E
E
'
'
current flow (schematically).
x
0
l
However, what about the geometry shown in Fig. 6b? In this case, the field distribution in the free space around the conductor is dramatically different, but according to the boundary problem defined by Eqs. (14) and (20), inside the conductor, the solution is exactly the same as it was in the former case. Now, the Laplace equation in the surrounding insulator has to be solved with the boundary values of the electrostatic potential, “dictated” by the distribution of the current (and hence potential) in the conductor. Note that as the result, the electric field lines are generally not normal to the conductor’s surface, because the surface is not equipotential – see Eq. (22) again.
Let us solve a problem in that this conduction hierarchy may be followed analytically to the very end. Consider an empty spherical cavity carved in a conductor with an initially uniform current flow with a constant density j0 = n zj 0 (Fig. 7a). Following the hierarchy, we have to solve the boundary problem in the conducting part of the system, i.e. outside the sphere (at r R), first. Since the problem is evidently axially symmetric, we already know the general solution of the Laplace equation – see Eq.
(2.172). Moreover, we know that in order to match the uniform field distribution at r , all coefficients al but one ( a 1 = – E 0 = – j 0/) have to be zero, and that the boundary conditions at r = R will give zero solutions for all coefficients bl but one ( b 1), so that
j
b
0 r cos 1 cos ,
r
for R .
(4.24)
r 2
In order to find the remaining coefficient b 1, we have to use the boundary condition (20) at r = R: Chapter 4
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j
2 b
0
1
.
(4.25)
r
cos 0
3
r
R
R
This gives b 1 = – j 0 R 3/2, so that, finally,
j
0
R 3
( r, )
r
cos ,
r
for R
.
(4.26)
2 r 2
(Note that this potential distribution corresponds to the dipole moment p = –E0 R 3/2. It is straightforward to check that if the spherical cavity was cut in a dielectric, the potential distribution outside it would be similar, with p = –E0 R 3( – 1)/( + 2). In the limit , these two results coincide, despite the rather different type of the problem: in the dielectric case, there is no current at all.)
(a)
(b)
j
0
z
R
0
Fig. 4.7. A spherical cavity carved in a uniform conductor: (a) the problem’s geometry, and (b) the equipotential surfaces as given by Eqs. (26) and (28).
Now, as the second step in the conductivity hierarchy, we may find the electrostatic potential distribution ( r,) in the insulator, in this particular case inside the empty cavity (at r R). It should also satisfy the Laplace equation with the boundary values at r = R, “dictated” by the distribution (26): 3 j
( R, )
0
R cos .
(4.27)
2
We could again solve this problem by the formal variable separation (keeping in the general solution (2.172) only the term proportional to a l, which does not diverge at r 0), but if we notice that the boundary condition (27) depends on just one Cartesian coordinate, z = R cos, the solution may be just guessed:
3 j 0
3 j
( r, )
z
0 r cos,
r
at R .
(4.28)
2
2
Indeed, it evidently satisfies the Laplace equation and the boundary condition (27), and corresponds to a constant electric field parallel to the vector j0, and equal to 3 j 0/2 – see Fig. 7b. Again, the cavity surface it not equipotential, and the electric field lines at r R are not normal to it at almost all points.
More generally, the conductivity hierarchy says that static electrical fields and charges outside conductors (e.g., electric wires) do not affect currents flowing in the wires, and it is physically very clear Chapter 4
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why. For example, if a charge in the free space is slowly moved close to a wire, it (in accordance with the linear superposition principle) only induces an additional surface charge (see Sec. 2.1) that screens the external charge’s field, without participating in the current flow inside the conductor.
Besides this conceptual issue, the two examples given above may be considered as applications of the first two methods discussed in Chapter 2 – the orthogonal coordinates (Fig. 6) and the variable separation (Fig. 7) – to dc current distribution problems. As the reader may recall, in that chapter we also discussed the method of charge images. It turns out that its analog may be also used for the solution of some dc conductivity problems. Indeed, let us consider a spherically-symmetric potential distribution of the electrostatic potential, similar to that given by the basic Eq. (1.35):
c
.
(4.29)
r
As we know from Chapter 1, this is a particular solution of the 3D Laplace equation at all points but r =
0. In free space, this distribution would correspond to a point charge q = 40 c; but what about a uniform Ohmic conductor? Calculating the corresponding electric field and current density, c
c
E
r,
j E
r,
(4.30)
3
3
r
r
we see that the total current flowing from the origin through a sphere of an arbitrary radius r does not depend on the radius:
I Aj 4
2
r j 4 .
c
(4.31)
Plugging the resulting coefficient c into Eq. (29), we get
I
.
(4.32)
4
r
Hence the Coulomb-type distribution of the electric potential in a conductor is possible (at least at some distance from the singular point r = 0), and describes the dc current I flowing out of a small-size electrode – or into such an electrode if the coefficient c is negative. Such current injection may be readily implemented experimentally; think for example about an insulated wire with a small bare end, inserted into a poorly conducting soil – an important method in geophysical research. Such point injection is even simpler in 2D situations – think about a wire attached, within a small spot, to a thin resistive layer, such as the thin films used for wiring in microelectronics.15
Now let the current injection point r ’ be close to a plane interface between the conductor and an insulator (Fig. 8). In this case, besides the Laplace equation, we should satisfy the boundary condition,
j E
0 ,
(4.33)
n
n
n
at the interface. It is clear that this can be done by replacing the insulator with an imaginary similar conductor with an additional current injection point, at the mirror image point r ” . Note, however, that in contrast to charge images, the sign of the imaginary current has to be similar, not opposite, to the initial one, so that the total electrostatic potential inside the conducting semi-space is
15 Note that in such layers, the current distribution near the injection point is different, j 1/ r rather than 1/ r 2.
Chapter 4
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I
1
1
r
( )
.
(4.34)
4 r - ' r
r - "
r
(The image current’s sign would be opposite at the interface between a conductor with a moderate conductivity and a perfect conductor (“electrode”), whose potential should be virtually constant.) I
'
r
jr
d
0
d
Fig. 4.8. Applying the method of images
"
r
for the current injection analysis.
I
This result may be readily used, for example, to calculate the current density at a plane surface of a uniform conductor, as a function of distance from point 0 – the surface’s point closest to the current injection site – see Fig. 8. At such surface, Eq. (34) yields
I
1
,
(4.35)
2 2
2
d 1/2
so that the current density is:
I
j E
.
(4.36)
2 2
2
d 3/2
Deviations from Eqs. (35) and (36) may be used to find and characterize conductance inhomogeneities, say, those due to mineral deposits in the Earth’s crust.16
So, the methods used in electrostatics to calculate the potential distribution in linear dielectrics may be also used to find such distributions in Ohmic conductors. Moreover, some of these methods are more valuable in this field. For example, in electrostatics the effective methods of solution of the 2D
Laplace equation, discussed in Secs. 2.3-2.6, could be only applied to cylindrical geometries. At Ohmic conduction, this equation is also valid in some 3D cases. A practically important example is the current flow in thin resistive layers where, due to the conductivity hierarchy principle, the 3D-distributed field outside a layer, induced by the 2D-distributed current in it, does not affect the flow and in many cases is not important. A few problems of this kind, formulated in Sec. 5, are left for the reader’s exercise.
