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Chapter 4
Magnetic Forces
4.1.
4.2.
4.3.
4.4.
The Magnetostatic and the Magnetic Induction Forces
Magnetic Forces on Moving Charged Particles
Magnetic Forces on Charged Particles bound in Moving Wires
Magnetic Forces on Current-Carrying Wires
A charged particle moving in the presence of a current will experience a force from the current.
This kind of electromagnetic force is called the magnetic force. The magnetic force can be
expressed in terms of the magnetic vector potential or the magnetic field discussed in the electric
induction force. It is seen that the magnetic force is always perpendicular both to the particles’s
velocity and the magnetic field. Thus the magnetic field can be used to deflect moving particles,
but not to change their speed. If the charged particles are bound in a thin conducting wire which
in turn is moving in a magnetic field, the magnetic force will exert a force on the charged particles
and tends to make them move along the wire to form a current. This phenomenon can be used
to implement a generator. Of the magnetic force exerted on the mobile charged particles bound
by a wire, the component transverse to the wire will transfer to the wire. This component will be
canceled by the corresponding component exerted on the ions forming the matrix of the wire, unless
when the wire carries a current such that the velocity of the mobile charges particles is different from
that of the immobile ions. Thus a wire carrying a current can acquire momentum from magnetic
forces. This phenomenon can be used to implement a motor.
4.1. The Magnetostatic and the Magnetic Induction Forces
Ordinarily, the mobile charged particles forming a conduction current drift slowly in a matrix (such as a metal wire) which tends to neutralize the charge. Consider a time-invariant conduction or magnetization current which is completely neutralized in electricity. The completeneutralization condition implies ∂ρn /∂t = 0, which in turn implies the validity of the magnetostatic condition.
Under the complete-neutralization condition, the magnetic force due to a time-invariant
current element of current I and directed length dl exerted on a particle of charge q moving at a
velocity v is postulated to be
F = qI
n
o
1
1
−
R̂(v
·
dl)
+
dl(v
·
R̂)
.
c2 ²0 4πR2
According to Quantum Electromagnetics, magnetic forces depend on the relative velocity between
the effector and the mobile electrons or the ions forming the matrix. The net contribution is then
related to the difference between the two relative velocities which in turn is given by the current I
or by the drift speed of the mobile electrons with respect to the ions or the matrix. Moreover, as
the drift speed is quite low, the velocity v is then can be taken as the relative velocity between the
effector and the neutralizing matrix carrying the current I.
Note that the factor q/²0 4πR2 also appears in the electrostatic force and the factor q/²0 c2 in
the electric induction force. The first part of the magnetic force, as well as the electrostatic force, is
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along the radial direction R̂ and is called the magnetostatic force. However, the force is attractive
when qv is parallel to the current element Idl. The second part of the magnetic force, as well as the
electric induction force, is along the direction of the current element and is called the magnetic
induction force. Note that the magnetic force F is always lies on the plane formed by dl and R̂
and that the component of v not parallel to this plane does not contribute to the magnetic force.
By using the vector identity a × (b × c) = b(a · c) − c(a · b), the magnetostatic force and the
magnetic induction force can be combined as
F = qIµ0
1
v × (dl × R̂).
4πR2
Note that the magnetic force F is always perpendicular to the velocity v. Thus the magnetic force
cannot change the speed of each individual charged particle, but just changes its traveling direction.
As a consequence, the magnetic force does no work on an individual charged particle.
The magnetic force due to the current in a thin wire is given by superposition. If the wire
forms a loop C and carries a uniform current I, the magnetic force is then given by the integral
qIµ0 I 1
F=−
v × (R̂ × dl),
4π C R2
For the ordinary case where the wire is stationary or moves as a whole uniformly and hence the
relative velocity v is a constant over the wire and hence can be taken out of the integral. Thereby,
the magnetic force is given by
I
1
qIµ0
F=−
v×
R̂ × dl,
4π
C R2
However, for some cases, such as the one where the wire is rotating, the velocity v may not remain
uniform over the wire and hence the preceding formula is no longer valid. It is noted that the
involved path integral (times −Iµ0 /4π) is just the magnetic flux density B given by the Biot-Savart
law.
By using more vector identities ∇(1/R) = −R̂(1/R2 ) and ∇ × (f g) = ∇f × g + f ∇ × g, the
magnetic force due to a current element Idl can be rewritten as
·
µ
¶
¸
"
Ã
1
1
1
Idl
F = qIµ0 v× ∇
× dl = qµ0 v× ∇ ×
4π
R
4π
R
!#
.
In writing this force formula, it is understood that the del ∇ operates on the field point r in the
distance R, but not on the current element Idl.
Due to the current of density J contained in a small volume 4v located at r0 , the magnetic
force exerted on a particle of charge q and velocity v and located at r is then given by
"
#
J(r0 )
F(r) = qµ0 v× ∇ ×
4v .
4πR
For a distribution of current of density J, the force is then given by superposition as
Z
F(r) = qµ0
v × ∇×
J(r0 ) 0
dv .
4πR
The del operator can be taken out of the integral. Further, if the velocity v is a constant with
respect to the various current elements, it can also be taken out. Thereby, the magnetic force can
be given by
Z
F(r) = qµ0 v × ∇ × G(R)J(r0 )dv 0 .
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This formula represents the general form of the magnetic force with a stationary or uniformly
moving matrix.
It is seen that the magnetic force is also related to the magnetic vector potential A encountered
in the electric induction force. That is, the magnetic force can be rewritten as
F(r) = qv × ∇ × A(r).
Further, by using the explicit definition B = ∇ × A discussed in the electric induction force, the
magnetic force can be written in terms of field B as
F(r) = qv × B(r).
Thus the magnetic vector potential A and the magnetic flux density B are involved in the electric
induction force, the magnetostatic force, and the magnetic induction force.
