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Chapter 3. Magnetostatics
Notes:
- Most of the material presented in this chapter is taken from Jackson, Chap. 5.
3.1 Introduction
Just as the electric field vector E is the basic quantity in electrostatics, the magnetic-flux
density or magnetic induction B plays a fundamental role in magnetostatics. Another
fundamental quantity is the magnetic dipole μ, which plays a role not unlike the electric
charge in electrostatics. It is, however important to realize that, as far as we know, there
does not exist any magnetic monopoles (or charges). The two quantities, B and μ, were
early on linked through simple relations. For example, the torque N exerted by a
magnetic-flux density on a test dipole (i.e., small enough not to alter the magnetic-flux) is
given by
N = μ! B. ( 3. 1 )
It was also established that there is a connection between electrical currents and magnetic
fields. As will soon be shown, a current density J (or simply a current I ) is a source of
magnetic-flux density. Since the current density is defined as the amount of charge that
flows through a cross-section per unit of time (units of Coulombs per square meter-
second), the conservation of charge requires that the so-called continuity equation be
satisfied
! t
+ # $ J = 0 ( 3. 2 )
Equation ( 3. 2 ) implies that any decrease (increase) in charge density within a small
volume must be accompanied by a corresponding flow of charges out of (in) the surface
delimiting the volume. Because magnetostatics is concerned with steady-state currents,
we will limit ourselves (at least in this chapter) to the following equation
! " J = 0. ( 3. 3 )
3.2 The Biot and Savart Law
The aforementioned link between the magnetic induction and the current, where the latter
is a source for the former, was first investigated experimentally by Biot and Savart, and
Ampère, and shown to be
d B =
μ 0
I ( d l " x )
x
3
μ 0
I d l " $
x
where μ 0
is the permeability of vacuum (units of Henry per meter;μ 0
" 7
H/m),
d l is a small length element from a wire pointing in the direction of the current I , and x
is the vector linking d l to the location where d B is evaluated (see Figure 3. 1 ). It is
apparent from equation ( 3. 4 ) that d B is oriented perpendicular to the plane formed by
d l and x. The lines of the magnetic induction are, therefore, concentric circles around the
wire. Let’s consider, as an example, the case of an infinitely long wire carrying a current
I. Integrating over the length of the wire, we find that
B =
μ 0
I
R dl
R
2
2
( )
3 2 "#
$
μ 0
I
R
d %
cos
2
( % ) 1 + tan
2
3 2 "! 2
! 2
$
μ 0
I
R
cos( % ) d %
"! 2
! 2
$
μ 0
I
R
where we made the change of variable l = R tan(! ). This result is known as the Biot and
Savart Law.
It was also determined experimentally that each of two current carrying wires experience
a force due to the presence of the other. Since we know that a current produces a
magnetic induction, this implies that there must exists a relation with the following form
d F = I 1
d l 1
(!^ B ).^ (^3.^6 )
If B is due to another current I 2
, we find the total force F 21
felt by the wire carrying I 1
using equation ( 3. 4 ), with integrations around the two loops (or circuits) made by the
wires (and x 12
= x 1
! x 2
, see Figure 3. 2 )
Figure 3. 1 – Magnetic induction d B from a current element I d l.
It follows from this that
F
12
μ 0
I
1
I
2
d l 1
" d l 2
( ) x
12
x 12
μ 0
I
1
I
2
d l 1
" d l 2
1
x 12
= ( F
21
as would be expected from Newton’s Third Law. For the case of two infinite parallel
wires a distance d apart, equations ( 3. 5 ) and ( 3. 6 ) yield
dF
dl
μ 0
I
1
I
2
d
In general, for a current density J ( x )and a magnetic induction B ( x )we have
F = J ( x )! B ( x ) d
3
x
N = x! # J ( x )! B ( x )
d
3
x
3.3 The Equations of Magnetostatics and Ampère’s Law
In general, when dealing with a current density J ( x !) instead of a current I equation
( 3. 4 ) for the magnetic induction B ( x )takes the form
B ( x ) =
μ 0
J ( x ")
x $ x "
x $ x (^) "
d
3
x ". ( 3. 13 )
Equation ( 3. 13 ) can be expressed differently, since
B ( x ) =!