4.4. Energy dissipation
Let me conclude this brief chapter with an ultra-short discussion of energy dissipation in conductors. In contrast to the electrostatic situations in insulators (vacuum or dielectrics), at dc 16 The current injection may be also produced, due to electrochemical reactions, by an ore mass itself, so that one need only measure (and correctly interpret :-) the resulting potential distribution – the so-called self-potential method – see, e.g., Sec. 6.1 in W. Telford et al., Applied Geophysics, 2nd ed., Cambridge U. Press, 1990.
Chapter 4
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conduction, the electrostatic energy U is “dissipated” (i.e. transferred to heat) at a certain rate P –
dU/ dt, with the dimensionality of power.17 This so-called energy dissipation may be evaluated by calculating the power of the electric field’s work on a single moving charge:
F v qE v
1
P
.
(4.37)
After the summation over all charges, Eq. (37) gives us the average dissipation power. If the charge density n is uniform, multiplying by it both parts of this relation, and taking into account that qnv
= j, for the energy dissipation in a unit volume we get the differential form of the Joule law 18
General
P
P N
Joule
p
1 P n qE v n E j.
(4.38)
law
V
V
1
In the case of the Ohmic conductivity (8), this expression may be also rewritten in two other forms: Joule law
2
for Ohmic
p 2
j
E
.
(4.39)
conductivity
With our electrostatics background, it is also straightforward (and hence left for the reader’s exercise) to prove that the dc current distribution in a uniform Ohmic conductor, at a fixed voltage distribution along its borders, corresponds to the minimum of the total dissipation in the sample,
P
d 3
p r
E 2 d 3 r
.
(4.40)
V
V
4.5. Exercise problems
R
R
R
1
1
1
4.1. DC voltage V 0 is applied to the end of a semi-infinite
chain of lumped Ohmic resistors, shown in the figure on the right.
Calculate the voltage across the j th link of the chain.
V
R
R
0
2
2
4.2. It is well known that properties of many dc current sources (e.g., batteries) may be reasonably well represented as a connection in series of a perfect voltage source and an Ohmic internal resistance. Discuss the option, and possible advantages, of using a different equivalent circuit that would include a perfect current source.
4.3. Prove the following Rayleigh-Lorentz-Carson reciprocity
relation: Results of two reciprocal experiments shown schematically in the
V 1
I 1
figure on the right, with an arbitrary Ohmic conductor with four
electrodes/terminals, are related as I 1 V 2 = I 2 V 1.
Hint: Try to apply the same approach as was used to prove Green’s
reciprocity relation in electrostatics in Problem 1.18, but with proper
I
V
2
2
modifications.
17 If this electric field and hence the electrostatic energy are time-independent, the energy is replenished at the same rate from the current source(s).
18 Named after James Prescott Joule, who quantified this effect in 1841.
Chapter 4
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4.4. Calculate the resistance between two large, uniform Ohmic
conductors separated by a very thin, plane, insulating partition, with a
circular hole of radius R in it – see the figure on the right.
Hint: You may like to use the oblate spheroidal coordinates which
2 R
were discussed in Sec. 2.4.
4.5. A very narrow plane crack inside a round conducting wire of
radius R does not reach its surface by a small distance w – see the figure j
on the right. Assuming that the Ohmic conductivity of the wire’s
material is otherwise constant, calculate the electric resistance of the
obstacle in the first approximation in small w/ R << 1.
Hint: You may like to use the same elliptic coordinates as were w
2 R
employed at the solution of Problem 2.12.
4.6. Calculate the effective (average) conductivity
ef of a
medium with many empty spherical cavities of radius R, carved at
random positions in a uniform Ohmic conductor (see the figure on the
R
right), in the limit of a low density n << R-3 of the spheres.
Hint: You may like to use the analogy with a dipole medium –
see, e.g., Sec. 3.2.
4.7. In two separate experiments, a narrow gap, possibly of irregular width, between two close, perfectly conducting electrodes is filled with some material: in the first case, with a uniform linear dielectric with an electric permittivity , and in the second case, with a uniform conducting material with an Ohmic conductivity . Neglecting the fringe effects, calculate the relation between the mutual capacitance C between the electrodes (in the first case) and the dc resistance R between them (in the second case).
I
4.8. Calculate the voltage V across a uniform, wide resistive
slab of thickness t, at distance l from the points of injection/pickup
t
l
V ?
of the dc current I passed across the slab – see the figure on the right.
I
4.9. Calculate the distribution of the dc current’s density in a thin, round, uniform resistive disk, if the current is inserted into some point at its rim, and picked up at the center.
I
I
4.10. DC current is passed between two point electrodes
connected to a wide, thin, uniform resistive sheet – see the figure on
the right. Use the model solution of the previous problem to prove,
without much new calculation, that cutting a round hole in the sheet
(outside of the current injection/extraction points) doubles the voltage
V
between any two points on it border.
Chapter 4
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4.11. The rim of a hemispherical thin shell, of radius R and
I
thickness t << R, made of a uniform Ohmic conductor, is connected to a plane grounded electrode. Calculate the distribution of the electrostatic
'
R
potential created in the shell by a dc current I injected into it through a
small-size electrode located at a polar angle ’ < /2 from the symmetry
axis – see the figure on the right.
Hint: You may like to use the variable substitution tan(/2) to map the hemisphere onto a unit circle.
4.12. A rectangle of area l w is cut from a resistive sheet of thickness t << l, w. Use two different approaches to calculate the voltage I V between its two adjacent corners, induced by the dc current I passed between the two other corners – see the figure on the right.
l
, t
V ?
Hint: Besides the charge/current image method, you may like to I
consider using the variable separation method, with due respect to the
w
current injection/extraction points.
4.13.* The simplest reasonable model of a vacuum diode consists of two plane, parallel metallic electrodes of area A, separated by a gap of thickness d << A 1/2: a “cathode” that emits electrons into the gap, and an “anode” that absorbs the electrons arriving from the gap at its surface. Calculate the dc I-V
curve of the diode, i.e. the stationary relation between the current I flowing between the electrodes and the voltage V applied between them, using the following simplifying assumptions: (i) due to the effect of the negative space charge of the emitted electrons, the current I is much lower than the emission ability of the cathode,
(ii) the initial velocity of the emitted electrons is negligible, and
(iii) the direct Coulomb interaction of electrons (besides the space charge effect) is negligible.
4.14.* Calculate the space-charge-limited current in a system with the same geometry as in the previous problem, and using the same assumptions besides that now the emitted charge carriers move not ballistically, but drift in accordance with the Ohm law, with the conductivity given by Eq. (13): =
q 2 n, with a constant mobility .19
Hint: In order to get a realistic result, assume that the medium in that the charge carriers move has a certain dielectric constant unrelated to the carriers.