When the electric field as well as the magnetic field is present, the force exerted on a particle
of charge q and velocity v is then given by
F(r) = q {E(r) + v × B(r)} ,
which is known as the Lorentz force law first formulated in exactly the present form in 1895,
about thirty years after the introduction of Maxwell’s equations. In terms of the scalar potential Φ
and the vector potential A, the Lorentz force law can be rewritten in the form
(
)
∂
F(r) = q −∇Φ − A + v × ∇ × A .
∂t
In Quantum Electromagnetics, it has been shown that the magnetostatic force is a modification
of the electrostatic force and the magnetic induction force is one form of the electric induction force.
Further, it has been shown that the magnetic vector potential is a modification of the electric scalar
potential.
4.2. Magnetic Forces on Moving Charged Particles
In this section we discuss the magnetic force exerted on moving charged particles. The phenomena discussed include the mass spectrometer, the magnetic deflection, the cyclotron, and the
Hall effect. Recall that the magnetic force F exerted on a moving charged particle is always perpendicular to the velocity v. Thus the magnetic force cannot change the speed of each individual
charged particle, but just deflects it.
Mass spectrometer
A particle of charge q and mass m can be accelerated by the electrostaticqforce by using a
voltage V . If the particle is stationary initially, the particle will gain a speed v = 2qV /m after the
acceleration. Then the particles are sent to a magnetic perpendicular to their path. These particles
will experience a magnetic force perpendicular both to the traveling path and the magnetic field.
Thereby, the particles tend to move around a circular path of radius r with a centrifugal force
F = mv 2 /r. The centrifugal force is counteracted by the magnetic force F = qvB.
Thus the mass m is related to radius r as
m=
qr2 B 2
qrB
=
.
v
2V
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For given q, B, and V , the radius r depends on the mass m. This mechanism can be used to
measure the mass m of a particle and has been used in isotope separation, where isotopes are
elements of the same atomic number but of different atomic weight. By using this
q mechanism, it
has been found by Bucherer in 1909 that the mass m increased by a factor of 1/ 1 − v 2 /c2 as v is
close to c.
Magnetic deflection
In a CRT (cathode-ray tube), the electron beam is accelerated by the electrostatic force due to
a voltage V . The electrostatic force can also be used to deflect the electron beam in the scanning
for forming a picture on a screen as discussed previously. In addition to the electrostatic force, the
electrons beam can also be deflected by using the magnetic force. Such a device is known as the
magnetic deflection yoke. As the accelerated electrons enter into the deflecting magnetic field, they
tend to travel around a circle of radius r given by
q
r=
mv
=
qB
2mV /q
B
.
It is seen that the radius is inversely proportional to the magnetic field.
Suppose the depth within which the particles are deflected by the magnetic field is w and
the depth from the screen to the far end of the magnetic field is D. Within the magnetic field the
particles move circularly. After exiting from the deflecting field, the electrons travel along a straight
line. The angle with which the particles travel in circular motion is known as the deflection angle
α and is given by
µ ¶
w
α = sin−1
.
r
In circular motion, the deflection d1 is associated with the deflection angle α as d1 = r(1 − cos α).
Thereafter, in the linear motion, the deflection d2 is also associated with the deflection angle α as
d2 = (D − w) tan α. The total deflection d is given by d = d1 + d2 .
It can be shown that deflection d depends on the deflecting field B, the ratio q/m, the accelerating voltage V , and on the depths w and D. The magnetic field can be produced by coils and
then its magnitude can be varied electronically.
Cyclotron
The magnetic force can also be used in cyclotron to confine charged particles which are subjecting to repeated acceleration within a more compact space. The cyclotron frequency ωc is defined as
2π divided by the time for the particle to circulate once in the magnetic field and thus is given by
ωc =
v
qB
=
,
r
m
where we have made used of r = mv/qB. Note that the cyclotron frequency is independent of the
radius r and linear velocity v, if the magnetic field B is a constant independent of radius r and if
the mass m is a constant independent of speed v.
The particles are accelerated by the electrostatic force due to a voltage applied across the gap
between two adjacent electrodes placed in the magnetic field. The accelerating electric field at the
gap is along the azimuthal direction and is perpendicular to the confining magnetic field.
The accelerating voltage cannot be static, since the electrostatic force is conservative and the
corresponding total work done on a charged particle after each circulation is zero. In other words,
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the electrostatic field due to a static voltage will eventually decelerate the same particles after they
exit the accelerating region between the plates. As a consequence, the accelerating voltage must
vary in time.
Further, in order to make the repeated acceleration constructive, the voltage should keep
synchronism with the circular motion of particles. That is, the voltage across the gap should vary
repeatedly at a frequency equal to the cyclotron frequency ωc . After many revolution of acceleration,
the particle can exit with a high energy.
Hall effect
Consider a semiconductor bar placed horizontally in the presence of a vertical (z-directed)
static magnetic field. The magnetic field tends to affect the particles moving in the semiconductor.
When the mobile particles are moving in the x direction along the bar, a y-directed magnetic force
is exerted on the charged particles. Then some of the mobile charges pile up on one side of the bar;
and a corresponding amount of the immobile charges of the opposite sign are left on the other side.
Thus an electrostatic force develops within the semiconductor bar due to the charge separation.
At equilibrium, the balance between the magnetic force and the electrostatic force leads to
F = ρv (E + v × B) = 0,
where ρv is the charge density of the mobile particles of velocity v.
It is known that the mobile charge density and the particles velocity v are related to the
corresponding current density J as J = ρv v. Then the electric field and the magnetic field exerted
on moving particles are related as
1
Ey = Jx Bz .
