic Field Produced by a Moving Charge
We know that a point charge, q, produces an electric field. Does it also produce a magnetic field?
The answer is yes!
P

r


v
q
A.
B.
Observations at point P a distance r from the charge that is traveling at a velocity v:
1.
the magnetic field is proportional to the charge q, the velocity of the charge v, the reciprocal of
the distance squared 1/r2, and the sine of the angle .
2.
the direction of the magnetic field is perpendicular to the plane containing the velocity vector


 
v , and the distance vector r -- in the direction of the cross product v  r .
Expression for the magnetic field:
B
or
 o qv sin 
,
4 r 2
  qv  rˆ
B o
,
4 r 2
where o = 4 x 10-7
C.
Tm
. Note that the units of the magnetic field is Tesla.
A
Interesting note: Remember
1
N  m2
= 9 x 109
. Combine with o :
4 o
C2
28-1
II.
Law of Biot-Savart
A.
Given a wire carrying a current I. What is the magnitude and direction of the magnetic field
produced by the wire at a point P in space?
Start with a small segment of the wire and think of it as having many charges, q, flowing in it, with
each charge contributing to the magnetic field at point P.
P

r
r̂
vd

dl

I
Consider a small segment of the wire of length dl that contains a total amount of charge dq. If n is the
number of charges per unit volume, A is the cross-sectional area of the wire, and vd represents the
drift velocity of the charges, then
dq = nAqdl.
The magnetic field produced by the small segment is like the magnetic field produced by a point
charge. (Why?)
dB 
or
 o dq vd sin  o nAqdl vd sin 
,

4
4
r2
r2
dB =
Using vector notation:

dB 
which is the Law of Biot-Savart.
Note the magnitude and the direction of magnetic field produced by the small segment.
The magnetic field produced by a finite length of wire can be found by integrating over the entire
length of the wire. Remember that the integration must be performed separately for the components
of the magnetic field vectors.
28-2
B.
Examples
1.
Find the magnetic field a perpendicular
distance a away from a “long” straight
wire carrying a current I. Also, draw the
lines of force of the magnetic field.
r
a

x
dl = dx
2.
y
Two long parallel straight wires are
carrying the same current I in the same
direction. Find the magnetic field a distance
y along the perpendicular bisector of the
line connecting the two wires.
P
y

I
28-3
a
a

x
I
3.
In question 2, find the magnitude and direction of the force exerted on the left wire due to the
magnetic field produced by the right wire.
y
4.
A long, flat metal ribbon of width w carries a
current I. What is the magnitude and direction
of the magnetic field at a distance y above the
center of the ribbon?
x
I
z
28-4
w
5.
y
Find the magnitude and direction of the magnetic field on
the axis of a circular loop of wire of radius R and carrying a
current I.
y
I
R
x

dl
z
6.
A flat disk of radius R has a charge Q
uniformly distributed over its surface. The
disk rotates about a vertical axis with an
angular velocity . What is the magnitude
and direction of the magnetic field at the
center of the disk?
28-5

y
7.
Find the magnitude and direction of the
magnetic field along the axis of a solenoid wire wound very close together in a helix.
The solenoid has n turns per unit length of
the solenoid. The radius of the helix is R and
the current in the wires is I.
xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
x
O
x
R
x
dx
...............................
x1
x2
If the solenoid is closely wound and is very long (the length >> radius), the solenoid is an ideal
solenoid. In this case, the magnetic field is constant across the cross-sectional of the solenoid.
a.
Find the magnetic field inside a long solenoid.
b.
Find the magnetic field at one end of a long solenoid.
28-6
III. Ampere's Law - another way of finding the magnetic field for simple wire arrangements.
A.
Come up with an expression for Ampere's law. Consider a long
wire carrying a current I. Take a closed path around the wire
 
and calculate B  dl along the path.

B

I
d
B.
r

dl
Examples
1.
A long wire of radius R carries a current I that is uniformly
distributed throughout the cross-section of the wire. Use
Ampere's law to find the magnetic field both outside and
inside the wire.
a.
R
outside, r > R
I
b.
inside, r < R
28-7
2.
A coaxial cable with an inner conductor of radius a and an
outside conductor of inside radius b and outside radius c. The
inner conductor carries a current I in one direction and the outer
conductor carries a current I in the opposite direction. Find the
magnetic field between the two conductors (a < r < b) and
outside the outer conductor (r > c).
c
b
I
a
I
3.
x
Use Ampere's law to find the magnetic field inside an ideal (B field is uniform across the crosssection of the solenoid) solenoid with n turns of wire per unit length of the solenoid. The wires
carry a current I.
xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
x
...............................
28-8
4.
Qualitative problem - introduce Maxwell's displacement
current Id .
path
Apply Ampere's law to a path around the left wire.
a.
b.
The current passes through the areas A1 and A2 whose
boundary is the path.
Is there a resolution to the problem?
28-9
I
I
A1
A2
IV. Magnetism in Matter - the magnetic effects in matter are due to magnetic moments. We already
know that a current loop has a magnetic moment, and that a current loop is simply charge that is
traveling in a circular path. So anytime we have charge that is traveling in a circular path, we have a
magnetic moment. There are three charge motions that contribute to the magnetic moment of an
atom.
A.
Orbital motion of the electron. The motion of the electron around the nucleus of an atom constitutes
a current loop and, therefore, has a magnetic moment,  . It also has an angular momentum, L.