4.15. Prove that the distribution of dc currents in a uniform Ohmic conductor with a given voltage distribution along its boundaries, corresponds to the minimum of the total energy dissipation rate (“Joule heat”).
19 As was mentioned in Sec. 2, the approximation of a constant (in particular, field- and charge-density-independent) mobility is most suitable for semiconductors.
Chapter 4
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Chapter 4
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Chapter 5. Magnetism
Even though this chapter addresses a completely new type of electric charge interaction, its discussion (for the stationary case) will take not too much time/space, because it recycles many ideas and methods of electrostatics, though with a twist or two.
5.1. Magnetic interaction of currents
DC currents in conductors usually leave them electroneutral, (r) = 0, with very good precision, because even a minute imbalance of positive and negative charge density results in extremely strong Coulomb forces that restore the electroneutrality by a very fast additional shift of free charge carriers.
This is why let us start the discussion of magnetism from the simplest case of two spatially-separated, dc-current-carrying, electroneutral conductors (Fig. 1).
' j
j
r
d 3
'
r
r'
dF
V'
V
d 3 r
Fig. 5.1. Magnetic
interaction of two
currents.
According to the Coulomb law, there is no electrostatic force between them. However, several experiments carried out in 18201 proved that there is a different, magnetic interaction between the currents. In the present-day notation, the results of all such experiments may be summarized with just one formula, in SI units expressed as
r '
r
Magnetic
0
3
3
F
d r d r' (
j r) ' j( '
r )
force
.
(5.1)
3
4
V
V '
r
'
r
Here the coefficient 0/4 (where 0 is called either the magnetic constant or the free space permeability) equals to almost exactly 10-7 SI units, with the product 00 equal to exactly 1/ c 2. 2
Note a close similarity of this expression to the Coulomb law (1.1) rewritten for the interaction of two continuously distributed charges, with the account of the linear superposition principle (1.4): Electric
1
r '
r
force
3
3
F
d r d r' (r) '
( '
r )
.
(5.2)
3
40
V
V '
r
'
r
1 Most notably, by Hans Christian Ørsted who discovered the effect of electric currents on magnetic needles, and André-Marie Ampère who extended this work by finding the magnetic interaction between two currents.
2 For details, see Appendix CA: Selected Physical Constants. In the Gaussian units, the coefficient 0/4 in Eq.
(1) and beyond is replaced with 1/ c 2.
© K. Likharev
EM: Classical Electrodynamics
Besides the different coefficient and a different sign, the “only” difference of Eq. (1) from Eq. (2) is the scalar product of the current densities, evidently necessary because of their vector character. We will see soon that this difference brings certain complications in applying the approaches discussed in the previous chapters, to magnetostatics.
Before going to their discussion, let us have one more glance at the coefficients in Eqs. (1) and (2). To compare them, let us consider two objects with uncompensated charge distributions (r) and
’(r), each moving parallel to each other as a whole, with certain velocities v and v ’, as measured in the same inertial (“laboratory”) reference frame. In this case, j(r) = (r)v, so that j(r)j ’(r) = (r) ’(r) vv’, and the integrals in Eqs. (1) and (2) become functionally similar, differing only by the factor F
vv'
1
vv'
magnetic
0
.
(5.3)
2
F
4
4
c
electric
0
(The last expression is valid in any consistent system of units.) We immediately see that the magnetism is an essentially relativistic phenomenon, very weak in comparison with the electrostatic interaction at the human scale velocities, v << c, and may dominate only if the latter interaction vanishes – as it does in electroneutral systems.3 The discovery and initial studies4 of such a subtle, relativistic phenomenon as magnetism were much facilitated by the relative abundance of natural ferromagnets: materials with a spontaneous magnetic polarization, whose strong magnetic field is due to relativistic effects (such as spin) inside the constituent atoms – see Sec. 5 below.
Also, Eq. (3) points to an interesting paradox. Consider two electron beams moving parallel to each other, with the same velocity v with respect to a lab reference frame. Then, according to Eq. (3), the net force of their total (electric plus magnetic) interaction is proportional to (1 – v 2/ c 2), tending to zero in the limit v c. However, in the reference frame moving together with the electrons, they are not moving at all, i.e. v = 0. Hence, from the point of view of such a moving observer, the electron beams should interact only electrostatically, with a repulsive force independent of the velocity v. Historically, this had been one of several paradoxes that led to the development of special relativity; its resolution will be discussed in Chapter 9 devoted to this theory.
Returning to Eq. (1), in some simple cases the double integration in it may be carried out analytically. First of all, let us simplify this expression for the case of two thin, long conductors (“wires”) separated by a distance much larger than their thickness. In this case, we may integrate the products j d 3 r and j ’d 3 r’ over the wires’ cross-sections first, neglecting the corresponding change of the factor (r – r ’). Since the integrals of the current density over the cross-sections of the wires are just the currents I and I’ flowing in the wires, and cannot change along their lengths (say, l and l’, respectively), they may be taken out of the remaining integrals, reducing Eq. (1) to
II'
r r
0
'
F
dr d ' r
.
(5.4)
3
4 l l'
r '
r
3 An important case when the electroneutrality may not hold is the motion of electrons in vacuum. (However, in this case, the electron speed is often comparable with the speed of light, so that the magnetic forces may be comparable in strength with electrostatic forces, and hence important.) Local violations of electroneutrality also play an important role in some semiconductor devices – see, e.g., SM Chapter 6.
4 The first detailed book on this subject, De Magnete by William Gilbert (a.k.a. Gilberd), was published as early as 1600.
Chapter 5
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As the simplest example, consider two straight, parallel wires (Fig. 2) separated by distance d, both with length l >> d.
F n F
y
dF
dF
y
I
r
d
x
r - '
r
d
Fig. 5.2. The magnetic force between
I'
two straight parallel currents.
x'
d '
r
Due to the symmetry of this system, the vector of the magnetic interaction force has to: (i) lie in the same plane as the currents, and
(ii) be normal to the wires – see Fig. 2.
Hence we may limit our calculations to just one component of the force – normal to the wires. Using the fact that with the coordinate choice shown in Fig. 2, the scalar product dr dr ’ is just dxdx’, we get
II'
sin
II'
d
0
0
F
dx dx'
dx dx'
.
(5.5)
2
2
4
d ( x x' )
4
2
2
d ( x x' ) 3/2
Now introducing, instead of x’, a new, dimensionless variable ( x – x’)/ d, we may reduce the internal integral to a table one, which we have already encountered in this course:
II'
0
d
II'
F
dx
0
dx .
(5.6)
4 d
1 3/ 2
2
2 d
The integral over x formally diverges, but it gives a finite interaction force per unit length of the wires: F
II'
0
.
(5.7)
l
2 d
Note that the force drops rather slowly (only as 1/ d) as the distance d between the wires is increased, and is attractive (rather than repulsive as in the Coulomb law) if the currents are of the same sign.