μ 0
J ( x #) $ %
x! x #
d
3
x # , ( 3. 14 )
but since
$! "( # a )
i
ijk
j
_a
k_
ijk
j
( #) a
k
ijk
j
a k
= [! # " a + #! " a ]
i
we get
B ( x ) =
μ 0
J ( x $)
x % x $
&
d
3
x $, ( 3. 16 )
if we set! = 1 x " x # and a = J ( x #), with! " J ( x #) = 0 , since the current density is
independent of x. It follows from equation ( 3. 16 ) that the magnetic induction is
divergence-less
! " B = 0 ( 3. 17 )
Equation ( 3. 17 ) is a mathematical statement on the inexistence of magnetic monopoles.
Taking the curl of the B field, remembering that! " (! " a ) =! (! # a ) $!
2
a , we get
! " B =
μ 0
J ( x $)
x % x $
&
d
3
x $
μ 0
! J ( x $) -!
x % x $
& ,
d
3
x $ % J ( x $)!
2
x % x $
& ,
d
3
x $
and using equation (1.74)
! " B =
μ 0
! J ( x %) &! %
x # x %
d
3
x % + μ 0
J ( x ). ( 3. 19 )
This equation can be further simplified by using the divergence theorem
J ( x !) " #!
x $ x!
d
3
x != # !"
J ( x !)
x $ x!
d
3
x !$
#! " J ( x !)
x $ x!
d
3
x!
J ( x !)
x $ x!
" n
d a!
since " !# J ( x !) = 0 (see equation ( 3. 3 )) and the surface of integration extends over all
space where the integrand vanishes. We therefore find that
! " B = μ 0
J ( 3. 21 )
But because of the freedom brought by the presence of! " in the equation defining the
potential vector (i.e., equation ( 3. 25 )), we can choose! " A to suit our needs. Using the
so-called Coulomb gauge , which sets! " A = 0 , we write
2
A = " μ 0
J. ( 3. 28 )
Just as
! ( x ) =
0
$ ( x %)
x & x %
d
3
x % '
,^ (^3.^29 )
is the solution to the Poisson equation (i.e.,!
2
" = # $ % 0
), the following is the solution
to equation ( 3. 28 )
A ( x ) =
μ 0
J ( x ")
x # x "
d
3
x " $
Equation ( 3. 30 ) is valid in general, as we can set! = cste due to the fact that! " A = 0
reduces to!
2
" = 0 from equation ( 3. 25 ). This is because
J ( x #)
x $ x #
%
d
3
x #=! "
J ( x #)
x $ x #
% +
d
3
x #
J ( x #)
x $ x #
% +
d
3
x #,
since "! # J ( x !) = 0. Equation ( 3. 31 ) vanishes from equation ( 3. 20 ).
3.5 The Magnetic Dipole Moment
Staying with the Coulomb gauge, and the resulting expression for the vector potential
(i.e., equation ( 3. 30 )), we want to determine the multipole term (just as we did for the
electrostatic potential) that will dictate the intensity of the vector potential at distances
large compared to that of the current source. Therefore, assuming that x! x !in equation
( 3. 30 ), we can use a Taylor series for the 1 x! x " term around x. Using equation (1.85)
we write
x! x "
x
x # x "
x
3
+!^ (^3.^32 )
Upon inserting this result in equation ( 3. 30 ), the components of the vector potential can
be approximated to
A
i
( x ) =
μ 0
x
J
i
( x ") d
3
x "
x
x
3
$ x " J i
( x ") d
3
x "
In order to transform this equation, consider the following expression
g J! #" f d
3
x "
= gJ i
i
f d
3
x "
i
J
i
( fg ) & fJ
i
i
g & fg %" i
J
i
d
3
x "
= '# "! ( J fg ) & f J! #" g & fg #"! J
d
3
x "
= fg J! n d a "
& [ f J! # " g + fg # "! J ]
d
3
x (^) ",
and if f and g are good functions, then the surface integral vanishes and we get
[ g J! #" f + f J! #" g + fg #"! J ]
d
3
x " = 0. ( 3. 35 )
We first set f = 1 and g = x! i
into equation ( 3. 35 ) to find that
J ( x !) d
3
x!
since "! # J ( x !) = 0. Second, if f = x!
i
and g = x! j
, then equation ( 3. 35 ) yields
x! i
J
j
J
i
( ) d
3
x!
Inserting this last equation in the second integral on the right-hand side of equation ( 3. 33 )
we have
x j
x! j
J
i
d
3
x!
x j
x! i
J
j
x!
j
J
( i )
d
3
x!
x j
im
jn
in
jm
( ) x!
m
J
n
d
3
x!
x j
kij
kmn
x! m
J
n
d
3
x!
x & ( x! & J ) d
3
x!
