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Notes on Magnetic Field and Magnetic Force
I am teaching this material in historical order, which means magnets first (Sects. 20.1 and 20.11),
then electromagnetism of circuits (Sects. 20.5-20.10), and lastly the electromagnetism of individual
moving charges (Sects. 20.2-20.4). Charges moving perpendicularly to a magnetic field B go in
circles whose radius R is proportional to their momentum over B; from this the mass of ions and
charged molecules can be determined. Before getting to the magnets notes, here is a brief summary.
1. Magnets were studied by Coulomb around 1785. The bulk of the notes below are about magnets.
Sects. 20.1 and 20.11 of the text are about magnets and magnetic materials.
2. In 1819 Oersted discovered that an electric current I can deflect a compass needle. We interpret
~ that circulates around the magnet according to Oersted’s rightthis in terms of a magnetic field B
hand-rule (see text Fig. 20.34).
~ as a function of I and of
3. Around 1820 Biot (“bee-oh”) and Savart (“s-a-var”) studied B = |B|
distance r from the axis of the wire. In modern language they found that the magnitude of the
field is proportional to I and grows inversely with r:
2km I
.
text(20.10)
r
4. Around 1821 Ampere found that a magnet can deflect a wire. In modern language he found
that if the magnet produces a uniform field of magnitude B then (a) the maximum force on a wire
of length l and carrying current I is given by
B=
Fmax = IlB,
text(20.6)
(b) that this occurs when the wire l and the field B are perpendicular, (c) that the direction of the
force F~ is normal to both l (or I) and B with sense given by text Figs. 20-23 and 20.24, (d) that
if l and B make an angle θ relative to one another, then
F = Fmax sin θ.
text(20.7)
5. Around 1821 Ampere found that a magnetic disk of magnetic moment µ, areas A and thickness
d (see below) produced and felt magnetic effects as if it was a current loop of the same dimension,
provided that
µ = IA.
text(20.9)
6. Also around Ampere also showed that the torque τ on such a loop has the same direction as
that on a magnet with the same µ and has a magnitude given by
τ = µB sin θ.
text(20.8, 20.9)
Items 2-5 are the basis of Sects. 20.5-20.10. You also must learn the directional rules.
Faraday around 1830 had the idea for anions and cations moving through voltaic cells and other
chemical solutions. He began to study charge flow in gases. But not until around 1870 or 1880 were
vacuum pumps good enough for electric currents flowing in “cathode ray tubes” to be noticeably
deflected by magnetic fields. By that time physicists knew that a wire of length l carrying N
identical charges q at velocity v would (1) cross the wire in time ∆t = l/v, (2) carry a charge
∆Q = N q, (3) carry a current I = ∆Q/∆t = N q/(l/v) = N qv/l, (4) feel a maximum force
of Fmax = IiB = N qvB. Therefore a single charge N = 1 would feel a maximum force of
Fmax = qvB. This occurs when v and B are perpendicular. When v and B make an angle θ
relative to one another, then
F = qvB sin θ.
text(20.1)
The direction of F~ is given in Figs. 20.9-11. This equation is the basis of Sects. 20.2-20.4.
In the magnets pages that follow, the [Optional] subsections and examples will not be discussed
in lecture and will not be tested on in the exams.
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1.1
Magnets
Inside a Magnet There Really Are No Poles
No one has yet observed any isolated magnetic pole: the sum of the pole strengths distributed on
any magnet sums to zero. Break a magnet into two or more pieces, and each piece will have zero
net pole strength, poles seeming to appear at the point of the break. See Fig. 1. Place the pieces
near one another, and they will spontaneously attract, tending to resume the original shape, with
the poles at the break cancelling one another.
N
N
S
before
S
N
S
after
Figure 1: Effect of breaking a bar magnet. (a) Before breaking it, it appears to have two poles.
(b) After breaking it, there appear to be two smaller magnets, each with the same pole strength as
the original magnet.