This is an important result,5 but again, the problems so simply solvable are few and far between, and it is intuitively clear that we would strongly benefit from the same approach as in electrostatics, i.e., from decomposing Eq. (1) into a product of two factors via the introduction of a suitable field. Such decomposition may be done as follows:
Lorentz
force:
F
current
(jr)B(r d 3
) r ,
(5.8)
V
5 In particular, until very recently (2018), Eq. (7) was used for the legal definition of the SI unit of current, the ampere (A), via the SI unit of force (the newton, N), with the coefficient 0 considered exactly fixed. (A brief description of the recent changes in legal metrology is given in Appendix CA.)
Chapter 5
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where the vector B is called the magnetic field. 6 In the case when it is induced by the current j ’:
r
'
r
Biot-
0
B(r)
' j( ' r)
3
d r' .
(5.9)
Savart
3
4
law
V '
r '
r
The last relation is called the Biot-Savart law,7 while the force F expressed by Eq. (8) is sometimes called the Lorentz force.8 However, more frequently the latter term is reserved for the full force, Lorentz
F qE v B,
(5.10) force:
particle
exerted by electric and magnetic fields field on a point charge q, moving with velocity v.9
Now we have to prove that the new formulation, given by Eqs. (8)-(9), is equivalent to Eq. (1).
At the first glance, this seems unlikely. Indeed, first of all, Eqs. (8) and (9) involve vector products, while Eq. (1) is based on a scalar product. More profoundly, in contrast to Eq. (1), Eqs. (8) and (9) do not satisfy the 3rd Newton’s law applied to elementary current components j d 3 r and j ’d 3 r’, if these vectors are not parallel to each other. Indeed, consider the situation shown in Fig. 3.
d 3
j r
r '
r
dB ' 0
dB ' 0 'd 3
j
r'
Fig. 5.3. The apparent violation of the
3rd Newton law in magnetism.
dF 0
dF ' 0
Here the vector j ’ is perpendicular to the vector (r – r ’), and hence, according to Eq. (9), produces a non-zero contribution dB ’ to the magnetic field directed (in Fig. 3) normally to the plane of the drawing, i.e. perpendicular to the vector j. Hence, according to Eq. (8), this field provides a non-zero contribution to F. On the other hand, if we calculate the reciprocal force F ’ by swapping the prime indices in Eqs. (8) and (9), the latter equation immediately shows that dB(r ’) j(r ’ – r) = 0, because the two operand vectors are parallel – see Fig. 3 again. Hence, the current component j ’d 3r’ does exert a force on its counterpart, while j d 3r does not.
6 The SI unit of the magnetic field is called tesla (T) – after Nikola Tesla, a pioneer of electrical engineering. In the Gaussian units, the already discussed constant 1/ c 2 in Eq. (1) is equally divided between Eqs. (8) and (9), so that in them both, the constant before the integral is 1/ c. The resulting Gaussian unit of the field B is called gauss (G); taking into account the difference of units of electric charge and length, and hence of the current density, 1 G
equals exactly 10-4 T. Note also that in some textbooks, especially old ones, B is called either the magnetic induction or the magnetic flux density, while the term “magnetic field” is reserved for the field H that will be introduced in Sec. 5 below.
7 Named after Jean-Baptiste Biot and Félix Savart who made several key contributions to the theory of magnetic interactions – in the same notorious 1820.
8 Named after Hendrik Antoon Lorentz, famous mostly for his numerous contributions to the development of special relativity – see Chapter 9 below. To be fair, the magnetic part of the Lorentz force was implicitly described in a much earlier (1865) paper by J. C. Maxwell, and then spelled out by Oliver Heaviside (another genius of electrical engineering – and mathematics!) in 1889, i.e. also before the 1895 work by H. Lorentz.
9 From the magnetic part of Eq. (10), Eq. (8) may be derived by the elementary summation of all forces acting on n >> 1 particles in a unit volume, with j = qnv – see the footnote on Eq. (4.13a). On the other hand, the reciprocal derivation of Eq. (10) from Eq. (8) with j = qv(r – r0), where r0 is the current particle’s position (so that dr0/ dt =
v), requires certain mathematical care and will be performed in Chapter 9.
Chapter 5
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Despite this apparent problem, let us still go ahead and plug Eq. (9) into Eq. (8):
r
'
r
0
3
3
F
d r
d r' (
j r) ' j( '
r )
.
(5.11)
4
3
V
V '
r '
r
This double vector product may be transformed into two scalar products, using the vector algebraic identity called the bac minus cab rule, a(bc) = b(ac) – c(ab).10 Applying this relation, with a = j, b
= j ’, and c = R r – r ’, to Eq. (11), we get
(
j r) R
R
0
3
3
0
3
3
F
d r' ' j( '
r )
d r
d r d r' (
j r) ' j( '
r )
.
(5.12)
3
3
4
R
4
R
V'
V
V
V '
The second term on the right-hand side of this equality coincides with the right-hand side of Eq. (1), while the first term equals zero because its internal integral vanishes. Indeed, we may break the volumes V and V’ into narrow current tubes – the stretched elementary volumes whose walls are not crossed by current lines (so that on their walls, jn = 0). As a result, the elementary current in each tube, dI = jdA =
jd 2 r, is the same along its length, and, just as in a thin wire, j d 2 r may be replaced with dIdr, with the vector dr directed along j. Because of this, each tube’s contribution to the internal integral in the first term of Eq. (12) may be represented as
R
1
1
dI dr
dI dr
dI dr
,
(5.13)
R 3
R
r
R
l
l
l
where the operator acts in the r-space, and the integral is taken along the tube’s length l. Due to the current continuity expressed by Eq. (4.6), each loop should follow a closed contour, and an integral of a full differential of some scalar function (in our case, of 1/ R) along such contour equals zero.
So we have recovered Eq. (1). Returning for a minute to the paradox illustrated in Fig. 3, we may conclude that the apparent violation of the 3rd Newton law was the artifact of our interpretation of Eqs.
(8) and (9) as the sums of independent elementary components. In reality, due to the dc current continuity, these components are not independent. For the whole currents, Eqs. (8)-(9) do obey the 3rd law – as follows from their already proved equivalence to Eq. (1).
Thus it is possible to break the magnetic interaction into two effects: the induction of the magnetic field B by one current (in our notation, j ’), and the effect of this field on the other current (j).
Now comes an additional experimental fact: other elementary components j d 3 r’ of the current j(r) also contribute to the magnetic field (9) acting on the component j d 3 r.11 This fact allows us to drop the prime sign after j in Eq. (9), and rewrite Eqs. (8) and (9) as
r r
0
'
B(r)
(j ' r)
d 3 r' ,
(5.14)
3
4 V'
r '
r
F (jr)B(r d 3
) r .
(5.15)
V
10 See, e.g., MA Eq. (7.5).
11 Just as in electrostatics, one needs to exercise due caution in transforming these expressions for the limit of discrete classical particles, and extended wavefunctions in quantum mechanics, to avoid the (non-existing) magnetic interaction of a charged particle with itself.