The magnetization or magnetic moment density is defined as
B =! " A
μ 0
m " x
x
3
μ 0
x
3
" ( m " x ) +
x
3
! " ( m " x )
μ 0
x " ( m " x )
x
5
x
3
! " ( m " x )
But since
x! ( m! x )
x
5
x
2
m " x ( m # x )
x
5
and
! " ( m " x ) = m (! # x ) $ x (! # m ) + ( x # !) m $ ( m # !) x
= 3 m $ m = 2 m ,
since m does not depend on the coordinates, then equation ( 3. 45 ) becomes
B ( x ) =
μ 0
3 n ( n " m ) # m
x
3
where n = x x. Just as we did for the derivation of the electric field due to an electric
dipole in section 2.3, we consider the volume integral of the magnetic induction. If the
volume (defined as a sphere of surface S and radius R ) encloses the dipole moment, it
can be seen, for the same reasons enumerated in the discussion leading to equation (2.97),
that
B ( x ) d
3
x
r < R
!
when equation ( 3. 48 ) is used. We can, however, also calculate
B ( x ) d
3
x
r < R
!
= " # A ( x ) d
3
x
r < R
!
= R
2
n # A ( x ) d $
S
!
where equation (1.31) was used, and n , this time, is the unit vector normal to the surface
S. Inserting equation ( 3. 30 ) into equation ( 3. 50 ) we get
B ( x ) d
3
x
r < R
!
μ 0
R
2
d
3
x $ J ( x $) %
n
x " x $
d &
S
!!
Using the results obtained from equations (2.101) to (2.106), we can write
B ( x ) d
3
x
r < R
!
μ 0
R
2
r <
r >
2
n " # J ( x ") d
3
x " !
μ 0
R
2
r <
r " r >
2
x " # J ( x ") d
3
x " !
Since we assumed that the current distribution is entirely contained in the sphere, then
r <
= r !and r >
= R and equation ( 3. 52 ) reduces to
B ( x ) d
3
x
r < R
!
2 μ 0
m. ( 3. 53 )
We combine this result with that from equation ( 3. 48 ) to write our final equation for the
magnetic induction due to a magnetic dipole moment
B ( x ) =
μ 0
3 n ( n " m ) # m
x
3
m $ ( x )
Going back to equation ( 3. 52 ), when the current distribution is entirely located outside
the sphere (i.e., r >
= r !and r <
= R ), we find from equation ( 3. 13 ) that
B ( x ) d
3
x
r > R
!
4 " R
3
B ( 0 ). ( 3. 55 )
3.6 The Force and Torque on, and the Potential Energy of, a Magnetic
Dipole Moment in an External Magnetic Induction
We now consider a current distribution that is subjected to an external magnetic
induction B. From equations ( 3. 12 ) we see that the force and the torque experienced by a
current distribution are proportional to the intensity of the magnetic induction. If the size
of the volume occupied by the current distribution is significantly smaller than the scale
over which B varies, it is to our advantage to use a Taylor expansion to approximate the
components of the magnetic induction around a suitable origin
B
k
( x ) =^ B
k
( 0 ) +^ x^!^ " B
k x = 0
Then, from the first of equations ( 3. 12 ), we can express the components of the force felt
by the current distribution as
f J! #" g d
3
x " $
= r " J! #" r " d
3
r " $
= r " J!
x "
r "
d
3
r " $
= J! x " d
3
r " $
and we find upon inserting this result in equation ( 3. 35 ) that
2 J! x (^) " d
3
r "
= $ r (^) "
2
%"! J d
3
x "
since "! # J = 0. The expression for the torque reduces to
N = m! B ( 0 ) ( 3. 65 )
It is interesting to note that if the magnetic dipole moment makes an angle! with the
magnetic induction, then from equation ( 3. 61 ) we have that
U =! mB cos( " ). ( 3. 66 )
If we now calculate the negative of the derivative of the potential energy with respect to
! we get
dU
d "
=! mB sin( " )
= m # B ( 0 )
= N.
We find that the dipole tends to settle (i.e., when dU d! = 0 and N = 0 ) itself parallel to
the field (i.e., when sin(! ) = 0 ). From equation ( 3. 66 ) we see that this is the lowest
energy configuration.