ρv
For a current in the x direction, the magnetic force tends to drive the mobile charged particles in
the −y direction, regardless of the sign of the charge. Thus the induced electric field will be of
different signs for mobile charge of different signs.
The Hall coefficient RH is defined as
RH =
Ey
1
= .
Jx B z
ρv
The electric field Ey and current density Jx can be determined by measuring the induced voltage
and the impressed current, respectively. Thus, under a given magnetic field Bz , the Hall coefficient
RH and hence the charge density ρv including its polarity can be measured.
4.3. Magnetic Forces on Charged Particles bound in Moving Wires
In the following two sections, we discuss the magnetic force exerted on the charged particles
bound in a thin conducting wire. Commonly, the mobile particles and immobile ions forming the
matrix of the wire are of opposite polarities in charge. As the mobile particles are bound, the force
exerted on the mobile charged particles will transfer to the binding wire, if the force is transverse
to the wire. Thus the wire may acquire an change in its momentum. On the other hand, the
electromagnetic force longitudinal to the wire can cause a slight redistribution in charge along the
wire, which leads to an emf or an open-circuit voltage. The emf is then associated with the magnetic
force longitudinal to the wire. To contribute an emf, the particles and hence the host wire segment
should move transversely to the segment.
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Motional emf
In a thin wire carrying no current, the observable effect of the magnetic force is to drive the
mobile charged particles along the wire. As these particles are moving along the wire, the charge
distribution and hence the electric scalar potential will be altered. Consequently, the electromotive
force can be induced in a wire which is moving in a static magnetic field. To distinguish this emf
from the stationary emf caused by the electric induction force, it is called the motional emf.
Quantitatively, consider a closed wire C which as a whole is moving at a velocity v across a
static or quasi-static magnetic field. The total work done on a particle of charge q and traveling
around the wire once is given by W = qVm , where the motional emf Vm is defined as
I
Vm =
C
(v × B) · dl.
In order for a wire element dl to contribute a motional emf, the element should have a component
perpendicular to the magnetic field and move at a velocity having a component perpendicular to
the magnetic field, too.
If the velocity is uniform over the loop, the motional emf Vm becomes
I
Vm = v·
C
B×dl,
where the vector identity a · b × c = c · a × b is made use of. On the other hand, if the magnetic
field is uniform over the loop, the motional emf Vm becomes
I
Vm = −B ·
C
v×dl.
If both velocity v and magnetic
field B are uniform, the motional emf Vm vanishes. This is because
H
that the path integral C dl = 0.
It is noted that the motional emf Vm is due entirely to the magnetic induction force. In other
words, the net contribution of the magnetostatic force to the emf Vm over an arbitrary closed wire
C is zero, just as that of the electrostatic force. This can be shown by considering the magnetostatic
force due to a current element Idl0 . For this case, we have
I
I
1
1
0
0
−Iµ0
R̂(v · dl ) · dl = Iµ0 (v · dl )
4π
C 4πR2
0
where v · dl is a constant over the path C.
µ
¶
1
∇
· dl = 0,
R
C
Open-circuit voltage
As in the electric induction force, when an open conducting wire is moving across a static
magnetic field, the magnetic induction force will drive the mobile electrons toward one end and leave
the other end positively charged. Thus a voltage will develop between the two ends. Again, one can
use the balance of forces to evaluate the voltage without the knowledge of charge distribution. At
equilibrium, the net driving force exerted on the mobile electrons will vanish. Thus the magnetic
force and the electrostatic force established by the accumulated charges will cancel each other.
Quantitatively, based on to the Lorentz force law and the balance of forces, one has
0 = (qE + qv × B) · dl,
where the directed path element dl is along the wire and the electric field E is due to the accumulated
charges. In terms of the potential Φ due to the accumulated charges, the preceding relation can be
written as
∇Φ · dl = (v × B) · dl.
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Then the voltage between the two terminals is given by the path integral
Z
Vab =
C
(v × B) · dl,
where the integration starts from the end point b and terminates at the other end point a.
When the integration of the magnetic force along the small gap can be neglected, the opencircuit voltage Vab becomes the motional emf Vm . That is,
I
Vab '
C0
(v × B) · dl
= Vm ,
where C 0 is a closed path composed of C and the gap.
Induction associated with two parallel wires
Consider the case with two long parallel wires of length l. Suppose the lines are parallel to the
z axis and are initially separated by a distance d in the x direction. A current I is impressed along
line 1and is in the +z direction. A magnetic field will establish around line 1. Then line 2 is made
to move away from line 1 slowly in the +x direction with a speed v. Then a magnetic induction
force is exerted on the mobile electrons in line 2 in the −z direction. Thus the induced current in
line 1 is parallel to the impressed current in line 1.
At the instant when the separation between the lines is d, the open-circuit voltage Voc between
the two end points of line 2 is given by the integral
Z
Voc =
(v × B) · ẑdz = vlI
µ0
,
2πd
where the fringe field is neglected. It is seen that the open-circuit voltage is proportional to the
speed v. Note that the current corresponding to the accumulation of the charge is of the sense of
the current generating the magnetic field.
Linear generator
Consider a metal bar sliding at a constant velocity x̂v over a pair of x-directed parallel conducting rails separated by a spacing h in the y direction. Such a structure is placed in a uniform
static magnetic field B = ẑB. During the movement of the metal bar, an open-circuit voltage Voc
develops between the terminals. The contributions comes only from the moving bar across the rails.
Thus the open-circuit voltage is then given by the path integral
Z
Voc =
bar
(v × B) · dl = −vBh,
where dl = ŷdy.