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Again, the field observation point r and the field source point r ’ have to be clearly distinguished. We immediately see that these expressions are close to, but still different from the corresponding relations of the electrostatics, namely Eq. (1.9) and the distributed-charge version of Eq. (1.6):
1
r '
r
E(r)
( ' r)
d 3 r' ,
(5.16)
3
40 V'
r '
r
F (r) E(r d 3
) r
.
(5.17)
V
(Note that the sign difference has disappeared, at the cost of the replacement of scalar-by-vector multiplications in electrostatics with cross-products of vectors in magnetostatics.)
For the frequent particular case of a thin wire of length l’, Eq. (14) may be re-written as
I
r r
0
'
B(r)
d ' r
.
(5.18)
3
4 l'
r '
r
Let us see how this formula works for the simplest case of a straight wire (Fig. 4a). The magnetic field contributions dB due to all small fragments dr ’ of the wire’s length are directed along the same line (perpendicular to both the wire and the shortest distance d from the observation point to the wire’s line), and its magnitude is
I dx'
I
dx'
d
0
0
dB
sin
.
(5.19)
2
4 r '
r
4 2
2
d x 2
2
d x 1/2
Summing up all such elementary contributions, we get
I
0
dx
I
B
0
.
(5.20)
4
( x 2 d 2 )3/ 2
2 d
(a)
dBz d
B
(b)
dB
z
r r '
dr '
d
I
0
R
r r '
dr '
I
Fig. 5.4. Calculating magnetic fields: (a) of a straight current, and (b) of a current loop.
This is a simple but important result. (Note that it is only valid for very long ( l >> d), straight wires.) It is especially crucial to note the “vortex” character of the field: its lines go around the wire, forming rings with the centers on the current line. This is in sharp contrast to the electrostatic field lines, which can only begin and end on electric charges and never form closed loops (otherwise the Coulomb force qE would not be conservative). In the magnetic case, the vortex structure of the field may be reconciled with the potential character of the magnetic forces, which is evident from Eq. (1), due to the vector products in Eqs. (14)-(15).
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Now we may readily use Eq. (15), or rather its thin-wire version
F I
r
d B(r)
,
(5.21)
l
to apply Eq. (20) to the two-wire problem (Fig. 2). Since for the second wire, the vectors dr and B are perpendicular to each other, we immediately arrive at our previous result (7), which was obtained directly from Eq. (1).
The next important example of the application of the Biot-Savart law (14) is the magnetic field at the axis of a circular current loop (Fig. 4b). Due to the problem’s symmetry, the net field B has to be directed along the axis, but each of its elementary components dB is tilted by the angle = tan-1( z/ R) to this axis, so that its axial component is
I
dr'
R
0
dB dB cos
.
(5.22)
z
2
2
4 R z 2
2
R z 1/2
Since the denominator of this expression remains the same for all wire components dr’, the integration over r ’ is easy ( dr’ = 2 R), giving finally
2
I
R
0
B
.
(5.23)
2 2
2
R z 3/2
Note that the magnetic field in the loop’s center (i.e., for z = 0),
I
B
0
,
(5.24)
2 R
is times higher than that due to a similar current in a straight wire, at distance d = R from it. This difference is readily understandable, since all elementary components of the loop are at the same distance R from the observation point, while in the case of a straight wire, all its points but one are separated from the observation point by distances larger than d.
Another notable fact is that at large distances ( z 2 >> R 2), the field (23) is proportional to z-3:
I R 2
0
0
2 m
B
,
m
with IA ,
(5.25)
2
z 3
4 z 3
where A = R 2 is the loop area. Comparing this expression with Eq. (3.13), for the particular case = 0, we see that such field is similar to that of an electric dipole (at least along its direction), with the replacement of the electric dipole moment magnitude p with the m so defined – besides the front factor.
Indeed, such a plane current loop is the simplest example of a system whose field, at distances much larger than R, is that of a magnetic dipole, with a dipole moment m – the notions to be discussed in much more detail in Sec. 4 below.
5.2. Vector potential and the Ampère law
The reader could see that the calculations of the magnetic field using Eq. (14) or (18) are still somewhat cumbersome even for the very simple systems we have examined. As we saw in Chapter 1, similar calculations in electrostatics, at least for several important highly symmetric systems, could be Chapter 5
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substantially simplified using the Gauss law (1.16). A similar relation exists in magnetostatics as well, but has a different form, due to the vortex character of the magnetic field.
To derive it, let us notice that in an analogy with the scalar case, the vector product under the integral (14) may be transformed as
j( '
r ) r
( '
r )
j( '
r )
,
(5.26)
r '
r 3
r '
r
where the operator acts in the r-space. (This equality may be readily verified by Cartesian components, noticing that the current density is a function of r ’ and hence its components are independent of r.) Plugging Eq. (26) into Eq. (14), and moving the operator out of the integral over r ’, we see that the magnetic field may be represented as the curl of another vector field – the so-called vector potential, defined as:12
B(r) A(r) ,
(5.27)
and in our current case equal to
Vector
(
j '
r )
potential
A(r) 0
d 3 r' .
(5.28)
4
r '
r
V'
Please note a beautiful analogy between Eqs. (27)-(28) and, respectively, Eqs. (1.33) and (1.38).13 This analogy implies that the vector potential A plays, for the magnetic field, essentially the same role as the scalar potential plays for the electric field (hence the name “potential”), with due respect to the vortex character of B. This notion will be discussed in more detail below.
Now let us see what equations we may get for the spatial derivatives of the magnetic field. First, vector algebra says that the divergence of any curl is zero.14 In application to Eq. (27), this means that No
B 0 .
(5.29) magnetic
monopoles
Comparing this equation with Eq. (1.27), we see that Eq. (29) may be interpreted as the absence of a magnetic analog of an electric charge, on which magnetic field lines could originate or end. Numerous searches for such hypothetical magnetic charges, called magnetic monopoles, using very sensitive and sophisticated experimental setups, have not given any reliable evidence of their existence in Nature.
Proceeding to the alternative, vector derivative of the magnetic field, i.e. to its curl, and using Eq. (28), we obtain
(
j r )
0
'
B(r)
d 3 r'
(5.30)
4
r r '
V'
This expression may be simplified by using the following general vector identity:15
c c
2
c ,
(5.31)
applied to vector c(r) j(r ’)/r – r ’: 12 In the Gaussian units, Eq. (27) remains the same, and hence in Eq. (28), 0/4 is replaced with 1/ c.
13 In Eq. (1.38), there was no real need for the additional clarification provided by the integration volume label V’.
14 See, e.g., MA Eq. (11.2).
15 See, e.g., MA Eq. (11.3).
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1
1
B 0 (j ' r)
d 3 r' 0 (j ' r 2
)
d 3 r' .