1.2
Outside a Magnet We Can Use Magnetic Charge (Poles)
It is an experimental fact that outside a magnet its magnetic properties can be obtained by treating
it as if it contained a distribution of magnetic poles, or charges, whose numerical value summed
over the magnet is zero. Moreover, these poles cannot be transferred from one magnet to another.
From Coulomb’s work on long, narrow, bar magnets, the interaction between two poles varies
as the inverse square of their separation. We choose units of magnetic charge, or pole strength,
µ0
using coupling constant (analogous to k of electricity) km ≡ 4π
= 1.0 × 10−7 N/A2 , exactly. Here
µ0 is called the permeability of free space. With this set of units, Table 1 summarizes the basic
correspondences between electric and magnetic poles. Example 1 shows how the units come about.
Quantity
Electricity
Magnetism
Charge
Field
q (C)
~ (N/C=V/m)
E
qm (A-m)
~ (T=N/A-m)
B
Force
~ (N)
qE
~ (N)
qm B
Coupling Constant
µ0
1
k ≡ 4π
≈ 9.0 × 109 N-m2 /C2 km ≡ 4π
= 1.0 × 10−7 N/A2
0
k|q|
k
|q
~ = 2
~ = m 2m|
|E|
|B|
r
r
Point Source
Sheet Source
σ(C/m2 )
~ = 2πk|σ|
|E|
σm (A/m)
~ = 2πkm |σm |
|B|
Dipole Moment
p = ql
µ = qm l
Charge/Area
Dipole Moment/Volume
Polarization P = p/V
~
Torque on Dipole in Field p~ × E
Magnetization M = µ/V
~
µ
~ ×B
Table 1: Electric Charge and Magnetic Pole Equivalences
A. Force Between Magnetic Poles
The magnetic poles of a magnet typically are not concentrated only at the ends; we can verify this
by sprinkling iron filings onto a piece of paper that covers the magnet. Nevertheless, for simplicity,
we will make this idealization. Thus, consider the force F~ on a pole of pole strength qm at position
~r, due to another pole qm1 at ~r1 , as in Fig.1.
2
R1 = r - r1
q m1
qm
r1
r
origin
Figure 2: Two idealized magnets and their nearest poles.
By Table 1, F~ satisfies a rule analogous to Coulomb’s Law, (3.4), for electric charges. That is,
km qm qm1
~ 1 = ~r − ~r1 )
F~ =
R̂1 .
(Force on pole qm , toward R̂1 , R
(10.1)
R12
Note: The magnetic poles of a magnet typically are not concentrated only at the ends; we can
verify this by sprinkling iron filings onto a piece of paper that covers the magnet. Nevertheless, for
simplicity, we will make this idealization.
Example 1: Interaction of Two Identical Long Bar Magnets. Estimating qm .
Consider two bar magnets with length l and cross-sectional area A, and with the same pole
√ strength
qm (determined by their ability to pick up the same number of nails, etc.). Let l A, so the
magnets are very long.
Q: (a) Give an equation for estimating the force between the magnets when they are collinear,
√
with two like poles so close that they provide the dominant force, but further
√ away than A. See
Fig.2. (b) For l = 20 cm= 0.2 m, and a square cross-section of side a = A = 0.5 cm= 0.005 m,
let |F~ | = 0.34 × 10−1 N when r = 0.04 m. Estimate qm .
S
l
−qm
N
r
qm
N
qm
S
l
− qm
Figure 3: Two collinear bar magnets, at a separation where their interaction is dominated by the
nearest poles, and those poles can be treated approximately as points.
A: (a) If the separation r between the two nearest poles greatly exceeds the width of the poles,
then it is a good approximation to consider only the interaction between these poles, treated as
points. Hence, from (10.1), with qm1 = qm ,
|F~ | ≈
km (qm )2
.
r2
(equal strength poles interacting)
(10.2)
(b) Solution of (10.2) yields qm = r(|F~ |/km )1/2 = 23.3 A-m. Including the forces between the other
poles (two attractive interactions at separation r + l ≈ l, and a repulsive interaction at separation
r + 2l ≈ 2l) would give a more accurate estimate for qm .