If a resistor R is connected to the terminals (via the conducting rails), a current will be induced
along the metal bar. This current tends to alter the distribution of magnetic field. It is noted that
the induced current always tends to make the total magnetic field decrease. This is associated with
Lenz’s law. Consequently, the resultant motional emf tends to be smaller than the open-circuit
voltage in magnitude and the load current tends to be smaller than |Voc |/R in magnitude. Thus
the physical origin of the loading effect in the linear generator connects to Lenz’s law. For the
case where the generated current is small and the magnetic field is not altered substantially, the
loading effect can be neglected.
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The linear generator may be not so practical, since it needs long rails in magnetic field and the
operation is not continuous. A more compact structure of generation is made possible by rotation.
Faraday disk generator
The Faraday disk generator consists of a thin circular metal disk rotating at a constant angular
frequency ω. This generator is placed in a uniform static magnetic field (B = ẑB0 ) parallel to the
axis of the rotation. During the rotation of the disk, the magnetic force will be exerted on charged
particles. At a radial distance ρ from the axis the magnetic force is given by
F = qv × B = q(ϕ̂ωρ) × ẑB0 = ρ̂qωρB0 .
It is seen that the force is in the radial direction. Thus the rim of disk tends to become positively
charged; and the shaft, negatively charged.
If the disk is a perfect conductor, the magnetic force and the electrostatic force due to accumulated charges should cancel. Thus, due to the balance of forces, the electric field due to
accumulated charge can be given by
E = −ρ̂ωρB0 ,
which is independent of ϕ. It is seen that the magnitude of the electric field increases with the
radial distance ρ. Thereby, one can find the open-circuit voltage Voc between the shaft and the rim
without the knowledge of charge distribution. It is noted that unlike the situation in electrostatics,
the electric field itself inside a metal does not vanish. It is also noted that the charge accumulation
in the presence of a vertical magnetic field is similar to that in the Hall effect. Thus the mechanism
for the Faraday generator can be viewed as the rotational Hall effect.
The voltage Voc at the disk center with reference to the rim of the disk of radius a is then given
by
Z
Z
Z 0
1
Voc = ∇Φ · dl = − E · dl = ωB0
ρdρ = − ωB0 a2 ,
2
a
where dl = −ρ̂|dρ| and the radius of the shaft is ignored. Note that the open-circuit voltage is a
constant independent of time. Thus Faraday disk is a DC (direct-current) generator. The disk can
be simplified by a or multiple radial metal arms.
By using Gauss’s law ∇·E = ρn /²0 , one can determine the distribution of the induced charges
over the disk. By so doing, it can be shown that the charge density within the thin disk is given by
ρn (ρ) = −2²0 ωB0 .
ρ<a
At the rim of the disk, the surface charge density is given by ρs = ²0 n̂·E = ²0 ωaB0 , where n̂ = −ρ̂.
Recall that the direction of n̂ is pointing from the region where the electric field is zero, not always
from the metal. The positive charge is distributed around the rim only and the total is ²0 ω2πa2 tB0 ,
where t is thickness of the disk. (Suppose that the electric field and charge density are uniform in
the z direction with thickness.) Meanwhile, the negative charge is distributed over the disk and the
total is −²0 ω2πa2 tB0 . Thus charge neutrality is maintained in the whole disk including the rim.
Rotary generator
A generator having a rotating shaft can be driven by an external mechanism, say a turbine, to
rotate about the ẑ axis at an angular frequency ω. One disadvantage of the Faraday disk generator
is that the output voltage may be too low, since a series connection seems not possible. Recall that
the magnetic field in the Faraday disk generator is along the shaft. Another approach to arrange
the magnetic field is to make it perpendicular to the shaft (say, B = x̂B0 ). In a rotary generator,
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the velocities of all the wires are along the azimuthal ϕ direction. Thus the corresponding magnetic
force has no azimuthal component, regardless of the arrangement of magnetic field. Consequently, a
motional emf cannot be induced around a circular ring centered at the shaft. Meanwhile, it cannot
be induced along a metal bar in the ρ̂ direction. In the Faraday disk generator the magnetic field is
along the shaft and hence the magnetic force can be along the radial ρ̂ direction. As the magnetic
field is perpendicular to the shaft, an open-circuit voltage can only be induced in a metal bar along
the shaft.
Thus the rotating generator can be implemented alternatively by attaching a conducting loop
with a small gap to the shaft without electrical contact, where the loop has two arms parallel to
the shaft and hence the normal of the loop is perpendicular to the shaft. Then the loop is placed
in a static magnetic field perpendicular to the shaft. The loop can be easily extended to a helical
structure equivalent to multiple loops in series connection. Thus the generator can be made more
useful by connecting more turns of wire in series or in parallel.
For simplicity, consider a single loop of rectangular shape of size h by w, where the two arms
of length h are parallel to the shaft and the other two perpendicular to it. The loop lies on the ρ-z
plane and the normal of the loop makes an angle ϕ with the x axis (n̂ = x̂ cos ϕ + ŷ sin ϕ), where
the angle changes with time as ϕ = ωt. It is noted that the velocities of all the four arms are along
the ϕ direction and are either parallel or antiparallel to n̂. Suppose B = x̂B0 . Thus the magnetic
force exerted on particles in all the arms is either parallel or antiparallel to the rotation z axis, since
both the velocity and magnetic field are in the x-y plane.
Consequently, for the two arms (of length w) perpendicular to the rotation axis, the magnetic
force is perpendicular to the wire and hence does not contribute to the emf. On the other hand, the
other two arms (of length h) are parallel to the shaft and hence contribute to the emf. In a structure
with symmetry, the contributions from these two arms are identical. Thus the open-circuit voltage
is given by the motional emf Vm as
I
Vm =
Z h
w
v × B · dl = −2 ωB0
(n̂ × x̂) · ẑdz = ωB0 wh sin ωt,
2
0
where v = ±n̂ωw/2 for the arm with dl = ∓ẑdz.
