UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 20.5
Prof. Steven Errede
LECTURE NOTES 20.5
Magnetostatic Boundary Value Problems in Magnetic Media:
Examples, Applications and Uses
Example # 1:
Use the magnetic scalar potential Vm for a magnetic sphere in a uniform external magnetic field.
Consider a sphere of radius R made of (arbitrary / unspecified) linear magnetic material of
magnetic permeability μ = μo (1 + χ m ) placed in the gap of a big electromagnet that produces a
uniform external magnetic field Bext = Bo zˆ as shown in the figure below:
Note that this magnetostatics problem of a linear magnetic sphere of radius R and magnetic
permeability μ = μo (1 + χ m ) = K m μo placed in an external uniform magnetic field Bext = Bo zˆ is
highly analogous to the electrostatics boundary value problem that we solved earlier (Griffiths
Example 4.7, pp. 186-8) with a linear dielectric sphere of radius R and dielectric permittivity
ε = ε o (1 + χ e ) = K eε o placed in an external uniform electric field Eext = Eo zˆ . Here, we will use
the magnetic scalar potential Vm to solve this magnetic boundary value problem. Note
(aforehand) that this problem (like that of the dielectric sphere) is manifestly azimuthally
symmetric – i.e. it has no explicit ϕ -dependence.
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
1
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 20.5
Prof. Steven Errede
Since there are no free currents/free current densities anywhere in the volume v′ of interest
(i.e. J free ( r ) = 0 ), then ∇ × H ( r ) = 0 , and thus we can write H ( r ) ≡ −∇Vm ( r )
where Vm ( r ) = magnetic scalar potential.
Again, note that since the SI units of H ( r ) are Amperes/meter → then the SI units of Vm ( r ) are
Amperes!!
n.b. This sort of makes “nice” sense, since for electrostatics, the SI units of VE ( r ) = Volts
Then since ∇ × H ( r ) = −∇ × ∇Vm ( r ) ≡ 0 here in this problem, and
∇i H ( r ) = −∇i∇Vm ( r ) = −∇ 2Vm ( r ) = −∇iΜ ( r ) = − ρ m ( r )
Now in the volume v′ , we have uniform magnetization: Μ ( r ) = Μ o zˆ (here)
∴ ∇iΜ ( r ) = ∇iΜ o zˆ = Μ o ∇i zˆ = 0 , i.e. ρ m ( r ) = 0 (here)
Again, extreme caution must be used here, for we know that differential relations will fail on the
boundaries / interfaces of dissimilar materials.
Nevertheless, away from these boundaries / interfaces:
∇i H ( r ) = −∇iΜ ( r ) = −∇ 2Vm ( r ) = 0
∇ 2Vm ( r ) = 0 is Laplace’s Equation for the Magnetic Scalar Potential Vm ( r )
We can/will use all the tools that we developed for solving electrostatics boundary-value
problems here too, for solving magnetostatics boundary-value problems!!!
We will use the magnetostatic boundary conditions (derived/obtained from) the integral relations
(given below) at the interface(s)/boundar(ies) of the magnetic material, in order to constrain the
allowed form of the magnetic scalar potential Vm ( r ) in various regions of v′ as well as at
boundaries / interfaces.
∫ ( ∇ × H ( r ) )ida = ∫ H ( r )id = I
∫ ( ∇ × Μ ( r ) )ida = ∫ Μ ( r )id = I
μ ∫ ( ∇ × B ( r ) )i da = μ ∫ B ( r )i d = I
S
S
1
o
C
enclosed
free
C
enclosed
Bound
1
S
o
C
∫ ( ∇i H ( r ) )idτ = ∫ H ( r )ida
= ∫ ( ∇iΜ ( r ) )idτ = ∫ Μ ( r )ida ≠ 0 in general
=0
∫ ( ∇i B ( r ) )idτ = ∫ B ( r )ida = Φ
and:
enclosed
Tot
v
S
v
S
v
S
enclosed
m
enclosed
enclosed
= I enclosed
+ I Bound
where ITot
free
2
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 20.5
Prof. Steven Errede
The most general solution for the magnetostatic version of Laplace’s Equation ( ∇ 2Vm ( r ) = 0 )
for the magnetic scalar potential, for problems with spherical symmetry and additionally ones
that also have manifest / explicit azimuthal symmetry (i.e. no ϕ -dependence), with cos Θ′ ≡ rˆirˆ′ ,
and choosing the origin (here) to be at the center of the magnetic sphere, then cos Θ′ → cos θ
(where θ = usual polar angle) and r = r − r ′ → r , then Vm ( r , θ ) is given by:
∞
B ⎞
⎛
Vm ( r , θ ) = ∑ ⎜ A r + +1 ⎟P ( cos θ )
r ⎠
=0 ⎝
where P ( cos θ ) is the “ordinary” Legendré Polynomial of order .