B. Measuring the Magnetic Field
~ is F~ = q E,
~ so for magnetism the
Just as for electricity the force on a charge q in an electric field E
force on a magnetic pole qm in a magnetic B-field is
~
F~ = qm B.
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(10.3)
~ with units of
Using this equation, with a known pole strength qm , a measurement of F~ gives B,
tesla (T). Example 2 shows that T = N/A-m. Hence the unit of magnetic charge has units of
A-m = N/T.
Example 2: Q: Find the field due to one pole at the other, in the previous example.
~ = |F~ /qm | = 0.34 × 10−1 N/(23.3 A-m)=1.46 × 10−3 T.
A: By (10.3), |B|
C. Field of a Monopole
Comparing (10.1) and (10.3), the field set up at ~r by qm1 at ~r1 is given by
~ = km qm1 R̂1 .
B
R12
~ 1 = ~r − ~r1 )
(R̂1 to observation point, R
(10.4)
~ the magnetic field B
~ points away from positive (north) poles,
See Fig.1. As for the electric field E,
and towards negative (south) poles. See Fig.3.
−
+
(a) North Pole
(b)
South Pole
Figure 4: Magnetic field direction and sign of poles. (a) Positive poles make outward magnetic
field. (b) Negative poles make inward magnetic field.
D. Field on Axis of Magnetic Dipole [Optional]
Consider a magnet of length l and pole strengths ±qm . It is called a magnetic dipole. Let us find
the field on its axis, a distance r from its center, as in Fig. 4a. By (10.4), at the center of the
compass needle the magnetic field due to the magnet points along its axis x̂, with magnitude due
to both qm and −qm , given by
Bmag =
k m qm
km (−qm )
(r + l/2)2 − (r − l/2)2
2km qm rl
+
= km qm
= 2
.
2
2
(r − l/2)
(r + l/2)
(r2 − l2 /4)2
(r − l2 /4)2
−qm
qm
BEh
(10.5)
θ
B
r
l
(a)
(b)
Bmag
Figure 5: (a) Compass needle along axis of bar magnet. (b) Decomposition of total magnetic field
acting on the compass needle into its components due to the earth and due to the bar magnet.
Effect of a bar magnet on the local magnetic field. Without the bar magnet, the needle would point
toward the top of the page.
Example 3: A Magnet Can Disturb A Compass Reading. [Optional]
Consider a magnet of the same material as in the previous example, so qm = 23.3 A-m, but let its
length be only l = 1 cm. Let the earth’s field point along true north and, as shown in Fig. 4a, let
the the magnet be 20 cm west of, and point toward, the compass needle.
Q: Determine by how much the magnet disturbs the orientation of a compass needle.
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A: Consider each pole separately, so that (10.5) applies. Let the in-plane (horizontal) component
of the earth’s magnetic field be BEh , and let it point along ŷ. Since r = 20.5 cm, Eq. (10.5) gives
Bmag = 0.541 × 10−5 T. The needle aligns along the direction of the net magnetic field (see Fig.
4b). Hence this field will cause the needle to rotate from ŷ toward x̂ by an angle θ satisfying
tanθ =
Bmag
.
BEh
(10.6)
Taking BEh = 5.0 × 10−5 T, this gives tanθ = 0.116, corresponding to an angular deflection of 6.7
degrees! In centuries past, no wonder sailors were warned to keep magnets and iron from the vicinity
of the navigator’s compass. Today, global navigation systems employing satellites has made the navigator’s compass obsolete; however, magnets don’t stop responding if the satellites stop functioning.
E. Magnetic Dipole Moment µ
In the numerator of Eq. (10.5) the combination qm l appears. It is a very useful quantity called the
magnetic dipole moment
µ = qm l.
