Download Chapter 28 Sources of Magnetic Field

Chapter 28
Sources of Magnetic Field
In this chapter we investigate the sources of magnetic of magnetic field, in particular, the magnetic field produced by moving charges (i.e., currents). Ampere’s Law
is introduced and plays a role analogous to Gauss’s Law. At the end of the chapter,
various forms of magnetism produced in materials at the atomic and particle level
are discussed (i.e., paramagnetism, diamagnetism, and ferromagnetism).
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Magnetic Field of a Moving Charge
Let’s investigate the magnetic field due to a single point charge q moving with a
constant velocity ~v . As we did for electric fields, the location of the moving charge
at a given instant will be called the source point, and the point P where we want
to calculate the magnetic field will be called the field point.
Figure 1: This figure shows the magnetic-field vectors due to a moving positive point charge q. At
~ is perpendicular to the plane of ~r and ~v . Notice that B
~ points in the direction of
each point, B
~v × ~r.
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1.1
Moving Charge: Vector Magnetic Field
~ = µo q ~v × r̂
B
4π r2
(1)
where µo = 4π × 10−7 T·m/A (exactly).
A point charge in motion also produces an electric field, with field lines pointing
radially outward from a positive charge. The magnetic field lines are completely
different. The magnetic field lines are circles centered on the line of ~v and lying in
planes perpendicular to this line.
When we study the propagation of electromagnetic waves we will see that their
speed of propagation is related to the electric permittivity and the magnetic permeability by the following relation:
c2 =
1
o µo
The Forces Between Two Moving Protons
Figure 2: This figure shows the electric and magnetic forces acting on a pair of protons moving in
opposite directions at their point of closest approach.
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F~E =
1 q2
̂
4πo r2
~ = µo q(v ı̂) × ̂ = µo qv k̂
B
4π
r2
4π r2
The velocity of the upper proton is −~v = −v î, so, the magnetic force on it is:
µo qv
µo q 2 v 2
~
~
FB = q(−~v ) × B = q(−vı̂) ×
k̂ =
̂
4π r2
4π r2
The magnetic interaction in this situation is also repulsive. The ratio of the force
magnitudes is
FB
v2
2
= o µo v = 2
FE
c
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2
Magnetic Field of a Current Element
Figure 3: This figure shows the magnetic-field vectors due to a current element d~`. The view along
the current axis is also shown.
dQ = nqA d`
The moving charges in this segment are equivalent to a single charge dQ traveling
with a velocity equal to the drift velocity ~vd .
dB =
µo |dQ| vd sin φ
µo n|q| vd A d` sin φ
=
4π
r2
4π
r2
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dB =
µo I d` sin φ
4π
r2
(2)
Current Element: Vector Magnetic Field
This can be written in a more compact vector notion:
~ =
dB
µo I d~` × r̂
4π
r2
(3)
The total magnetic field due to a current-carrying conductor is the sum of the
~ contributions:
individual dB
~ = µo
B
4π
3
Z
I × d~` × r̂
r2
(Biot-Savart Law)
(4)
Magnetic Field of a Straight Current-Carrying Conductor
Figure 4: This figure shows the magnetic field produced by a straight current-carrying conductor
of length 2a.
Using the Bio-Savart Law, we can calculate the magnetic field produced by a
straight current-carrying conductor.
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µo I
B =
4π
a
x dy
−a
(x2 + y 2 )3/2
Z
Integrating this by trigonometric substitution (y = x tan θ), we obtain:
B=
µo I
2a
√
4π x x2 + a2
(due to a finite wire of length 2a)
(5)
If we extend this solution for an infinitely long wire by taking the limit that a → ∞,
we find the following magnetic field a distance r perpendicular the wire is:
B =
µo I
2πr
(due to an infinitely long wire)
The magnetic field B has an axial symmetry about the y-axis (see Fig. 5 below)
Magnetic Field of a Single Wire
B =
µo I
2πr
(6)
Figure 5: This figure shows the magnetic-field around a long, straight, current-carrying conductor.
The field lines form concentric circles, with directions determined by the right-hand rule.
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Magnetic Field of Two Wires
What is the magnetic field resulting from two long, straight current-carrying wires?
~ total = B
~1 + B
~ 2.
We just use the principel of superposition. That is, B
Figure 6: This figure shows the magnetic-field lines for two long, straight conductors carrying equal
currents in opposite directions. The magnetic field is strongest in the region between the conductors.
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4
Force Between Parallel Conducting Wires Carrying Current
~ caused by the current in the lower conductor
Figure 7: This figure shows how the magnetic field B
exerts a force F~ on the upper conductor. The force between two parallel wires carrying current in
the same direction are attractive, and consistent with Newton’s 3rd law.