The boundary conditions for this magnetostatics problem parallel those (not identically
though!!) for the analogous electrostatics problem – that of a linear dielectric sphere of radius R
and linear electric permittivity ε = ε o (1 + χ e ) = K eε o - in uniform electric field Eext = Eo zˆ
(see / refer to P435 Lecture Notes 11).
The boundary conditions that we have here for the magnetic sphere of radius R and linear
magnetic permeability μ = μo (1 + χ m ) = K m μo in a uniform magnetic field Bext = Bo zˆ are:
0) Vm ( r ) = finite ∀ r in the volume v′
1) Vminside ( r = R ) = Vmoutside ( r = R ) ⇐ Vm is continuous / single-valued at / across interface /
boundary at r = R.
2) Vminside ( z = 0 ) = Vmoutside ( z = 0 ) = 0 (i.e. the x-y mid-plane = magnetic scalar equipotential due to
the symmetry of problem (see picture on page 1))
π⎞
π⎞
⎛
⎛
Because z = r cos θ this BC also says: Vminside ⎜ r ,θ = ⎟ = Vmoutside ⎜ r ,θ = ⎟ = 0
2⎠
2⎠
⎝
⎝
upper (south!)
3) Vmoutside ( z = ± Lgap ) = Vmoutside ( r cos θ = ± Lgap ) = ±Vmo { lower
(north!) } magnetic poles of the external
magnet
4) Far away from the magnetic sphere (r >> R) we demand:
B outside ( r R ) = Bext = Bo zˆ where zˆ = ⎡ rˆ cos θ − θ sin θ ⎤ in spherical polar coordinates.
⎣
⎦
In the region exterior to the magnetized sphere (r > R):
1 outside
B
( r > R ) = H outside ( r > R ) = −∇Vmout ( r > R )
μo
Thus:
1
μo
B outside ( r
= H outside ( r
R) =
1
μo
Bext =
1
μo
Bo zˆ =
1
μo
Bo ⎡ rˆ cos θ − θ sin θ ⎤
⎣
⎦
R ) = H ext = H o zˆ = H o ⎡ rˆ cos θ − θ sin θ ⎤ = −∇Vmout ( r
⎣
⎦
R)
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
3
UIUC Physics 435 EM Fields & Sources I
Then: H outside ( r
R) =
= H ext
=
= H o zˆ
=
1
μo
1
μo
1
μo
Fall Semester, 2007
B outside ( r
Lecture Notes 20.5
Prof. Steven Errede
R)
Bext
Bo zˆ
⎛
⎞
1
Bo ⎟
⎜ Ho ≡
μo ⎠
⎝
←
1
= H o ⎡ rˆ cos θ − θ sin θ ⎤ =
B ⎡ rˆ cos θ − θ sin θ ⎤
⎣
⎦ μ o⎣
⎦
o
= −∇Vmoutside ( r
∴
R)
−∇Vmoutside ( r
We see that: Vmoutside ( r
1
R ) = H o ⎡ rˆ cos θ − θ sin θ ⎤ =
B ⎡ rˆ cos θ − θ sin θ ⎤
⎣
⎦ μ o⎣
⎦
o
R ) = − H o r cos θ satisfies this requirement / condition.