Alternatively, as the field is uniform, the emf can be evaluated by
I
Vm = −B ·
µ
¶
w
v × dl = −2B0 x̂ · −n̂ω × ẑh = ωB0 wh sin ωt.
2
Moreover, as the velocity is uniform over each arm (but not over the whole loop), the emf can be
evaluated alternatively by
Z
Vm = 2v ·
arm
B × dl = −n̂ωw · (B0 x̂ × ẑh) = ωB0 wh sin ωt.
It is noted that only the y component of the velocity contributes to the emf. It is seen the results
of the emf calculated by the three approaches are identical as they should be.
It is seen that the open-circuit voltage varies sinusoidally with time. Thus this kind of generator
is an AC (alternating-current) generator. The emf Vm reaches a maximum in magnitude when
ϕ = ±π/2, that is, when the normal n̂ of the loop comes to be perpendicular to the magnetic field.
On the other hand, when the normal n̂ is parallel to the magnetic field, the linked magnetic flux
reaches a maximum but the emf is zero.
Generator with time-varying magnetic field
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Consider a loop C moving in a time-varying magnetic field B. Recall that the stationary emf
due to the electric induction force is given by
Ã
I
Vs =
C
!
Z
∂
− A · dl = −
∂t
Ã
!
Z
∂
∇×
A · ds = −
∂t
S
Ã
S
!
∂
B · ds.
∂t
Then the total emf Vt is given by the sum of the stationary emf Vs and the motional emf Vm . That
is,
!
Z Ã
I
∂
Vt = −
B · ds +
v × B · dl,
∂t
S
C
where S is a surface bounded by path C. In this case, both the electric induction force and the
magnetic induction force are involved. The open-circuit voltage Voc induced across a small gap in
the loop is then given approximately by the total emf Vt .
Consider the case where the rotor generator is placed in a time-varying magnetic field given by
B = x̂B0 sin ωm t. Thus ∂B/∂t = x̂ωm B0 cos ωm t. Then the total emf is
Vt = B0 wh(−ωm cos ωm t cos ωt + ω sin ωm t sin ωt).
By applying a time-varying magnetic field of ωm = ω, the total emf becomes Vt = −ωwhB0 cos 2ωt.
Thus a rotor generator can serve as a second-harmonic generator.
Alternative formulation of total emf in total time derivative
The motional emf can be related to the time rate of change of the linked magnetic flux. Consider
a cylinder formed by a surface S bounded by the path C which moves at velocity v. Aside from the
top and bottom ends, the cylinder has a lateral surface Sl composed of the directed surface element
dl × v4t, where the element dl is along the path C and 4t is a short time interval.
The motional emf can then be given by
I
Vm =
I
C
(v × B) · dl =
1 I
B · (dl × v) = lim
4t→0 4t
C
C
B · (dl × 4ζ̄),
where the differential displacement 4ζ̄ = v4t. Thus the motional emf can be given by
1 Z
Vm = lim
4t→0 4t
Sl
B · n̂l ds,
where n̂l ds = dl × 4ζ̄ and n̂l is the unit outward normal vector at the lateral surface of the cylinder.
Further, by using the divergence relation ∇ · B = 0, one has
Z
Z
Sl
B · n̂l ds = −
Z
Stop
B · n̂t ds +
Sbottom
B · n̂t ds = −4ψ,
where n̂t is the unit outward normal vector at the top surface of the cylinder, the magnetic flux
Z
ψ=
S
B · ds,
and ds = n̂t ds.
The term −4ψ can be interpreted as the decrease in the linked magnetic flux when the surface
S is moving at velocity v in a static magnetic field. Thus the motional emf can be given in a form
of time derivative as
4ψ
.
Vm = −
4t
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This relation states that the motional emf is equal to minus the time rate of change in the
linked magnetic flux due to the movement of the loop in the static magnetic field. The
movement includes translation and rotation. It is of interest to note that a change of position or
orientation of the loop in space with respect to the current generating the magnetic field actually
results in a change of the linked flux with time. In order for the motional emf Vm to be nonvanishing
for a loop in translation, the magnetic field should be nonuniform spatially.
It is noted that the preceding formula for the motional emf Vm is similar to that for the one
for the stationary emf Vs which in turn is due to the electric induction force and is related to minus
the time rate of change in the linked magnetic flux due to the change of the field itself with time.
The total emf induced on a loop moving in a time-varying magnetic field is then given by
Ã
Z
Vt = Vs + Vm = −
S
!
I
∂
B · ds +
∂t
C
v × B · dl = −
∂ψ 4ψ
−
.
∂t
4t
It is of essence to note that Vs is due to a change of the current with time, while Vm is due to a
change of of position or orientation of the loop relative to the current. Both of them result in a
change of the linked flux. Thus the total emf can be represented by the total time derivative of the
flux linkage as
d Z
dψ
Vt = −
B · ds = − ,
dt S
dt
which involves both the electric induction force and the magnetic induction force. Thereby, for a
loop moving in a time-varying magnetic field, the total electromotance is given by the time rate of
change of the flux experienced by the loop.
The total emf in a rotor generator placed in the aforementioned time-varying magnetic field
can thus be given by
Vt = −
d Z
dt
S
B · n̂ds = −
d
[wh(B0 sin ωm t) cos ωt]
dt
= B0 wh(−ωm cos ωm t cos ωt + ω sin ωm t sin ωt),
where the derivative with sin ωm t is associated with the electric induction force and that with cos ωt
is associated with the magnetic induction force. It is seen that this results agrees with the one
obtained previously.
To help imagine the total time derivative, consider a fly is moving through a space where the
temperature is both spatially nonuniform and temporally varying. Then the total time derivative
of temperature is related to the rate of change in the temperature experienced by the fly.