~ produced by the lower conductor is:
The magnetic field B
B =
µo I
2πr
~ × B,
~ where
The force this field exerts on a length L of the upper wire is F~ = I 0 L
~ is in the direction of the current I 0 .
the vector L
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µo II 0 L
F = I LB =
2πr
0
This can be written in terms of the force per-unit-length:
µo I I 0
F
=
L
2πr
(7)
Magnetic Forces and Defining the Ampere
One ampere is that unvarying current that, if present in each
of two parallel conducting wires of infinite length and one meter apart in empty space, causes each conductor to experience
a force of exactly 2 × 10−7 newtons per meter of length.
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Magnetic Field of a Circular Current Loop
~ produced by a current element I d~` with
Figure 8: This figure shows the magnetic-field vector dB
components resolved along the x and y axes.
~ produced by a current element
We use Eq. 3 to determine the “magnitude of dB”
I d~` and find the following:
dB =
µo I
dl
2
4π (x + a2 )
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~ vector are resolved along the x and y directions,
Next, the components of the dB
and we find the following:
dBx = dB cos θ =
µo I
dl
a
4π (x2 + a2 ) (x2 + a2 )1/2
dBy = dB sin θ =
µo I
dl
x
2
2
4π (x + a ) (x2 + a2 )1/2
Integrating dBy gives a null result. Meanwhile, integrating the x-component of dB
gives:
Z
Z
Bx =
The integration
dBx =
R
µo I
a d`
µo I
a
=
4π (x2 + a2 )3/2
4π (x2 + a2 )3/2
Z
d`
d~` = 2πa gives the follow result:
Bx =
µo Ia2
2 (x2
+
a2 )3/2
(Magnetic field on the axis)
(8)
Magnetic Field on the Axis of a Coil
Figure 9: This figure shows how to apply the right-hand rule for the direction of the magnetic field
produced on the axis of a current-carrying coil.
The above equation can be written in terms of the magnetic dipole moment µ,
where µ = N Iπa2 where N is the number of turns in the coil, and πa2 is the area
of the coil.
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Bx =
µo µ
2π (x2 + a2 )3/2
Figure 10: This figure shows the magnetic field strength along the axis of a circular coild with N
turns. When x is much larger than a, the field magnitude decreases approximately as 1/x3 .
Figure 11: This figure shows the magnetic field lines produced by the current in a circular loop.
~ field along the x-axis has the same direction as the magnetic moment of the current loop µ
The B
~.
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6
Ampère’s Law
~ through a closed surface, and it
Gauss’s law for electric fields involves the flux of E
states that the flux is equal to the total charge enclosed within the surface, divided
by o . Strictly speaking–Gauss’s law for magnetic fields does not result in a useful
relationship between magnetic fields and current distributions because it states that
~ through any closed surface is always zero, whether or not there are
the flux of B
currents within the surface.
Ampère’s law is not formulated in terms of the magnetic flux, but rather in terms
~ around a closed path:
of the line integral of B
I
~ · d~`
B
C
Ampere’s Law for a Long, Straight Conductor
In order to develop Ampère’s law, we revisit the magnetic field caused by a long,
straight conductor carrying a current I. We found in a previous section that the
magnetic field can be written as:
B =
µo I
2πr
This results in circular concentric magnetic field lines centered on the conductor
~ around a circle with radius r,
(see Fig. 7 above). If we take the line integral of B
we find the following:
I
~ · d~` =
B
I
I
Bk d` = B
C
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d` =
µo I
(2πr) = µo I
2πr
~ in the vicinity of
Figure 12: This figure shows three integration paths for the line integral of B
a long, straight conductor carrying current I “out” of the plane of the page. The wire is viewed
end-on.
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~ around a
Figure 13: This figure shows a more general integration path for the line integral of B
long, straight conductor carrying current I “out” of the plane of the page. The wire is viewed
end-on.
Ampere’s Law: General Statement
Suppose we have several long, straight conductros passing through the surface
~ at any point on the
bounded by the integration path C. The total magnetic field B
path is the vector sum of the fields produced by the individual currents. Thus the
~ equals µo times the algebraic sum of the currents.
line integral of the total B
If the integration path does not enclose a particular wire, the line integral of the
~ field due to that wire is zero. Thus we can replace I in our equation with Iencl ,
B
the algebraic sum of the currents enclosed by the integration path, with the sum
evaluated by using the sign rule.
I
~ · d~` = µo Iencl
B
C
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(9)
Figure 14: This figure shows the application of Ampere’s law for multiple currents. You must
have a high degree of symmetry in order to apply Ampere’s law. In this case, the currents are
approximated to be infinitely long.
Figure 15: This figure shows two long, straight conductors carrying equal currents in opposite
directions.