Explicit check:
−∇Vmoutside ( r
⎡∂
1 ∂
1
R ) = − ⎢ rˆ +
θ+
r ∂θ
r sin θ
⎣ ∂r
⎡∂
1 ∂
1
= + ⎢ rˆ +
θ+
r ∂θ
r sin θ
⎣ ∂r
∂ ⎤ outside
ϕˆ Vm
(r
∂ϕ ⎥⎦
R)
∂ ⎤
ϕˆ H o r cos θ
∂ϕ ⎥⎦
= H o ⎡ rˆ cos θ − θ sin θ ⎤ = H o zˆ
⎣
⎦
⊥
⊥
5) Boutside
( r = R ) = Binside
( r = R ) ⇒ Broutside ( r = R ) = Brinside ( r = R )
( ⊥= r̂ direction at r = R interface / boundary)
6) H outside ( r = R ) − H inside ( r = R ) = K free × nˆ surface = K free × rˆ surface = 0 ( K free = 0 here)
7) Boutside ( r = R ) − Binside ( r = R ) = μo BTot × nˆ
8) ( H
surface
= μo KTOT × rˆ
surface
= μo K bound × rˆ
( r = R ) − H ( r = R )) = − (Μ ( r = R ) − Μ ( r = R ))
( r = R ) − Μ inside
( r = R ))
( H routside ( r = R ) − H rinside ( r = R ) ) = − ( Μ outside
r
r
⊥
outside
⊥
inside
⊥
outside
⊥
inside
surface
( ⊥= r̂ direction at r = R
interface / boundary)
Note also that because of the manifest / intrinsic odd reflection symmetry associated with this
problem (as we saw for the dielectric sphere in uniform external electric field problem) about the
z = 0 midplane (i.e. z → -z), namely that Vm ( − z ) = −Vm ( + z ) {i.e. because of the corresponding
θ → −θ reflection symmetry properties associated with the Legendré Polynomials themselves –
Ρ ( −θ ) = ( −1) Ρ (θ ) } we anticipate / know in advance / expect that all even- P ( cos θ )
terms must vanish – i.e. only odd- P ( cos θ ) terms will be present in Vm ( r , θ ) due to the
manifest / intrinsic odd reflection symmetry associated with this problem!
4
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 20.5
Prof. Steven Errede
Again, the general solution for the magnetostatic version of Laplace’s Equation (in sphericalpolar coordinates) is:
∞
B ⎞
⎛
Vm ( r , θ ) = ∑ ⎜ A r + +1 ⎟P ( cos θ )
r ⎠
=0 ⎝
Apply BC 0): Inside the magnetic sphere ( r ≤ R ) we demand:
Vminside ( r ≤ R ) must be finite ∀ r ≤ R
→ All B = 0 ∀
in the region r ≤ R (inside the magnetic sphere)
∞
Thus: Vminside ( r ≤ R ) = ∑ A r P ( cosθ )
=0
Apply BC 0): Outside the magnetic sphere ( r ≥ R ) we must allow both r and
1
terms,
r +1
because the region r = ∞ is formally excluded in this problem (when x, y → ∞ at
π
π⎞
⎛
the midplane region, simultaneously θ → and Vm ⎜ θ = ⎟ = 0 , automatically
2
2⎠
⎝
satisfied for all odd- P ( cos θ ) terms)!!!
∞
B′ ⎞
⎛
Thus: Vmoutside ( r ≥ R ) = ∑ ⎜ A′ + +1 ⎟P ( cos θ )
r ⎠
=0 ⎝
Next, we apply BC 4), namely that for r
B out ( r R ) = Bext = Bo zˆ
R , i.e. far from the magnetic sphere:
H outside ( r R ) = H ext = H o zˆ = −∇Vmoutside ( r
B outside ( r
R ) = μo H outside ( r
= Bext ( r
R ) = μo H ext ( r
= Bo zˆ
We showed that Vmout ( r
R)
R)
R)
= μo H o zˆ
R ) = − μ1o Bo r cos θ = − μ1o Bo z = − H o r cos θ = − H o z satisfies this
boundary condition ( z = r cos θ in spherical coordinates).
Apply BC 3): We also want: Vmout ( z = ± Lgap ) = ±Vmo on
{
upper (south!)
lower (north!)