(5.32)
4
r '
r
4
r '
r
V'
V'
As was already discussed during our study of electrostatics in Sec. 3.1,
2
1
4
(r '
r ) ,
(5.33)
r '
r
so that the last term of Eq. (32) is just 0j(r). On the other hand, inside the first integral we can replace
with (- ’), where prime means differentiation in the space of the radius vectors r ’. Integrating that term by parts, we get
1
' j r '
0
2
( )
B
j (r ' )
d r'
3
d r' (
j r)
.
(5.34)
n
4
0
r r '
r r '
S '
V '
Applying this equality to the volume V’ limited by a surface S’ either sufficiently distant from the field concentration, or with no current crossing it, we may neglect the first term on the right-hand side of Eq.
(34), while the second term always equals zero in statics, due to the dc charge continuity – see Eq. (4.6).
As a result, we arrive at a very simple differential equation16
B j .
(5.35)
0
This is (the dc form of) the inhomogeneous Maxwell equation – which in magnetostatics plays a role similar to Eq. (1.27) in electrostatics. Let me display, for the first time in this course, this fundamental system of equations (at this stage, for statics only), and give the reader a minute to stare, in silence, at their beautiful symmetry – which has inspired so much of the later development of physics:
E ,
0
B , j
Maxwell
0
equations:
(5.36)
statics
E
,
B .
0
0
Their only asymmetry, two zeros on the right-hand sides (for the magnetic field’s divergence and electric field’s curl), is due to the absence in the Nature of magnetic monopoles and their currents. I will discuss these equations in more detail in Sec. 6.7, after the first two equations (for the fields’ curls) have been generalized to their full, time-dependent versions.
Returning now to our current, more mundane but important task of calculating the magnetic field induced by simple current configurations, we can benefit from an integral form of Eq. (35). For that, let us integrate this equation over an arbitrary surface S limited by a closed contour C, and apply to the result the Stokes theorem.17 The resulting expression,
Ampère
B dr
j d 2 r I
law
,
(5.37)
0 n
0
C
S
where I is the net electric current crossing surface S, is called the Ampère law.
16 As in all earlier formulas for the magnetic field, in the Gaussian units, the coefficient 0 in this relation is replaced with 4/ c.
17 See, e.g., MA Eq. (12.1) with f = B.
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As the first example of its application, let us return to the current in a straight wire (Fig. 4a).
With the Ampère law in our arsenal, we can readily pursue an even more ambitious goal than that achieved in the previous section, namely to calculate the magnetic field both outside and inside of a wire of an arbitrary radius R, with an arbitrary (albeit axially-symmetric) current distribution j() – see Fig. 5.
z
j
R
C
C
B
R
R
Fig. 5.5. The simplest application of the Ampère
law: the magnetic field of a straight current.
Selecting the Ampère-law contour C in the form of a ring of some radius in the plane normal to the wire’s axis z, we have B dr = B d, where is the azimuthal angle, so that Eq. (37) yields:
2 j( '
) 'd ' ,
for
R,
2 B
0
(5.38)
0
R
2 j( ' ) ' d ' I, for
.
R
0
Thus we have not only recovered our previous result (20), with the notation replacement d , in a much simpler way but also have been able to calculate the magnetic field’s distribution inside the wire.
In the most common particular case when the current is uniformly distributed along its cross-section, j() = const, the first of Eqs. (38) immediately yields B for R.
Another important system is a straight, long solenoid (Fig. 6a), with dense winding: n 2 A >> 1, where n is the number of wire turns per unit length, and A is the area of the solenoid’s cross-section.
I
(a)
(b)
B
C
1
B
I
l
N
C 2
Fig.
5.6.
Calculating
magnetic fields of (a) straight
and (b) toroidal solenoids.
I
From the symmetry of this problem, the longitudinal (in Fig. 6a, vertical) component Bz of the magnetic field may only depend on the distance of the observation point from the solenoid’s axis. First taking a plane Ampère contour C 1, with both long sides outside the solenoid, we get Bz(2) – Bz(1) = 0, Chapter 5
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because the total current piercing the contour equals zero. This is only possible if Bz = 0 at any outside of the solenoid, provided that it is infinitely long.18 With this result on hand, from the Ampère law applied to the contour C 2, we get the following relation for the only ( z-) component of the internal field: Bl NI ,
(5.39)
0
where N is the number of wire turns passing through the contour of length l. This means that regardless of the exact position of the internal side of the contour, the result is the same:
N
B
I nI .
(5.40)
0 l
0
Thus, the field inside an infinitely long solenoid (with an arbitrary shape of its cross-section) is uniform; in this sense, a long solenoid is a magnetic analog of a wide plane capacitor, explaining why this system is so widely used in physical experiment.
As should be clear from its derivation, the obtained results, especially that the field outside of the solenoid equals zero, are conditional on the solenoid length being very large in comparison with its lateral size. (From Eq. (25), we may predict that for a solenoid of a finite length l, the close-range external field is a factor of ~ A/ l 2 lower than the internal one.) A much better suppression of such
“fringe” fields may be obtained using toroidal solenoids (Fig. 6b). The application of the Ampère law to this geometry shows that in the limit of dense winding ( N >> 1), there is no fringe field at all (for any relation between the two radii of the torus), while inside the solenoid, at distance from the system’s axis,
NI
0
B
.
(5.41)
2
We see that a possible drawback of this system for practical applications is that the internal field does depend on , i.e. is not quite uniform; however, if the torus is relatively thin, this deficiency is minor.
Next let us discuss a very important question: how can we solve the problems of magnetostatics for systems whose low symmetry does not allow getting easy results from the Ampère law? (The examples are of course too numerous to list; for example, we cannot use this approach even to reproduce Eq. (23) for a round current loop.) From the deep analogy with electrostatics, we may expect that in this case, we could calculate the magnetic field by solving a certain boundary problem for the field’s potential – in our current case, the vector potential A defined by Eq. (28). However, despite the similarity of this formula and Eq. (1.38) for , which was noticed above, there is an additional issue we should tackle in the magnetic case – besides the obvious fact that calculating the vector potential distribution means determining three scalar functions (say, Ax, Ay, and Az), rather than just one ().
To reveal the issue, let us plug Eq. (27) into Eq. (35):
A j ,
(5.42)
0
and then apply to the left-hand side of this equation the same identity (31). The result is 18 Applying the Ampère law to a circular contour of radius , coaxial with the solenoid, we see that the field outside (but not inside!) it has an azimuthal component B, similar to that of the straight wire (see Eq. (38) above) and hence (at N >> 1) much weaker than the longitudinal field inside the solenoid – see Eq. (40).
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( A
2
) A j .
(5.43)
0
On the other hand, as we know from electrostatics (please compare Eqs. (1.38) and (1.41)), the vector potential A(r) given by Eq. (28) has to satisfy a simpler (“vector-Poisson”) equation Poisson
2
A j ,
(5.44) equation
0
for A
which is just a set of three usual Poisson equations for each Cartesian component of A.