H The conductors are seen end-on, and the integration path is counterclockwise. The line
~ · d~` gets zero contribution from the upper and lower segments, a positive contribution
integral B
from the left segment, and a negative contribution from the right segment; the net integral is zero.
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7
Applications of Ampère’s Law
Ampère’s law is useful when we can exploit symmetry to evaluate the line integral
~ We illustrate this with the following examples.
of B.
Field of a Long, Straight, Current-Carrying Conducting Wire
I
~ · d~` =
B
I
Bk d` = B(2πr) = µo I
Field of a Long Cylindrical Conductor
Figure 16: This figure shows how the magnetic field can be calculated inside a conductor r < R.
The current through the gray area is (r2 /R2 )I. To find the magnetic field outside the conductor
r > R, we apply Ampere’s law to the circle enclosing the entire conductor.
Inside the long cylindrical conductor
we find that the current enclosed is (r2 /R2 )I.
H
~ · d~`, we find
Evaluating the line integral B
B =
µo I r
2π R2
(inside the conductor, r < R)
Outside the conductor, we calculate again
we calculated before:
B =
µo I
2πr
H
~ · d~` = µo Iencl , and we find the result
B
(outside the conductor, r > R)
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Figure 17: This figure shows the strength of the magnetic field inside and outside the conducting
wire carrying a current I and having radius r = R
Field of a Solenoid
A solenoid is produced by wrapping an insulated conducting wire many times (N
turns) around a form (e.g., a cylinder or toroid) in order to confine the magnetic
field to particular region of space.
Figure 18: This figure shows the magnetic field lines produced by the current in a solenoid.
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Figure 19: This figure shows how to calculate the strength of the magnetic field in a solenoid using
Ampere’s Law.
I
I
~ · d~` = B L
B
~ · d~` = B L = µo N I
B
C
B = µo n I
(solenoid)
(10)
where n = N/L, the number of turns per unit length.
Figure 20: This figure shows the magnitude of the magnetic field at points along the axis of a
solenoid with length 4a, equal to four times its radius a. The field magnitude at each end is about
half its value at the center.
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Field of a Toroidal Solenoid
Figure 21: This figure shows a toroidal solenoid with only a few turns shown. In real applications
there are many more turns. In figure (b) there are three integration paths (black circles) used to
~ set up by the current (shown as dots and crosses).
compute the magnetic field B
The figure above shows a doughnut-shaped toroidal solenoid, tightly wound with
N turns of wire carrying a current I. Let’s use Ampère’s law to find the B-field
inside and outside the toroid.
Inside the toroid the line integral (path 2) encloses N times the current I.
I
~ · d~` = N µo I
B
C
B · 2πr = N µo I
or
B =
µo N I
2π r
(toroidal solenoid)
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(11)
The Bohr Magneton
In the Bohr model of the atom, electrons move in circular orbits about the nucleus.
In some atoms, an external field can cause the current loops of valence electrons
to become oriented preferentially with the field, so their magnetic fields add to the
external field. We then say that the material is magnetized.
Figure 22: This figure illustrates how an electron moving with speed v in a circular orbit of radius
~ and an oppositely directed orbital magnetic dipole moment µ
r has an angular moment L
~ . The
~ and an oppositely directed spin magnetic dipole
electron also has a spin angular momentum S
moment µ
~ s (not shown in the figure).
~ and a magnetic dipole
The electron’s orbit produces both angular momentum L
moment µ
~.
~ = ~r × p~
L
(angular momentum)
~
µ
~ = IA
(magnetic dipole moment)
Let’s see how the magnetic dipole moment µ is related to the angular momentum L.
As the electron orbits, it produces a current I = e/T = ev/2πr.
µ = IA =
ev
evr
e
e
πr2 =
=
(mvr) =
L
2πr
2
2m
2m
Strictly speaking this should be a vector equation, so, we write:
µ
~ = −
21
e ~
L
2m
(12)
If the magnetic field defines the z direction, then
µz =
e
Lz
2m
In quantum mechanics, the angular momentum is observed to be quantized along
the z-axis in units of h-bar where ~ = h/2π = 1.054 × 10−34 J·s.
µB =
e
(~)
2m
(Bohr magneton)
where µB = 9.274 × 10−24 A·m2 .
Paramagnetism
Diamagnetism
Ferromagnetism
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(13)
Figure 23: This figure shows the magnetic domains produced in a single crystal of nickel. The
material becomes magnetized as the external magnetic field is increased. The size of the domains
change as the external magnetic field increasingly “flips” more of the dipole moments in each region
~ thus increasing the magnetization of the material.
to point in the direction of B,
Figure 24: This figure shows the magnetization curve for a ferromagnetic material. The magnetization M approaches its saturation value Msat as the magnetic filed Bo (caused by external currents)
becomes large.
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