4.4. Magnetic Forces on Current-Carrying Wires
The emf is associated with the force longitudinal to the wire. Then we discuss the magnetic
force transverse to the wire. For the case where a neutralized wire carries no current and moves
as a whole, the velocities of the mobile and immobile particles are identical. The immobile ions
as well as the mobile electrons also experience the magnetic force, when they are moving with the
mobile electrons. Thus the transverse force exerted on the immobile ions tends to cancel the one
transferred from the mobile particles. Consequently, the net transverse magnetic force exerted on
the wire vanishes under the condition of complete neutralization. This situation of cancellation is
similar to that in the electric induction force discussed in the preceding Chapter. Thus the wire
carrying no current cannot acquire momentum in a magnetic force.
Magnetic Forces by ccsu
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On the other hand, for a wire carrying a current, the velocity of the mobile particles is different
from that of the immobile particles. The difference in velocity is the drift velocity of the mobile
charged particles with respect to the wire. Note that it is just this velocity that determines the current density in a neutralized wire. Thereby, the magnetic force exerted on the wire can be different
from zero, depending on the drift velocity and the current. As the drift velocity is longitudinal to
the wire, the resultant magnetic force is transverse to the wire.
In a thin conducting wire carrying a current I formed by the mobile charged particles of line
charge density ρl , the drift velocity is given by vd = ˆlI/ρl , where ˆl is pointing along the direction
of the wire segment. Ordinarily, the drift speed of the mobile charged particles in a conductor is
extremely low. A typical value is of the order of 10−4 m/sec. However, the particle concentration
is high. Thus the magnetic force exerted on a wire carrying a current can still be considerable.
In a magnetic field B, the force exerted on a differential segment dl of the wire is then given by
F = ρl vd × Bdl = −IB × dl.
It is seen that the force is proportional to the current I and is perpendicular to the direction of the
current.
Force between two parallel wires
Consider two long parallel wires of length l, each of which carries a uniform I1 or I2 . The lines
are parallel to the z axis and are separated by a distance d in the x direction. The force exerted on
wire 2 due to the current on wire 1 is given by
1
.
2πd
This force is due entirely to the magnetostatic force. It is seen that if I1 I2 > 0, this force tends to
attract wire 2 toward wire 1. Obviously, F12 = −F21 . That is, this force, as well as the electrostatic
force between two charged particles, is in accord with Newton’s third law of motion.
F21 = −x̂I2 By1 l = −x̂I1 I2 lµ0
Magnetic levitation
Consider a square loop of circumference l and mass m which carries a uniform current I and is
placed on (but insulated from) a ground plane. Then charges along with currents will be induced
on this plane. The induced current distributed on the ground plane can be replaced by an image
current, which can be envisaged as due to the motion of the image charges. As the image charge is
of the opposite polarity, the image current flowing on the plane is in the reverse direction. Thus the
vector potential at the location of the removed ground plane vanishes due to cancellation. Thereby,
the current induced on the ground plane tends to lift the loop by the magnetostatic force.
Assume the wire is long enough, such that the magnetic field can be approximated to the one
due to an infinitely long wire. Thus the magnetic force exerted on the loop placed at a vertical
distance d from the ground plane (2d from the respective image current) is then given by
1
.
4πd
This force is repelling and tends to lift the wire. It is seen that the lift force decreases when the
distance d increases.
The final levitation d is determined by the balance between the magnetic force and the gravitational force −ẑmg. Thus one has
I 2 lµ0
,
d=
4πmg
F = ẑI 2 lµ0
Magnetic Forces by ccsu
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where the acceleration of gravity g = 9.8 m·sec−2 on earth’ surface. Note that the levitation d
increases quadratically with current.
Linear accelerator
Consider two long x-directed parallel conducting wires of radius a separated by a center-tocenter distance d in the y direction. A conducting bar is put perpendicularly on the two rails. A
uniform current I is made to flow in the circuit composed of the rails and the bar. The currents in
the rails are in the ±x direction and a z-directed magnetic field is generated in the region between
the parallel wires. Meanwhile, the current in the bar is in the y direction. Then a magnetic force
will be exerted on the bar. The force is in the x direction and tends to accelerate the bar to make
it move along the rails.
Quantitatively, the magnetic field at the bar due to the U-shape (rail-bar-rail) structure can
be given by
Ã
!
1
I
I
B(y) = ẑµ0
+
,
2 2πy 2π(d − y)
where the y coordinates of the rails are respectively 0 and d at the respective centers, the factor
1/2 comes from the fact that only one half of each rail carries the current I, the current on the bar
does not contribute to the magnetic field right at the bar, and the fringe field is ignored.
Thus the force exerted on the bar is given by
Z
F=−
bar
IB × dl = x̂Iµ0
Z d−a
a
B(y)dy = x̂
µ0 2 d − a
I ln
,
2π
a
where dl = ŷdy. The force is due entirely to the magnetic induction force, since the mobile charges
in the bar move toward and away from the currents flowing on the rails.
Linear motor with external magnetic field
Reconsider the metal bar put perpendicularly over the pair of x-directed parallel conducting
rails separated by a spacing h in the y direction. However, the structure is placed in a uniform
magnetic field B = ẑB from an external source and a direct current I is made to flow only in the
bar. Again, a magnetic force will be exerted on the bar. Suppose that the current I is in the y
direction and the external magnetic field is much stronger than the one due to the current. Thus
the force is given by
Z
F = −I
B×dl = x̂IBh,
bar
where dl = ŷdy. A load can be attached to the bar and then be driven by the magnetic force.
Counter emf and force in motor and generator
Note that as the metal bar starts to move along the rails in the linear motor, an additional
magnetic force will be exerted on the mobile charge in the bar, just as in the linear generator. Recall
that in the linear generator as the metal bar sliding at a constant velocity x̂v along the conducting
rails, the induced motional emf is given by Vm = vBh in magnitude. Thus, in the motor, the
additional magnetic force exerted on the sliding bar tends to impede the current as a counter emf.