To resolve the difference between these results, let us note that Eq. (43) is reduced to Eq. (44) if
A = 0. In this context, let us discuss what discretion we have in the choice of the potential. In electrostatics, we may add, to the scalar function ’ that satisfies Eq. (1.33) for the given field E, not only an arbitrary constant but even an arbitrary function of time:
'
f t() '
E .
(5.45)
Similarly, using the fact that the curl of the gradient of any scalar function equals zero,19 we may add to any vector function A ’ that satisfies Eq. (27) for the given field B, not only any constant but even a gradient of an arbitrary scalar function (r, t), because
(A ' ) A '
( ) A ' B .
(5.46)
Such additions, which keep the fields intact, are called gauge transformations.20 Let us see what such a transformation does to A ’:
(A '
2
) '
A .
(5.47)
For any choice of such a function A ’, we can always choose the function in such a way that it satisfies the Poisson equation 2 = –A ’, and hence makes the divergence of the transformed vector potential, A A ’ + , equal to zero everywhere,
A 0 ,
(5.48) Coulomb
gauge
thus reducing Eq. (43) to Eq. (44).
To summarize, the set of distributions A ’(r) that satisfy Eq. (27) for a given field B(r), is not limited to the vector potential A(r) given by Eq. (44), but is reduced to it upon the additional Coulomb gauge condition (48). However, as we will see in a minute, even this condition still leaves some degrees of freedom in the choice of the vector potential. To illustrate this fact, and also to get a better gut feeling of the vector potential’s distribution in space, let us calculate A(r) for two very basic cases.
First, let us revisit the straight wire problem shown in Fig. 5. As Eq. (28) shows, in this case the vector potential A has just one component (along the axis z). Moreover, due to the problem’s axial symmetry, its magnitude may only depend on the distance from the axis: A = n zA(). Hence, the gradient of A is directed across the z-axis, so that Eq. (48) is satisfied at all points. For our symmetry (/ = / z = 0), the Laplace operator, written in cylindrical coordinates, has just one term,21 reducing Eq. (44) to
19 See, e.g., MA Eq. (11.1).
20 The use of the term “gauge” (originally meaning “a measure” or “a scale”) in this context is purely historic, so the reader should not try to find too much hidden sense in it.
21 See, e.g., MA Eq. (10.3).
Chapter 5
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1 d dA
j()
.
(5.49)
0
d
d
Multiplying both sides of this equation by and integrating them over the coordinate once, we get
dA
j( ' ) 'd ' const .
(5.50)
0
d
0
Since in the cylindrical coordinates, for our symmetry, B = – dA/ d,22 Eq. (50) is nothing else than our old result (38) for the magnetic field.23 However, let us continue the integration, at least for the region outside the wire, where the function A() depends only on the full current I rather than on the current distribution. Dividing both parts of Eq. (50) by , and integrating them over this argument again, we get
I
R
A 0 ln const,
where I
2
j() d
, for R
.
(5.51)
2
0
As a reminder, we had similar logarithmic behavior for the electrostatic potential outside a uniformly charged straight line. This is natural because the Poisson equations for both cases are similar.
Now let us find the vector potential for the long solenoid (Fig. 6a), with its uniform magnetic field. Since Eq. (28) tells us that the vector A should follow the direction of the inducing current, we may start with looking for it in the form A = n A(). (This is especially natural if the solenoid’s cross-section is circular.) With this orientation of A, the same general expression for the curl operator in cylindrical coordinates yields A = n z(1/) d( A)/ d. According to Eq. (27), this expression should be equal to B – in our current case to n zB, with a constant B – see Eq. (40). Integrating this equality, and selecting such integration constant that A(0) is finite, we get
B
B
A
,
A
i.e.
n .
(5.52)
2
2
Plugging this result into the general expression for the Laplace operator in the cylindrical coordinates,24
we see that the Poisson equation (44) with j = 0 (i.e. the Laplace equation) is satisfied again – which is natural since, for this distribution, the Coulomb gauge condition (48) is satisfied: A = 0.
However, Eq. (52) is not the unique (or even the simplest) vector potential that gives the same uniform field B = nz B. Indeed, using the well-known expression for the curl operator in Cartesian coordinates,25 it is straightforward to check that each of the vector functions A ’ = n yBx and A ” = –n xBy also has the same curl, and also satisfies the Coulomb gauge condition (48).26 If such solutions do not look very natural because of their anisotropy in the [ x, y] plane, please consider the fact that they represent the uniform magnetic field regardless of its source – for example, regardless of the shape of the long solenoid’s cross-section. Such choices of the vector potential may be very convenient for some 22 See, e.g., MA Eq. (10.5) with / = / z = 0.
23 Since the magnetic field at the wire’s axis has to be zero (otherwise, being normal to the axis, where would it be directed?), the integration constant in Eq. (50) has to equal zero.
24 See, e.g., MA Eq. (10.6).
25 See, e.g., MA Eq. (8.5).
26 The axially-symmetric vector potential (52) is just a weighted sum of these two functions: A = (A ’ + A ” )/2.
Chapter 5
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problems, for example for the quantum-mechanical analysis of the 2D motion of a charged particle in the perpendicular magnetic field, giving the famous Landau energy levels.27
5.3. Magnetic energy, flux, and inductance
Considering the currents flowing in a system as generalized coordinates, the magnetic forces (1) between them are their unique functions, and in this sense, the energy U of their magnetic interaction may be considered the potential energy of the system. The apparent (but somewhat deceptive) way to derive an expression for this energy is to use the analogy between Eq. (1) and its electrostatic analog, Eq. (2). Indeed, Eq. (2) may be transformed into Eq. (1) with just three replacements:
(i) (r) ’(r ’) should be replaced with [j(r)j ’(r ’)], (ii) 0 should be replaced with 1/0, and
(iii) the sign before the double integral has to be replaced with the opposite one.
Hence we may avoid repeating the calculation made in Chapter 1, by making these replacements in Eq.
(1.59), which gives the electrostatic potential energy of the system with (r) and ’(r ’) describing the same charge distribution, i.e. with ’(r) = (r), to get the following expression for the magnetic potential energy in the system with, similarly, j’(r) = j(r):28
1
0
3
3
j r
( ) j( '
r )
U
d r d r'
.
(5.53)
j
4 2
r '
r
But this is not the unique answer! Indeed, Eq. (53) describes the proper potential energy of the system (in particular, giving the correct result for the current interaction forces) only in the case when the interacting currents are fixed – just as Eq. (1.59) is adequate when the interacting charges are fixed.
Here comes a substantial difference between electrostatics and magnetostatics: due to the fundamental fact of electric charge conservation (already discussed in Secs. 1.1 and 4.1), keeping electric charges fixed does not require external work, while the maintenance of currents generally does. As a result, Eq.
(53) describes the energy of the magnetic interaction plus of the system keeping the currents constant –
or rather of its part depending on the system under our consideration. In this situation, using the terminology already used in Sec. 3.5 (see also a general discussion in CM Sec. 1.4.), Uj may be called the Gibbs potential energy of our magnetic system.