} poles of external magnet
⎛ Vo ⎞
⎛ Vo ⎞
⎛ Vo ⎞
R ) = ⎜ m ⎟ z = ⎜ m ⎟ r cos θ = ⎜ m ⎟ rP1 ( cos θ )
⎜ Lgap ⎟
⎜ Lgap ⎟
⎜ Lgap ⎟
⎝
⎠
⎝
⎠
⎝
⎠
o
⎛V ⎞ 1
Bo
Vmo = − H o Lgap
or:
Thus: H o = − ⎜⎜ m ⎟⎟ =
⎝ Lgap ⎠ μo
⎛ Vmo ⎞
1
Vmo = − Bo Lgap (we’ll need these later…)
Bo = − μo ⎜
or:
⎜ Lgap ⎟⎟
μo
⎝
⎠
Thus: Vm ( r
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
5
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 20.5
Prof. Steven Errede
∞
B′ ⎞
⎛
So, if Vmoutside ( r ≥ R ) = ∑ ⎜ A′r + +1 ⎟P ( cos θ )
r ⎠
=0 ⎝
⎛ Vmo ⎞
outside
V
r
R
If for r >> R, m
(
) = ⎜⎜ ⎟⎟ rP1 ( cos θ )
⎝ Lgap ⎠
then we know that / demand that all A ′ vanish (i.e. all A ′ = 0)
except the
So now: V
outside
m
Vmo
′
= 1 term: A1 =
Lgap
⎛ Vmo
( r ≥ R ) = ⎜⎜
⎝ Lgap
∞
⎞
⎛ B′ ⎞
⎟⎟ r cos θ + ∑ ⎜ +1 ⎟P ( cos θ )
⎠
=0 ⎝ r
⎠
Now apply BC 1): Vminside ( r = R ) = Vmoutside ( r = R ) ⇐ Vm ( r ) is continuous across the boundary /
interface
⎛ Vmo
∴ at r = R : ∑ A R P ( cos θ ) = ⎜
⎜L
=0
⎝ gap
∞
∞
⎞
B′
⎟⎟ R cos θ + ∑ +1 P ( cos θ )
⎠ P1 ( cosθ ) =0 R
Thus by the method of inspection, we see that, because of the orthogonality properties of the
P ( cos θ ) , all A and B ′ coefficients must vanish except the = 1 terms:
⎛ Vmo
i.e. A1 R cos θ = ⎜⎜
⎝ Lgap
or:
⎛ Vo
A1 = ⎜ m
⎜L
⎝ gap
⎞
B′
⎟⎟ R cos θ + 12 cos θ
R
⎠
⎞ B1′
⎟⎟ + 3
⎠ R
all other A =B ′ = 0
Then: Vminside ( r ≤ R ) = A1r cos θ
outside
m
V
⎛ Vmo
( r ≥ R ) = ⎜⎜
⎝ Lgap
⎞
B′
⎟⎟ r cos θ + 21 cos θ
r
⎠
We still have one remaining unknown – e.g. B1′ . Thus, we need to apply one more boundary
condition in order to obtain an independent relationship between A1 and B1′ .
Let’s choose BC 5):
Broutside ( r = R ) = Brinside ( r = R )
(i.e. here radial normal components of B are continuous across an interface)
Now: B outside ( r ≥ R ) = μo H outside ( r ≥ R )
B inside ( r ≤ R ) = μ H inside ( r ≤ R )
6
and
μ = μo (1 + χ m ) = μo K m
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 20.5
Prof. Steven Errede
∴ BC 5) also says: μo H routside ( r = R ) = μ H routside ( r = R )
But H ( r ) = −∇Vm
∂Vm
in spherical polar coordinates
∂r
∂Vminside ( r )
= −μ
∂r
r =R
H r = −∇ rVm = −
∂Vmoutside ( r )
∴ BC 5) also says: − μo
∂r
r=R
⎛ Vo ⎞
B′
Vmoutside ( r ≥ R ) = ⎜ m ⎟ r cos θ + 12 cos θ
⎜L ⎟
r
⎝ gap ⎠
⎡⎛ V o ⎞
⎤
⎛ Vo ⎞
2B ′
2B ′
∴ − μo ⎢⎜ m ⎟ cos θ − 31 cos θ ⎥ = − μ A1 cos θ
or:
μo ⎜ m ⎟ − μo 31 = μ A1
⎜L ⎟
R
R
⎢⎣⎜⎝ Lgap ⎟⎠
⎥⎦
⎝ gap ⎠
⎛ Vmo ⎞ B1′
μo ⎡⎛ Vmo ⎞ 2 B1′ ⎤
∴ A1 =
A1 = ⎜
+
(from BC 1))
⎢⎜
⎥ but:
⎟−
⎜ Lgap ⎟⎟ R 3
μ ⎣⎢⎜⎝ Lgap ⎟⎠ R 3 ⎦⎥
⎝
⎠
⎛ V o ⎞ B′ ⎛ μ ⎞ ⎛ V o ⎞ ⎛ μ ⎞ B′
∴ ⎜ m ⎟ + 13 = ⎜ o ⎟ ⎜ m ⎟ − 2 ⎜ o ⎟ 13 Solve for B1′ .