3.7 The Macroscopic Equations, the Magnetic Field H, and the
Boundary Conditions
Just as we introduced macroscopic equations and the electric displacement D when
studying electrostatics, we do the same here with the introduction of the magnetic field
H and the corresponding macroscopic equations of magnetostatics. So, we again define a
new set of fields and average quantities. For example, if we have a microscopic magnetic
induction B μ
that we average over a volume element! V , then the resulting
macroscopic field is
B =
! V
B
μ
( x + x ") d
3
x "
! V
Taking the divergence on both sides of this equation reveals that the macroscopic
magnetic induction is, like the its microscopic counterpart, divergence-less (see equations
(2.128) and (2.129)). That is,
! " B = 0 , ( 3. 69 )
and we can define a macroscopic vector potential A such that
B =! " A. ( 3. 70 )
If we now consider a medium made up of a large number of molecules or atoms of type
i , and each with its own magnetic moment m i
, we can then define the average
macroscopic magnetization or magnetic moment density as
M ( x ) = N
i
m i
i
!
where N i
is the average number per unit volume of particles of type i , and m i
is the
average magnetic moment at location x. The vector potential resulting from the bulk
magnetization expressed through equation ( 3. 71 ) (multiplied by the element of volume
! V containing the medium) can be calculated with equation ( 3. 43 ). In general, however,
there will not only exist a magnetization, but also a current density J ( x ) that can give
rise to a vector potential, as shown in equation ( 3. 30 ). Therefore, the total vector potential
from a small volume! V is given by
! A ( x ) =
μ 0
J ( x #)! V
x $ x #
M ( x #)! V % ( x $ x #)
x $ x #
3
Taking the limit when! V " 0 and integrating, then equation ( 3. 72 ) is transformed to
A ( x ) =
μ 0
J ( x ")
x # x "
M ( x ") $ ( x # x ")
x # x "
3
d
3
x "
μ 0
J ( x ")
x # x "
+ M ( x ") $ ,"
x # x "
d
3
x "
and upon using the general relation! "( # a ) =! # " a + #! " a and equation (1.31) to
modify the second integral on the right-hand side we have
where μ is the permeability of the medium under consideration. The ratio μ μ 0
for paramagnetic substances and < 1 for diamagnetic substances) typically differs from
unity by only a few parts in 10
5
. This simple picture is, however, not valid for every
substances. For example, ferromagnetic media have a permeability such that μ μ 0
with a non-linear relationship between B and H. The phenomenon known as hysteresis
(i.e., the lack of a one-to-one correspondence between B and H ) is commonly observed
in their response.
The boundary conditions for the components of the magnetic induction and the magnetic
field can be derived using Figure 1.3, and the divergence and Stokes’ theorems. The
continuity of the normal components of the magnetic induction at the boundary between
two media (labeled 1 and 2) can be verified using the small pillbox of Figure 1.3, which
straddles the interface between regions 1 and 2. The volume integral of the last of
equations ( 3. 79 ) reduces to (using the divergence theorem)
! " B d
3
x
V
= B " d a
S
where n is a unit vector normal to the boundary that extends from region 1 to region 2.
In the limit where the height of the box goes to zero, the surface integral of the magnetic
induction reduces to
B
2
! B
1
( ) "^ n^ =^0 ,^ (^3.^82 )
which verifies the aforementioned continuity of the normal components of the magnetic
induction at the boundary between two media. If we now consider the integral of the first
of equations ( 3. 79 ) over an open surface of contour C that also straddles the boundary
between the two media, as shown in Figure 1.3, we find (using Stokes’ theorem) that
(! " H ) # t da
S
$
= H # d l
C
!$
= J # t da
S
$
,^ (^3.^83 )
where t is a unit vector normal to the open surface and parallel to the boundary. In the
limit where the two segments of C that are parallel to n become infinitesimally short,
we get
H! d l
C
!"
= ( t # n )! H
2
$ H
1
( ) % l , ( 3. 84 )
and
J! t da
S
"
= K! t # l , ( 3. 85 )
where! l is the length of the segments of C that are parallel to the boundary, and K is
the idealized surface current density, which exist at the interface between the two regions.
Since a! ( b " c ) = b! ( c " a ), equations ( 3. 83 ), ( 3. 84 ), and ( 3. 85 ) can be shown to yield
n! H 2
" H
1
( ) = K. ( 3. 86 )
That is, the tangential components of the magnetic field are discontinuous at the
boundary, and are related to each other through the surface current density in the manner
shown in equation ( 3. 86 ).