(10.7)
The vector µ
~ points from the negative to the positive pole. It has units of A-m2 . See Fig. 5a. Some
field lines for it are drawn in Fig. 5b.
−qm
+qm
S
µ
N
S
A
N
l
(b)
(a)
Figure 6: Bar magnet. (a) Relationship between poles and direction of magnetic moment. (b)
Field-line pattern.
Example 4: The Distant Magnetic Field Along the Axis of a Magnet. [Optional]
Consider (10.5) when r l, so r/(r2 − l2 /4)2 ≈ r/r4 = 1/r3 . Then, using (10.7), Eq. (10.5)
becomes
2km µ
Bmag ≈
.
(on axis)
(10.8)
r3
The field of a magnetic dipole falls off with distance as r−3 , just as for an electric dipole. Applied
to example 3, where qm = 23.3 A-m, l = 1 cm, (10.7) yields µ = 0.233 A-m2 . With r = 20.5 cm,
(10.8) then yields 5.41 × 10−6 T, in three-decimal place agreement with the exact calculation of
example 3.
~
Example 5: Suspended Magnet in a Horizontal B-field.
[Optional]
Consider a magnet of mass m, length l and magnetic moment µ
~ , where µ = |~
µ|. It is suspended
from the ceiling by a string, its N -pole toward ground, due to gravity. A horizontal magnetic field
~ is now applied, where B = |B|.
~ See Fig. 6.
B
Q: Find the condition that determines the equilibrium angle.
~ at angle θ with respect to the vertical. Since
A: The field acts to make µ
~ point rightward, along B,
~
~ is counterclockwise, and of magnitude
the angle between µ
~ and B is π/2 − θ, the torque from µ
~ ×B
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T
S
µ
θ
(π/2)−θ
N
B
mg
Figure 7: A bar magnet suspended by a string in the earth’s gravity and a uniform horizontal
magnetic field.
µBsin(π/2 − θ) = µBcosθ. By taking the torque with respect to the point of contact with the
string, the torque from the string tension T can be neglected. However, the torque from gravity is
clockwise, and of magnitude mg(l/2)sinθ. When the magnet is in equilibrium, these two torques
are equal, or
µBcosθ = mg(l/2)sinθ,
leading to a condition that enables us to deduce B if we know the properties of the magnet:
tanθ =
1.3
2µB
.
mgl
Magnetization and Magnetic Dipole Moment
~.
A magnet is characterized by its magnetization M
~ is Magnetic Dipole Moment per Unit Volume
A. Magnetization M
~ , where
The magnetic dipole moment per unit volume is called the magnetization M
~
~ = µ
M
.
V
(10.9)
~ has units of magnetic pole strength per unit area, or N/(T-m2 )=A/m. Because M
~ is independent
M
of the volume of the system, for some purposes it is a more fundamental quantity than µ
~.
B. “Magnetic Charge” Per Unit Area σ Equals Magnetization M
~ |, length l and cross-sectional area A (so
Consider a bar magnet of uniform magnetization M = |M
its volume V = Al), as in Fig. 6a. By (10.9) it has
µ = M V = M Al.
(10.10)
Comparison of (10.7) and (10.10) yields the length-independent “magnetic charge,” or pole strength,
qm =
µ
MV
=
= M A.
l
l
(10.11)
Hence the charge densities on the surfaces are ±σm = ±qm /A. Use of (10.11) then yields
σm =
qm
= M.
A
(“magnetic charge” per unit area)
Example 6: From Magnetic Moment to Magnetization.
Consider the bar magnets of example 1 (qm = 23.3 A-m, A = 0.25 × 105 A/m, l=0.2 m).
Q: Find their magnetic moment and magnetization.
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(10.12)
A: By (10.7) µ = qm l=(23.3 A-m)(0.2 m)=4.66 A-m2 . By (10.11), M = qm /A = 9.32 × 105 A/m.
This is slightly less than for the alloy Alnico V (often used for loudspeakers).