⎜ Lgap ⎟ R ⎝ μ ⎠ ⎜ Lgap ⎟ ⎝ μ ⎠ R
⎝
⎠
⎝
⎠
Now: Vminside ( r ≤ R ) = A1r cos θ
and
⎞ ⎤ ⎛ Vmo ⎞
⎟
⎟ − 1⎥ ⎜⎜
⎠ ⎦ ⎝ Lgap ⎟⎠
⎡
⎛ μo ⎞ ⎤ B1′ ⎡⎛ μo ⎞ ⎤ ⎛ Vmo ⎞
1
2
+
⎟⎟
⎢
⎜ ⎟ ⎥ 3 = ⎢⎜ ⎟ − 1⎥ ⎜⎜
⎝ μ ⎠⎦ R
⎣
⎣⎝ μ ⎠ ⎦ ⎝ Lgap ⎠
⎛ μo ⎞
⎜ μ − 1⎟ ⎛ V o ⎞
⎛ μ − μ ⎞ ⎛ Vmo
⎝
⎠
3
B1′ =
⎜⎜ m ⎟⎟ R = ⎜ o
⎟⎜
μ + 2 μo ⎠ ⎜⎝ Lgap
⎛
μo ⎞ ⎝ Lgap ⎠
⎝
⎜1 + 2 μ ⎟
⎝
⎠
⎛ μ ⎞ B′ ⎡⎛ μ
B1′
+ 2 ⎜ o ⎟ 13 = ⎢⎜ o
3
R
⎝ μ ⎠R
⎣⎝ μ
⎞ 3
⎛ μ − μo
⎟⎟ R = − ⎜
⎝ μ + 2 μo
⎠
⎞ ⎛ Vmo
⎟ ⎜⎜
⎠ ⎝ Lgap
⎞ 3
⎟⎟ R
⎠
We assume μ > μo {doesn’t really matter...}
⎛ μ − μo ⎞ ⎛ Vmo
B1′ = − ⎜
⎟ ⎜⎜
⎝ μ + 2μo ⎠ ⎝ Lgap
⎞ 3
⎟⎟ R
⎠
⎛ Vmo ⎞ B1′ ⎛ Vmo ⎞ ⎡
( μ − μo ) ⎤
1−
+ 3 =⎜
Then: A1 = ⎜
⎟
⎟
⎢
⎥
⎜ Lgap ⎟ R ⎜ Lgap ⎟
⎝
⎠
⎝
⎠ ⎣ ( μ + 2 μo ) ⎦
⎡ μ + 2 μo − μ + μo ⎤ ⎛ Vmo ⎞ ⎛ 3μo
A1 = ⎢
=
⎥ ⎜⎜
⎟⎟ ⎜
μ + 2 μo
⎣⎢
⎦⎥ ⎝ Lgap ⎠ ⎝ μ + 2μo
⎛ 3μo ⎞ ⎛ Vmo
A1 = ⎜
⎟ ⎜⎜
⎝ μ + 2μo ⎠ ⎝ Lgap
⎞ ⎛ Vmo
⎟ ⎜⎜
⎠ ⎝ Lgap
⎞
⎟⎟
⎠
⎞
⎟⎟
⎠
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
7
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 20.5
Prof. Steven Errede
Thus, finally we now have the fully-specified magnetic scalar potentials:
⎛ 3μo
Vminside ( r ≤ R ) = ⎜
⎝ μ + 2 μo
outside
m
V
⎞ ⎛ Vmo
⎟ ⎜⎜
⎠ ⎝ Lgap
⎞
⎛ 3 ⎞ ⎛ Vmo ⎞
⎟⎟ r cos θ = ⎜
⎟⎟ z SI Units = Amperes for Vm? Yes!