On the other hand, if a resistor R is connected to the terminals of a generator, a current will be
induced along the sliding bar. The induced current incurs an additional magnetic force which tends
to impede the motion of the bar as a counter force. The counter emf or force is associated with
Lenz’s law.
Magnetic Forces by ccsu
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In a generator, in order to maintain the bar with a fixed velocity, a mechanical force due to
an external source is needed. This external mechanical force is used to balance the counter force.
Thus the mechanical power is converted through the generator into the electric power of the resistive
load connected to the terminals. On the other hand, in a motor, in order to maintain a fixed current
I, an electrostatic force due to an external source is needed. This external force is used to balance
the magnetic force contributing to the counter emf. In other words, an externally applied voltage
is used to balance the counter emf. Thus the electric power is converted through the motor into
the mechanical power of the load attached to the bar. From this viewpoint, the generator and the
motor provide the conversion between the mechanical and electric powers.
If a generator is open-circuit, there is no induced current and hence the bar is free to move. On
the other hand, if a motor is over loaded, the movement of the bar may be impeded and hence
there is no induced emf. A unbalanced input voltage tends to bring forth a large current. Thus a
motor without protection circuits may be damaged by over-heating.
To discuss the power conversion quantitatively, reconsider the linear generator loaded with a
resistor R. Suppose the resistance R is high and the loading effect can be neglected. Thus the load
current is simply given by I = Voc /R. Then the electric power Pe due to ohmic dissipation is
Pe = IVoc = −IvBh.
After the current I is induced, the bar experiences a magnetic force F = x̂IBh. Since I as well
as Voc is negative, the magnetic force is in the −x̂ direction. That is, this magnetic force tends to
impede the movement of the metal bar.
To maintain the metal bar with a uniform velocity x̂v, a force balance is yet to establish. This
can be achieved by an external mechanical force. The required mechanical force is Fmech = −F and
the required mechanical power Pmech is
Pmech = Fmech · x̂v = −IBhv = Pe ,
which is equal to the electric dissipation power. That is, the mechanical power done on the bar
is converted to the electric power dissipated in the load. In various types of power generator, the
mechanical power can be supplied by gravity or by steam. The gravity with water is utilized in the
hydroelectric power. The steam in turn can be generated by using coal, oil, gas, or nuclear reaction.
Torque on a magnetic dipole
As in the linear generator, the linear motor may not be practical, since a magnetic field over
a long distance is needed. In what follows, the motors in a rotational motion are discussed. To
being with, the torque is discussed. The torque is associated with a force which produces or tends
to produce rotation or torsion. From classical mechanics, the torque T exerted on a small body
rotating about an axis is given by
T = r × F,
where r is the distance from the rotation axis to the body.
The magnetic force is given by Idl × B, the torque T exerted on a magnetic dipole formed by
the current I is given by
·I
I
¸
1
(r×dl) × B,
T=−I
r × (B × dl) = I
2
C
C
where we have made of the situation that the magnetic field B is uniform over the loop. Thereby,
the torque T exerted on a magnetic dipole of moment m in a uniform field B is given by
T = m × B.
Magnetic Forces by ccsu
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The torque tends to align the moment to parallel to the magnetic field. Once this situation is
achieved, the torque vanishes.
The motors utilize such a torque due to magnetic force to drive a heavy load, in addition to a
variety of other applications discussed below. A permanent magnet is composed of numerous tiny
magnetic dipoles oriented in an identical direction. A small permanent magnet bar tends to deflect
to align its dipole moment with earth’s magnetic field. This device is known as a compass. In a
current meter, a current under test is impressed in a coil placed in the magnetic field due to a
permanent magnet. The torque tends to rotate the magnetic dipole formed with the coil. Then
this torque is balanced by that due to the electrostatic force developed in a spiral spring. From
the deflection of the pointer, the torque and hence the current can be read from a scale suitably
calibrated.
The direction of the magnetic dipoles in a permanent magnet is maintained by the magnetic
field due to the dipoles themselves. Meanwhile, the dipole direction can be changed by a strong
external magnetic field. Thus the permanent magnetization can be used in magnetic memory. In a
magnetic tape or disk, the direction of the applied magnetic field and hence that of the data-storing
dipoles depends on the polarity of the current impressed in a coil in the read/write head. Thus
digital data can be written magnetically in a memory cell which contains data-storing dipoles and is
distributed along a track on the disk. Then digital information is stored by preserving the direction
of the dipole moment in a cell. In reading, the head is moving through the data-storing dipoles. A
motional emf will develop in the head due to the magnetic field of the dipoles. The polarity of the
voltage depends on the direction of the dipole moment. Thus digital information can be retrieved.
In short, the writing process involves a mechanism like a motor; and the reading process like a
generator.
DC motor
Reconsider the rectangular conducting loop of size h by w and placed in a uniform static
magnetic field given by B = x̂B0 . Suppose the normal of the loop makes an angle ϕ with the x
axis, that is, n̂ = x̂ cos ϕ + ŷ sin ϕ. The loop carrying a uniform current I forms a magnetic dipole of
moment n̂Iwh. Then, as the magnetic field is uniform over the armature loop, the torque exerted
on the loop is given by
T = n̂Iwh × x̂B0 = −ẑIwhB0 sin ϕ.
Note that the torque depends on the angle ϕ. The torque will reach its maximum in magnitude
when ϕ = π/2. That is, the moment is perpendicular to the magnetic field. And the magnetic
force tends to align the moment in the x direction. When the moment comes in this direction, the
torque vanishes.
The torque tends to deflect the dipole clockwise with a decreasing ϕ when 0 < ϕ < π, while
the torque tends to deflect the dipole counterclockwise with an increasing ϕ when −π < ϕ < 0.