C. Magnetic Field due to a Sheet of “Magnetic Charge” qm
~ by B,
~ and q
In the analogy between electricity charges and magnetic poles we replace k by km , E
~ = 2πkσ, has
by qm . Hence the electric field due to a sheet of uniform charge per unit area σ, |E|
as its magnetic analog,
~ = 2πkm σm .
|B|
(field due to sheet of “magnetic charge”)
(10.13)
By (10.12), this may be rewritten as
~ = 2πkm M.
|B|
(10.14)
~ Just Outside the End of a Long Magnet of Alnico V. [Optional]
Example 6: B
Consider a long bar magnet made of the alloy Alnico V, of length l = 20 cm and square cross-section
of area A = 1 cm2 . Let its M be along its axis. See Fig. 6a. At room temperature in zero applied
field, Alnico V has M = 9.95 × 105 A/m.
Q: (a) Estimate the field on the axis of the magnet, 1 mm outside the north pole (treated as a
sheet of magnetic pole), and (b) find the magnetic field of the distant pole.
A: (a) At 1 mm outside the North pole, which is nearly 20 cm from the South pole, the total
magnetic field is dominated
by the contribution from the North pole. Moreover, since the pole faces
√
each have width a = A = 1 cm, which is much larger than 1 mm, the North pole appears to be a
sheet of magnetic charge density σm = Mr = 9.95×105 A/m. From (10.13), B = 2πkm σm = .625 T,
which is more than 10,000 times the Earth’s magnetic field. (b) By (10.4), the magnetic field of the
distant pole of this bar magnet is km qm /r2 = km σm A/l2 = 2.49×10−4 T, which adds a contribution
much smaller than that due to the near pole, but about five times that of the earth.
D: Lifting Strength of a Magnet – Parallel Plate Capacitor Analogy [Optional]
Consider a permanent magnet of magnetization M along its axis. When it is placed against the
surface of a refrigerator (a “soft” magnet), it is attracted by a force we will call its lifting strength.
Only if we pull harder than the lifting strength will the magnet come off. It can be estimated as
follows. When a permanent magnet is brought up to a refrigerator door, the refrigerator responds
as if it had an image magnet of the opposite polarity within. See Fig. 8. Hence there is an
attraction between the permanent magnet and the refrigerator. (The dominant effect arises from
the near end of the magnet and its image.) However, if you look inside the refrigerator there will
be, perhaps, a tuna sandwich, but certainly no image magnet. This attraction is similar to what
happens with the amber effect, discussed in Sect.2.2; a more precise analogy is to a small electric
charge that is brought up to a large sheet of electrical conductor.
"soft" iron
N
S
N
S
image magnet
Figure 8: Permanent magnet in shape of a bar, placed very near the face of a large “soft” iron
magnet. This is related to the geometry of a magnet on a refrigerator. The “soft” iron responds
as if there were an image bar magnet of opposite polarity and (nearly) equal strength.
Let the magnet have its N pole, a flat surface of area A, against the refrigerator. Then there is
an image magnet (a S pole) just against the magnet; for Fig. 8 consider that the gap between the
actual N pole and the image S pole is very small. This geometry is just like that for two nearby
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capacitor plates. (Here the poles can’t move, so contact can’t cause discharge). Neglecting edge
effects, the magnetic field acting on the image pole, due to the actual magnet, is given by (10.14).
Use of (10.3), (10.12) and (10.14) yields
~ = (σm A)(2πkm M ) = (2πkm )M 2 A.
|F~ | = | − qm ||B|
(lifting strength)
(10.15)
For the Alnico V magnet of the previous example, qm = σm A = (9.95 × 105 A/m)(10−4 m2 ) =
99.5 A-m, giving a force |F~ | = (99.5A − m)(.625T) = 62.2 N. In terms of the force per unit area, this
amounts to F/A = 62.2 N/10−4 m2 = 6.22 × 105 N/m2 , or about six times atmospheric pressure.
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