⎟ ⎜⎜
K
2
L
+
m
gap
⎝
⎠
⎠
⎝
⎠
since z = r cos θ and K m ≡ μ μo = (1 + χ m )
⎛ Vmo ⎞
⎛ μ − μo ⎞ ⎛ Vmo ⎞ ⎛ R 3 ⎞
( r ≥ R ) = ⎜⎜ ⎟⎟ r cos θ − ⎜
⎟⎟ ⎜ 2 ⎟ cos θ
⎟ ⎜⎜
⎝ μ + 2μo ⎠ ⎝ Lgap ⎠ ⎝ r ⎠
⎝ Lgap ⎠
SI Units = Amperes for Vm? Yes!
⎛ Vmo ⎞ ⎡ ⎛ K m − 1 ⎞⎛ R 3 ⎞
⎤
cos θ ⎥
z−
=⎜
⎜ L ⎟⎟ ⎢ ⎜ K + 2 ⎟⎜⎝ r 2 ⎟⎠
gap
m
⎝
⎠
⎣
⎦
⎝
⎠
Thus, we see that inside the magnetic sphere the magnetic scalar potential Vminside ( r ≤ R )
increases linearly with z, whereas outside the magnetic sphere the magnetic scalar potential
Vmoutside ( r ≥ R ) is the sum {i.e. linear superposition} of two terms, one which increases linearly
with z, and another term which corresponds to the {magnetic scalar} potential associated with a
{magnetic} dipole. The linear dependence of the magnetic scalar potential arises from the
uniform external magnetic field Bext = Bo zˆ , and the dipole term in the external magnetic scalar
potential arises simply from the magnetic dipole moment m = 43 π R 3Μ associated with the
magnetized sphere! Note that for z = r cos (θ = π 2 ) = 0 that Vminside ( z = 0 ) = Vmoutside ( z = 0 ) = 0 ,
i.e. the magnetic scalar potential Vn ( z = 0 ) on the horizontal x-y plane in the middle of the gap of
the electromagnet is an equi-“potential” of 0 Amperes. We also see that on the surface of the
⎛ 3 ⎞ ⎛ Vmo ⎞
sphere, Vminside ( r = R ) = Vmoutside ( r = R ) = ⎜
⎟⎟ R cos θ
⎟ ⎜⎜
⎝ K m + 2 ⎠ ⎝ Lgap ⎠
and that Vmoutside ( z = ± Lgap ) = ±Vmo for r R .
⎛ Vo
Now recall that: ⎜⎜ m
⎝ Lgap
And since:
⎞
1
⎟⎟ = − H o = − Bo ⇐ (for Bext = Bo zˆ ) SI Units = Amps/meter for H
μo
⎠
⎧∂
1 ∂
1
∂ ⎫
H ( r ) ≡ −∇Vm ( r ) = − ⎨ rˆ +
θ+
ϕˆ ⎬Vm ( r )
r ∂θ
r sin θ ∂ϕ ⎭
⎩ ∂r
Then:
⎛ 3 ⎞ ⎛ Vmo ⎞
⎛ 3 ⎞ ⎛ Vmo ⎞
⎡
⎤
ˆ
θ
θ
θ
H inside ( r ≤ R ) = −∇Vminside ( r ≤ R ) = − ⎜