L
e


B.
Spin of the electron - the electron can be thought of as spinning around an axis (a
classical description) and this motion constitutes charge traveling around a
circular path. Therefore, there is a magnetic moment associated with the spin.

s
electron


C.
Spin of the nucleons - again the nucleons can be thought of as a charged objects spinning, and they
also have magnetic moments
D.
The primary magnetic effects of atoms is a result of the spin magnetic moment of the electron. The
orbital magnetic moments generally cancel, and the spin magnetic moments may or may not pair off
and cancel. Generally, those atoms with an even number of electrons tend to have no magnetic
moment while those atoms with an odd number of electrons have a net magnetic moment.
28-10
V.
Magnetic Properties
A.
Examine two solenoids, one without any material inside and one with a material inside.
without material
with material
current I
current I
xxxxxxxxxxxxxxxxxxxxxxxxxxx
xxxxxxxxxxxxxxxxxxxxxxxxxx
x
Bo
Bo
Bm
. . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . .
. .

The total magnetic field inside a solenoid (without or with a material) is written as B , where
 
  
B  Bo withouta material, and B  Bo  Bm with a materal.
B.
C.
D.

Bo is the magnetic field due to the current in the wires of the solenoid. It can be written as



Bo  o H , where H = the magnetic field strength in A/m and is dependent on the current in the
wires (e.g., remember for a solenoid B = onI = oH, hence H = nI).



Bm is the magnetic field contribution due to the material and can be written as o M, whereM is
called the magnetization of the material and equals sum of the magnetic moments per unit volume

 
 
M  
.


volume


Therefore, in general,



B  o H  o M .
E.

The magnetization M should also be dependent on the magnetic field produced by the wires of the
  
solenoid because the magnetic field will tend to align the magnetic moments (remember     B ).




Therefore, M is dependent on the magnetic intensity H , and we can then write M  H. The total
magnetic field in the solenoid can be rewritten as





B  o H  oH  o 1  H  H .
The constants are  = magnetic susceptibility and  = permeability of the material. A third constant
called the relative permeability is defined as
m 
B
 1  .
Bo
This is the magnetic equivalent of the electrical dielectric constant. Some values of the magnetic
susceptibility and the relative permeability are given in the table below:
28-11
material
vacuum
aluminum
copper
iron

0
2.3 x 10-5
-9.8 x 10-6
10 to 100,000
m =
B

Bo
1
1.000023
0.9999902
10 to 100,000
Note how large (or small) the field with the material is in comparison to the field without the
material.
VI. Types of Magnetic Materials - paramagnetic, diamagnetic, ferromagnetic materials
A.
Paramagnetic materials exhibit paramagnetism. These are materials where m > 1, or the total
magnetic field is greater than the magnetic field due to just the solenoid current (B > Bo). This
means that the material produces a small magnetic field that is in the same direction as the applied
field.
1.
Examples: aluminum, lithium, platinum, etc. All these atoms have a net magnetic moment.
2.
Apply an external magnetic field to a paramagnetic material:
a.
The magnetic torque tends to align magnetic moments with the external field. This
alignment causes an increase in the total magnetic field.
b.
We expect to get all the magnetic moments aligned and as a result have a large increase in
the magnetic field. However, thermal effects prevent this from occurring. Curie's Law
describes the magnetization as a function of the absolute temperature.
28-12
B.
Ferromagnetic materials exhibit ferromagnetism. These are materials where m >> 1, which means
that these materials produce a very large increase in the total magnetic field (B >> Bo ).
1.
Examples: iron, cobalt, nickel, etc. The atoms in these materials have a magnetic moment, but
the strong effect is due to many of these atoms forming "domains" where their magnetic
moments all point in the same direction (by "exchange coupling"). The size of these "domains"
is about 10-5 cm.
2.
Ferromagnetic materials that are not yet magnetized have domains where their magnetic
moments point in random directions.
3.
When placed in an external magnetic field, the magnetic moments of the domains tend to line
up with the external magnetic field by domain growth and domain rotation.
no external field, Bo = 0
small external field
Bo
stronger external field
Bo
domain growth
domain rotation
Mechanism of domain growth: What is happening at the boundary?
no BO
field
Bo
field
original
boundary
new
boundary
28-13
4.
Magnetization M vs Magnetic Intensity H (which is proportional to the current I )
M
H
M, Bo , B, m
H
5.
Hysteresis: Place a hunk of ferromagnetic material in a solenoid. Gradually increase the
current in the solenoid to some value, then gradually decrease the current back to zero. Then
gradually increase the current again, but in the opposite direction (this reverses the B field in
the solenoid). Then decrease it back to zero. This process can be repeated over and over again.
B
H
28-14
The changing direction of the magnetic field changes the direction of the magnetic moments in
the material and thus, requires energy. This energy is converted into heat energy and is directly
related to the area of the “hysteresis” curve.
Problem: Draw a hyteresis curve for a material that 1) could be a good candidate for a
permanent magnet, 2) might make a strong permanent magnet but can be easily demagnetized,
3) can produce a magnetic field (even though it may be weak) but cannot be easily
demagnetized, and 4) minimizes the losses of energy in a transformer.
1)
2)
3)
4)
C.. Diamagnetic materials exhibit diamagnetism. These are materials where m < 1, or the total
magnetic field is less than the field produced by the solenoid current (B < Bo ). This means that
these materials produce a small magnetic field that is in the opposite direction to the applied field.
1.
Examples: copper, gold, mercury, etc. These atoms do not have a net magnetic moment.
2.
When placed in an external magnetic field, a weak magnetic moment is set up in the opposite
direction to the applied field, thus resulting in a total field that is less than the field due just to
the current. The effect, however, is very small and is masked by paramagnetic and
ferromagnetic effects (B = Bo - Bm ). The oppositely directed magnetic field is an induced
magnetic field (Faraday’s Law).
VIII. Gauss's Law for Magnetism - Reprise
S
N
28-15