Now to exclude from Uj the contribution due to the interaction with the current-supporting system(s), i.e. calculate the potential energy U of our system as such, we need to know this contribution.
The simplest way to do this is to use the Faraday induction law that describes this interaction and will be discussed at the beginning of the next chapter. This is why let me postpone the derivation until that point, and for now ask the reader to believe me that the removal of the interaction leads to an expression similar to Eq. (53), but with the opposite sign:
1
Magnetic
0
3
3
j r
( ) j( '
r )
U
d r d r'
,
(5.54) interaction
4 2
r '
r
energy
27 See, e.g., QM Sec. 3.2.
28 Just as in electrostatics, for the interaction of two independent current distributions j(r) and j ’(r ’), the factor ½
should be dropped.
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I will prove this result in Sec. 6.2, but actually, this sign dichotomy should not be quite surprising to the attentive reader, in the context of a similar duality of Eqs. (3.73) and (3.81) for the electrostatic energies including and excluding the interaction with the field source.
Due to the importance of Eq. (54), let us rewrite it in several other forms, convenient for various applications. First of all, just as in electrostatics, it may be recast into a potential-based form. Indeed, with the definition (28) of the vector potential A(r), Eq. (54) becomes
1
U (jr) A(r d 3
) r .
(5.55)
2
This formula, which is a clear magnetic analog of Eq. (1.60) of electrostatics, is very popular among field theorists, because it is very handy for their manipulations; it is also useful for some practical applications. However, for many calculations, it is more convenient to have a direct expression of the energy via the magnetic field. Again, this may be done very similarly to what had been done for electrostatics in Sec. 1.3, i.e. by plugging into Eq. (55) the current density expressed from Eq. (35) and then transforming it as29
1
1
1
1
U
j d 3
A r
A d 3 r
B
d 3 r
(AB) d 3 r . (5.56)
2
2
2
2
0
0
0
Now using the divergence theorem, the second integral may be transformed into a surface integral of (AB) n. According to Eqs. (27)-(28) if the current distribution j(r) is localized, this vector product drops, at large distances, faster than 1/ r 2, so that if the integration volume is large enough, the surface integral is negligible. In the remaining first integral in Eq. (56), we may use Eq. (27) to rewrite A as B. As a result, we get a very simple and fundamental formula.
1
U
B 2 d 3 r .
(5.57a)
20
Just as with the electric field, this expression may be interpreted as a volume integral of the magnetic energy density u:
Magnetic
field
U u
1
r 3
d r, with r
B2 r
u
,
(5.57b)
energy
20
clearly similar to Eq. (1.65).30 Again, the conceptual choice between the spatial localization of magnetic energy – either at the location of electric currents only, as implied by Eqs. (54) and (55), or in all regions where the magnetic field exists, as apparent from Eq. (57b), cannot be done within the framework of magnetostatics, and only the electrodynamics gives a decisive preference for the latter choice.
For the practically important case of currents flowing in several thin wires, Eq. (54) may be first integrated over the cross-section of each wire, just as was done at the derivation of Eq. (4). As before, since the integral of the current density over the k th wire's cross-section is just the current Ik in the wire, and cannot change along its length, it may be taken from the remaining integrals, giving 29 For that, we may use MA Eq. (11.7) with f = A and g = B, giving A(B) = B(A) – (AB).
30 The transfer to the Gaussian units in Eqs. (57) may be accomplished by the usual replacement 0 4, thus giving, in particular, u = B 2/8.
Chapter 5
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EM: Classical Electrodynamics
1
dr dr
U 0
I I
' ,
(5.58)
k k '
k
k
4 2 k, k'
r
l l
r
k
k '
k k '
where lk is the full length of the k th wire loop. Note that Eq. (58) is valid if all currents Ik are independent of each other, because the double sum counts each current pair twice, compensating the coefficient ½ in front of the sum. It is useful to decompose this relation as
1
U I I L ,
(5.59)
k k '
kk '
2 k, k'
where the coefficients Lkk’ are independent of the currents:
dr dr
Mutual
L 0
' ,
(5.60) inductance
kk'
k k
4
r
l l
r
coefficients
k
k '
k k '
The coefficient Lkk’ with k k’, is called the mutual inductance between current the k th and k’ th loops, while the diagonal coefficient Lk Lkk is called the self-inductance (or just inductance) of the k th loop.31 From the symmetry of Eq. (60) with respect to the index swap, k k’, it is evident that the matrix of coefficients Lkk’ is symmetric:32
L L ,
(5.61)
kk'
k'k
so that for the practically most important case of two interacting currents I 1 and I 2, Eq. (59) reads 1
1
2
2
U L I MI I L I ,
(5.62)
1 1
1 2
2 2
2
2
where M L 12 = L 21 is the mutual inductance coefficient.
These formulas clearly show the importance of the self- and mutual inductances, so I will demonstrate their calculation for at least a few basic geometries. Before doing that, however, let me recast Eq. (58) into one more form that may facilitate such calculations. Namely, let us notice that for the magnetic field induced by current Ik in a thin wire, Eq. (28) is reduced to
dr
A r
( ) 0 I
,
(5.63)
k
k
k
4
r r
l'
k
so that Eq. (58) may be rewritten as
1
U I A r dr .
(5.64)
k
k ' k
k'
2 k, k' lk
But according to the same Stokes theorem that was used earlier in this chapter to derive the Ampère law, and Eq. (27), the integral in Eq. (64) is nothing else than the magnetic field’s flux (more frequently called just the magnetic flux) through a surface S limited by the contour l : 31 As evident from Eq. (60), these coefficients depend only on the geometry of the system. Moreover, in the Gaussian units, in which Eq. (60) is valid without the factor 0/4, the inductance coefficients have the dimension of length (centimeters). The SI unit of inductance is called the henry, abbreviated H – after Joseph Henry, who in particular discovered the effect of electromagnetic induction (see Sec. 6.1) independently of Michael Faraday.
32 Note that the matrix of the mutual inductances Ljj’ is very much similar to the matrix of reciprocal capacitance coefficients pkk’ – for example, compare Eq. (62) with Eq. (2.21).
Chapter 5
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EM: Classical Electrodynamics
Magnetic
Ar dr A 2
2
d r B d r Φ
flux
(5.65)
n
n
l
S
S
– in the particular case of Eq. (64), the flux kk’ of the field induced by the k’th current through the loop of the k th current.33 As a result, Eq. (64) may be rewritten as
1
U I Φ .
(5.66)
k
kk '
2 k, k'
Comparing this expression with Eq. (59), we see that
Φ
d r L I
kk
B
,
(5.67)
k'
2
'
n
kk ' k '
Sk
This expression not only gives us one more means for calculating the coefficients Lkk’, but also shows their physical sense: the mutual inductance characterizes what part of the magnetic flux (colloquially, “what fraction of field lines”) induced by the current Ik' pierces the k th loop’s area Sk – see Fig. 7.
Sk"
S
B
k'''
k '
I
kk'
k '
Fig. 5.7. The physical sense of
S