r
cos
sin
−
=
−
⎜
⎟
⎟⎟ zˆ
⎟⎜
⎜
⎟ ⎜⎜
⎦
⎟⎣
⎝ K m + 2 ⎠ ⎝ Lgap ⎠
⎝ K m + 2 ⎠ ⎝ Lgap ⎠
⎛ 3 ⎞ ⎛ Vmo ⎞
⎛ 3 ⎞
1 ⎛ 3 ⎞
H inside ( r ≤ R ) = − ⎜
⎟⎟ zˆ = + ⎜
⎟ ⎜⎜
⎟ H o zˆ = ⎜
⎟ B zˆ K m ≡ μ μo = (1 + χ m )
μo ⎝ K m + 2 ⎠ o
⎝ K m + 2 ⎠ ⎝ Lgap ⎠
⎝ Km + 2 ⎠
8
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 20.5
Prof. Steven Errede
And:
H outside ( r ≥ R ) = −∇Vmoutside ( r ≥ R )
⎛ Vo ⎞
⎛ K − 1 ⎞ ⎛ Vmo ⎞ ⎡ −2 R 3
⎤
R3
ˆ
r
cos
θ
θ sin θ ⎥
= − ⎜ m ⎟ ⎡ rˆ cos θ − θ sin θ ⎤ + ⎜ m
−
⎜
⎟
⎟
⎢
3
3
⎦
⎜ L ⎟⎣
⎜
⎟
r
⎦
⎝ K m + 2 ⎠ ⎝ Lgap ⎠ ⎣ r
⎝ gap ⎠
3
⎫⎪
⎛ V o ⎞ ⎧⎪ ⎛ K − 1 ⎞ ⎛ R ⎞
⎡
⎤
ˆ
θ
θ
θ
H outside ( r ≥ R ) = − ⎜ m ⎟ ⎨ zˆ + ⎜ m
r
−
2
cos
sin
⎬
⎟⎜ ⎟ ⎣
⎦
⎜ Lgap ⎟ ⎪
⎝
⎠ ⎩ ⎝ Km + 2 ⎠ ⎝ r ⎠
⎭⎪
⎛ Vmo ⎞
1
But:
⎜⎜
⎟⎟ = − H o = − Bo
μo
⎝ Lgap ⎠
3
⎛ K −1 ⎞ ⎛ R ⎞
H outside ( r ≥ R ) = + H o zˆ + ⎜ m
⎟ ⎜ ⎟ H o ⎡⎣ 2rˆ cos θ − θ sin θ ⎤⎦
⎝ Km + 2 ⎠ ⎝ r ⎠
or:
3
⎛ 1 ⎞
⎛ K −1 ⎞ ⎛ R ⎞ ⎛ 1 ⎞
H outside ( r ≥ R ) = ⎜ ⎟ Bo zˆ + ⎜ m
⎟ ⎜ ⎟ ⎜ ⎟ Bo ⎡⎣ 2rˆ cos θ − θ sin θ ⎤⎦ SI units: Amps/meter
⎝ μo ⎠
⎝ K m + 2 ⎠ ⎝ r ⎠ ⎝ μo ⎠
Thus, we see that the H-field inside the magnetized sphere is constant/uniform, whereas the
H-field outside the magnetized sphere is the linear superposition of the H-field associated with
the constant/uniform externally-applied magnetic field H ext ( r ) = μ1o Bext ( r ) = μ1o Bo zˆ and the Hfield associated with the magnetic dipole moment m = 43 π R 3Μ of the magnetized sphere!
⎛ 3 ⎞⎛ μ ⎞
⎛ 3K m ⎞
Then: B inside ( r ≤ R ) = μ H inside ( r ≤ R ) = ⎜
⎟ ⎜ ⎟ Bo zˆ = ⎜
⎟ Bo zˆ but note that:
⎝ K m + 2 ⎠ ⎝ μo ⎠
⎝ Km + 2 ⎠
3 (1 + χ m )
⎛ 3K m ⎞
⎛ 1 + χm ⎞
B inside ( r ≤ R ) = ⎜
Bo zˆ = ⎜
⎟ Bo zˆ =
⎟ Bo zˆ SI units = Teslas
(3 + χm )
⎝ Km + 2 ⎠
⎝ 1 + χm 3 ⎠
n.b. This is the same answer as Griffiths Problem 6.18 – it better be the same!!!
