Download Chapter 27 Sources of Magnetic Field

Chapter 27
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
ĵ
4πo r2
~ = µo q(v î) × ĵ = 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:
2 2
~ = q(−v î) × µo qv k̂ = µo q v ĵ
F~B = q(−~v ) × B
4π r2
4π r2
The magnetic interaction in this situation is also repulsive. The ratio of the force
magnitudes is
v2
FB
= o µo v 2 = 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
2πx
The magnetic field B has an axial symmetry about the y-axis.
Magnetic Field of a Single Wire
B =
µo I
2πr
(5)
Figure 5: This figure shows the magnetic-field around a long, straight, current-carrying conductor.
The field lines are 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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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.
~ 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:
F
µo I I 0
=
L
2πr
(6)
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
We use the definition of the differential magnetic field dB
element I d~`:
dB =
µo I
dl
4π (x2 + a2 )
Next, resolve the components of the dB vector along the x and y directions
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dBx = dB cos θ =
dl
a
µo I
4π (x2 + a2 ) (x2 + a2 )1/2
dBx = dB sin θ =
dl
x
µo I
4π (x2 + a2 ) (x2 + a2 )1/2
Integrating dBy gives a null result. Meanwhile, integrating the x-component of dB
gives:
Z
Bx
Z
dBx =
Bx =
a d`
µo I
a
µo I
=
3/2
4π (x2 + a2 )
4π (x2 + a2 )3/2
µo Ia2
2 (x2
+
a2 )3/2
Z
d`
(Magnetic field on the axis)
(7)
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.
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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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Ampere’s Law
Ampere’s Law for a Long, Straight Conductor
Ampere’s Law: General Statement
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Applications of Ampere’s Law
Field of a Long, Straight, Current-Carrying Conducting Wire
Field of a Long Cylindrical Conductor
Field of a Solenoid
Field of a Toroidal Solenoid
The Bohr Magneton
Paramagnetism
Diamagnetism
Ferromagnetism
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~ 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.
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.
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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.
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.
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
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Figure 18: This figure shows the magnetic field lines produced by the current in a solenoid.
Figure 19: This figure shows how to calculate the strength of the magnetic field in a solenoid using
Ampere’s Law.
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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Figure 21: This figure shows a toroidal solenoid with only a few turns shown. In parb (b) the
~ set up by the current
integration paths (black circles) are used to computer the magnetic field B
(shown as dots and crosses).
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
~
electron also has a spin angular momentum S and an oppositely directed spin magnetic dipole
moment µ
~ s (not shown in the figure).
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Figure 23: This figure shows the magnetic-field vectors due to a moving positive point charge q.
~ is perpendicular to the plane of ~r and ~v . Notice that B
~ points in the direction of
At each point, B
~v × ~r.
Figure 24: This figure shows the magnetic-field vectors due to a moving positive point charge q.
~ is perpendicular to the plane of ~r and ~v . Notice that B
~ points in the direction of
At each point, B
~v × ~r.
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