Introduction to Magnetostatics. The
Magnetization and Magnetic Boundary Conditions
1 Magnetization of Materials
Materials are composed of atoms. Atoms consist of electrons orbiting around a positive nu-
cleus. The electrons also spin about their axes. The combined orbital and spin motion of
electrons create a magnetic field Bi due to each electron. This is similar to the magnetic field
produced by a current loop. The equivalent loop has a magnetic dipole moment µ = ib Sn̂,
where S is the area of the loop and ib is the current.
In the absence of an external field, the individual electronic dipoles will be oriented in such
a way that the net dipole moment is zero. When an external magnetic field is applied, the
magnetic moments of the electrons themselves align themselves in the direction of the field B
so that the magnetic moment is not zero.
The magnetization M is defined as the magnetic dipole moment per unit volume. Its unit is
Ampere/meter.
If there are N atoms in a given volume ∆v, and the jth atom has a magnetic dipole moment µ j ,
then the magnetization
N
∑ µj
j=1
M = lim (1)
∆v→0 ∆v
The medium for which M is not zero everywhere is said to be magnetized.
µ = M dv0 . So the magnetic vector
For a differential volume dv0 , the magnetic moment is dµ
potential due to dµµ is
µ0 M × R̂ 0 µ0 M × R 0
dA = dv = dv (2)
4πε0 R2 4πε0 R3
R is the vector from source element to field point.
R = |r − r0 | = [(x − x0 )2 + (y − y0 )2 + (z − z0 )2 ]1/2
Taking the gradient with respect to the source co-ordinates (x0 , y0 , z0 ), we have
0 1 R
∇ = 3 (3)
R R
1
, Substituting this, we get the expression for magnetic vector potential as
0 1
Z
µ0
A= M ×∇ dv0 (4)
4π R
Now,
M
1 1
M ×∇ = ∇0 ×M
0
M − ∇0 × (5)
R R R
Substituting this result,
∇0 ×MM 0 µ0 M 0
Z Z
µ0
A= dv − ∇0 × dv (6)
4π v0 R 4π v0 R
Using the vector identity,
Z I
∇0 × Fdv0 = − F × dS (7)
v0 S0
to the second term in the expression for A, we have,
∇0 ×MM 0 µ0 M × n̂ 0
Z I
µ0
A= dv + dS (8)
4π v0 R 4π S0 R
where n̂ is a unit vector normal to the surface.
This can be written as
Jb dv0 µ0 Kb dS0
Z Z
µ0
A= + (9)
4π v0 R 4π S0 R
where Jb = ∇0 ×MM is the bound volume current density with the unit Ampere/meter2 (A/m2 )
and Kb = M × n̂ is the bound surface current density with the unit Ampere/meter (A/m).
In free space, M = 0 and
∇ × H = Jf (10)
or
∇ × B/µ0 = Jf (11)
where J f is the free volume current density.
In a material medium, since M 6= 0, this become
M = ∇ × (H +M
∇ × B/µ0 = Jf + Jb = ∇ × H + ∇ ×M M) (12)
i.e.,
M)
B = µ0 (H +M (13)
2
Magnetization and Magnetic Boundary Conditions
1 Magnetization of Materials
Materials are composed of atoms. Atoms consist of electrons orbiting around a positive nu-
cleus. The electrons also spin about their axes. The combined orbital and spin motion of
electrons create a magnetic field Bi due to each electron. This is similar to the magnetic field
produced by a current loop. The equivalent loop has a magnetic dipole moment µ = ib Sn̂,
where S is the area of the loop and ib is the current.
In the absence of an external field, the individual electronic dipoles will be oriented in such
a way that the net dipole moment is zero. When an external magnetic field is applied, the
magnetic moments of the electrons themselves align themselves in the direction of the field B
so that the magnetic moment is not zero.
The magnetization M is defined as the magnetic dipole moment per unit volume. Its unit is
Ampere/meter.
If there are N atoms in a given volume ∆v, and the jth atom has a magnetic dipole moment µ j ,
then the magnetization
N
∑ µj
j=1
M = lim (1)
∆v→0 ∆v
The medium for which M is not zero everywhere is said to be magnetized.
µ = M dv0 . So the magnetic vector
For a differential volume dv0 , the magnetic moment is dµ
potential due to dµµ is
µ0 M × R̂ 0 µ0 M × R 0
dA = dv = dv (2)
4πε0 R2 4πε0 R3
R is the vector from source element to field point.
R = |r − r0 | = [(x − x0 )2 + (y − y0 )2 + (z − z0 )2 ]1/2
Taking the gradient with respect to the source co-ordinates (x0 , y0 , z0 ), we have
0 1 R
∇ = 3 (3)
R R
1
, Substituting this, we get the expression for magnetic vector potential as
0 1
Z
µ0
A= M ×∇ dv0 (4)
4π R
Now,
M
1 1
M ×∇ = ∇0 ×M
0
M − ∇0 × (5)
R R R
Substituting this result,
∇0 ×MM 0 µ0 M 0
Z Z
µ0
A= dv − ∇0 × dv (6)
4π v0 R 4π v0 R
Using the vector identity,
Z I
∇0 × Fdv0 = − F × dS (7)
v0 S0
to the second term in the expression for A, we have,
∇0 ×MM 0 µ0 M × n̂ 0
Z I
µ0
A= dv + dS (8)
4π v0 R 4π S0 R
where n̂ is a unit vector normal to the surface.
This can be written as
Jb dv0 µ0 Kb dS0
Z Z
µ0
A= + (9)
4π v0 R 4π S0 R
where Jb = ∇0 ×MM is the bound volume current density with the unit Ampere/meter2 (A/m2 )
and Kb = M × n̂ is the bound surface current density with the unit Ampere/meter (A/m).
In free space, M = 0 and
∇ × H = Jf (10)
or
∇ × B/µ0 = Jf (11)
where J f is the free volume current density.
In a material medium, since M 6= 0, this become
M = ∇ × (H +M
∇ × B/µ0 = Jf + Jb = ∇ × H + ∇ ×M M) (12)
i.e.,
M)
B = µ0 (H +M (13)
2