In order to make the device useful, the torque of such a device is made to be an identical sense
(say, clockwise with decreasing ϕ) by using an electric brush as a contact between the loop and the
DC power supply. Thereby, the direction of the current and hence that of the dipole are reversed
when the loop is rotating through a given orientation (say, when the angle ϕ is crossing 0). Thus
this kind of motor can be operated by a direct current and is known as a DC motor.
Synchronous motor
The torque of a DC motor varies sinusoidally. To make the torque a constant independent of
angle ϕ and hence of time, one can use a rotating magnetic field at a given angular frequency.
The kind of rotating magnetic field can be implemented by using a three-phase alternating current
Magnetic Forces by ccsu
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imposed on the three windings wound on the stator symmetrically in the azimuthal direction. The
current on each winding is also varying with time as cos(ωm t + ϕ), where ϕ is the phase angle. The
phase angle in the three winding is made to be ϕ = 0, −120, and −240◦ , respectively. Then the
magnetic field due to the space- and time-varying currents is given by superposition as
√ !
√ !
(
Ã
µ
¶ Ã
µ
¶)
2
1
3
2π
1
3
2π
B = B0 x̂ cos ωm t + −x̂ + ŷ
cos ωm t −
+ −x̂ − ŷ
cos ωm t +
.
3
2
2
3
2
2
3
An easy manipulation leads to that the magnetic field is given by
B = B0 (x̂ cos ωm t + ŷ sin ωm t).
It is seen that the magnetic field rotates with respect to the stator.
The rotor is the armature loop carrying a direct current I0 imposed by an external source.
Suppose the rotor is rotating at an angular frequency ω with a phase ϕ0 , that is,
n̂ = x̂ cos(ωt + ϕ0 ) + ŷ sin(ωt + ϕ0 ).
Assume the magnetic field is uniform over the armature loop, the torque can be given by
T = n̂I0 wh × B = −ẑI0 whB0 sin[(ω − ωm )t + ϕ0 ].
Note that the torque varies with time and its time average is zero, except at synchronism with
ω = ωm . At synchronism the torque then becomes static and is determined by the phase angle ϕ0 .
Thus this kind of motor is known as the synchronous motor. As the load is varied, the rotating
speed will be changed momentarily until the phase angle was properly adjusted by some mechanism.
As synchronism is required, this motor has a start-up problem, that is, it cannot be started by
itself as the DC motor. To overcome this problem, an auxiliary shorted-circuit winding known as
the damper is provided. A time-varying current will be induced on the armature loop. This current
can change the time dependence of the torque to make its time average nonzero at standstill, as
shown in what follows.
Induction Motor
In the synchronous motor the current on the rotor loop is generated from an external DC source
and is a constant. In the induction motor the rotor loop is short circuited. And the current on it
is induced by the time-varying magnetic field due to the alternating current imposed on the stator
windings. It will be shown that if the armature loop has a finite resistance, the current induced
on the armature loop has an additional term of the form sin[(ω − ωm )t + ϕ0 ], which is in phase
with the torque in the synchronous motor. Then the time-average torque will be nonzero out of
synchronism.
Suppose the magnetic field B and the normal n̂ of the armature loop are identical to those
given in the synchronous motor. Thus the magnetic flux linkage over the surface S bounded by the
armature loop is given by
Z
ψ =
S
B · ds = B0 wh[cos ωm t cos(ωt + ϕ0 ) + sin ωm t sin(ωt + ϕ0 )]
= B0 wh cos(ωs t − ϕ0 ),
where the slip frequency ωs = ωm − ω. The time derivative of this flux is associated with the emf
and is given by
dψ
= ωs B0 wh sin(ωs t − ϕ0 ).
Vt = −
dt
Magnetic Forces by ccsu
em4–17
It is noted that the time dependence is identical to the one of the torque in the synchronous motor.
At synchronism ωm = ω, the linked flux becomes static and hence the emf vanishes.
Suppose the armature winding has a resistance Ra and an inductance La . Then the current Ia
induced on the armature can be given by
Ia = ωs B0 wh[Ra sin(ωs t − ϕ0 ) − ωs La cos(ωs t − ϕ0 )]/(Ra2 + ωs2 L2a ).
Assume the magnetic field is uniform over the armature loop, the torque is given by
T = m × B = ẑIa wh sin(ωs t − ϕ0 ).
By replacing Ia with I0 , the torque reduces to that in the synchronous motor. Written explicitly,
the torque is as complicated as
T = ẑωs B02 w2 h2 sin(ωs t − ϕ0 )[Ra sin(ωs t − ϕ0 ) − ωs La cos(ωs t − ϕ0 )]/(Ra2 + ωs2 L2a ).
It is noted that the armature resistance Ra also introduces the same sinusoidal time dependence as
that of the torque in the synchronous motor. Thus the torque has second-harmonic terms.
Due to the in-phase second-harmonic term, the time-average torque Tav becomes
1
Ra
Tav = ẑ ωs B02 w2 h2 2
.
2
Ra + ωs2 L2a
It is important to see that the time-average torque is nonzero even when the operating frequency
ω is zero. That is, the induction motor has a nonzero time-average starting torque. Unlike in the
synchronous motor, the time-average torque in the induction motor is independent of the phase
angle ϕ0 . On the other hand, after the start-up, this additional torque vanishes at synchronism,
since the induced current then disappears.
Note that the torque vanishes in time average when the armature resistance Ra is zero. This
is because the in-phase second-harmonic term in the torque disappears. On the other hand, when
the resistance is too large, the torque becomes weak. This is because the induced current is small
accordingly. The maximum torque occurs at Ra = ωs La . If the inductance La can be omitted, then
the torque is inversely proportional to the resistance Ra .