3.8 Faraday’s Law of Induction
We now leave the domain of strictly static phenomena to consider systems where there
can be a time-dependency for some quantities. In particular we study the results obtained
by Faraday and contained in the law famously named after him. If we consider a circuit
C bounded by an open surface S threaded by a magnetic induction B , as shown in
Figure 3. 4 , we know from Faraday’s experiment that if the flux of B changes with time,
then an electric field is induced along the circuit according to Faraday’s law of
induction
E = E! " d l
C
!#
= $ k
d
dt
B " n da
S
where E is the electromotive force that causes a current to flow around the circuit, E !is
the induced electric field at the element d l , and k is a constant. The surface integral on
the right-hand side of equation ( 3. 87 ) is the magnetic flux. This law of induction can
apply to many experimental situations. For example, it could be that:
- The circuit C is unchanging but the magnetic induction is time varying (i.e.,
! B! t " 0 ).
- The circuit C is unchanging but dragged across an inhomogeneous magnetic induction
(i.e.,! B! t = 0 ).
- The circuit is deformed in time and the magnetic induction can be homogeneous and
time-independent.
In considering the different choices enumerated above, it should become clear that one
must be careful when calculating the (total) time derivative present in Faraday’s law.
Figure 3. 4 – Magnetic flux linking a circuit C.
Furthermore, if we combine to this last equation the fact that! B j
! x j
= 0 , we can also
write that! B j
v
( i )
! x j
= 0 and subtract this term to the right-hand side of equation ( 3. 92 )
without changing its content. Therefore,
dB i
dt
! B
i
! t
! x j
B
i
v
( j )
! x j
B
j
v
( i )
! B
i
! t
im
jn
in
jm
! x j
B
m
v n
! B
i
! t
kij
kmn
! x j
B
m
v n
or alternatively
d B
dt
! B
! t
+ " # ( B # v ). ( 3. 95 )
Using this result to evaluate the surface integral of equation ( 3. 87 ), we find
d
dt
B! n da
S
# B
t
+ $ % ( B % v )
! n da
S
# B
t
! n da
S
+ ( B % v )
C
! d l ,
where Stokes’ theorem was used for the last step, and the velocity v is defined as that of
the circuit. Inserting equation ( 3. 96 ) into equation ( 3. 87 ), we can transform Faraday’s law
to
$ E!^ "^ k^ ( v^ #^ B )
C
) d l = " k
* B
) n da
S
It is seen that the left-hand side of equation ( 3. 97 ) contains the contribution due to the
motion of the circuit. However, for a stationary observer not moving with the circuit, the
only term that can be responsible for the induction of an electric field E will be that due
to the possible time dependency of the magnetic induction. This is exactly the term
present on the right-hand side of equation ( 3. 97 ). Therefore, we can also write
E
C
" d l = # k
$ B
$ t
" n da
S
Equating the left-hand sides of the last two equations we get
E! = E + k ( v " B ). ( 3. 99 )
To evaluate the constant k we go back to more restrictive conditions where the magnetic
induction is homogeneous and time independent (i.e.,! B! t = 0 ), but we still consider a
circuit moving with a velocity v relative to a stationary observer. From equation ( 3. 97 ),
we see that force F !experienced by a charge q at rest in the moving circuit is given by
F! = q E! = kq ( v " B ). ( 3. 100 )
On the other hand, since the charge is moving at velocity v relative to the frame of
reference of the stationary observer, it is seen by this observer to be equivalent to a
current density J = q v! x " x 0
(^) ( ) ( x 0
is the instantaneous position of the charge). Then,
using the first of equations ( 3. 12 ) for the force F measured by this observer, we find that
F = q v! x " x 0
( ) #^ B^ ( x ) d
3
x $^0
= q v # B x 0
( ).
However, because the position of the charge in two frames of reference is related through
a coordinate transformation that is a linear function of time (i.e., x 0
= x! 0
the force is proportional to the second derivative of the position, therefore it must be that
F = F!. ( 3. 102 )
Equating equations ( 3. 100 ) and ( 3. 101 ), we find that k^ =^1 (in SI units). Faraday’s law of
induction then becomes
E! " d l
C
!#
d
dt
B " n da
S
and we also obtain, as a result of this analysis, an expression for the force F acting on a
charge q moving at a velocity v , and subjected to an electric field E and a magnetic
induction B. That is, we have the so-called Lorentz force
F = q " E + ( v! B )
We can put Faraday’s law in a differential form using equation ( 3. 98 ) (with k = 1 ), since
E and B must be evaluated in the same frame of reference, and Stokes’ theorem. Then,
! " E +
# B
t
S
but since the circuit C and the unit normal vector n are arbitrary, we can write