3
And:
⎛ K −1 ⎞ ⎛ R ⎞
B outside ( r ≥ R ) = μ H outside ( r ≥ R ) = Bo zˆ + ⎜ m
⎟ ⎜ ⎟ Bo ⎡⎣ 2rˆ cos θ + θ sin θ ⎤⎦
K
+
2
⎝ m
⎠⎝ r ⎠
3
⎛ K −1 ⎞ ⎛ R ⎞
B outside ( r ≥ R ) = Bo zˆ + ⎜ m
⎟ ⎜ ⎟ Bo ⎡⎣ 2rˆ cos θ + θ sin θ ⎤⎦
⎝ Km + 2 ⎠ ⎝ r ⎠
⎛ μ
but: K m = ⎜
⎝ μo
⎞
⎟ = (1 + χ m )
⎠
3
Or:
⎛ 1 ⎞ ⎛ χm ⎞ ⎛ R ⎞
B outside ( r ≥ R ) = Bo zˆ + ⎜ ⎟ ⎜
⎟ ⎜ ⎟ Bo ⎡⎣ 2rˆ cos θ + θ sin θ ⎤⎦ SI units = Teslas
⎝ 3 ⎠ ⎝ 1 + χm 3 ⎠ ⎝ r ⎠
Thus, again we see that the B-field inside the magnetized sphere is constant/uniform, whereas
the B-field outside the magnetized sphere is the linear superposition of the B-field associated
with the constant/uniform externally-applied magnetic field Bext ( r ) = Bo zˆ and the B-field
associated with the magnetic dipole moment m = 43 π R 3Μ of the magnetized sphere!
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
9
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 20.5
Prof. Steven Errede
Then since:
H inside ( r ) =
1
μo
B inside ( r ) − Μ ( r ) or: Μ ( r ) =
1
μo
B inside ( r ) − H inside ( r ) and: B inside ( r ) = μ H inside ( r )
⎛ μ
⎞
∴ Μ ( r ) = ⎜ − 1⎟ H inside ( r ) = ( K m − 1) H inside ( r ) = χ m H inside ( r )
⎝ μo ⎠
i.e. Μ ( r ≤ R ) = χ m H inside ( r ≤ R )
But:
{Note that Μ ( r > R ) ≡ 0 }
⎛ Vo
⎛ 3 ⎞
1 ⎛ 3 ⎞
1
Bo = − ⎜ m
H inside ( r ≤ R ) = + ⎜
⎟ H o zˆ =
⎜
⎟ Bo zˆ since H o =
⎜ Lgap
μo ⎝ K m + 2 ⎠
μo
⎝ Km + 2 ⎠
⎝
⎞
⎟⎟
⎠
⎛ 1 ⎞ ⎛ 3χ m ⎞
⎛ 1 ⎞ ⎛ 3χ m ⎞
Thus: Μ ( r ≤ R ) = ⎜ ⎟ ⎜
⎟ Bo zˆ = ⎜ ⎟ ⎜
⎟ Bo zˆ = Μ o zˆ SI units = Amps/meter
⎝ μo ⎠ ⎝ K m + 2 ⎠
⎝ μo ⎠ ⎝ χ m + 3 ⎠
⎛ 1 ⎞ ⎛ 3χ m ⎞
⎛ 1 ⎞ ⎛ χm ⎞
⎛ 1 ⎞ ⎛ Km −1 ⎞
i.e. Μ o = ⎜ ⎟ ⎜
⎟ Bo = ⎜ ⎟ ⎜
⎟ Bo = 3 ⎜ ⎟ ⎜
⎟ Bo
⎝ μo ⎠ ⎝ χ m + 3 ⎠
⎝ μo ⎠ ⎝ 1 + χ m 3 ⎠
⎝ μo ⎠ ⎝ K m + 2 ⎠
Thus, we see that the magnetization (magnetic dipole moment per unit volume) of the
magnetized sphere is constant/uniform, and is aligned parallel (anti-parallel) with the applied
external magnetic field for χ m > 0 ( χ m < 0 ) respectively.
The magnetic dipole moment of the magnetized sphere is therefore:
⎛ 1 ⎞ ⎛ χm ⎞
3 ⎛ 1 ⎞ ⎛ Km −1 ⎞
m = 43 π R 3Μ = 43 π R 3Μ o zˆ = 43 π R 3 ⎜ ⎟ ⎜
⎟ Bo zˆ = 4π R ⎜ ⎟ ⎜
⎟ Bo zˆ
+
+
μ
1
χ
3
μ
K
2
m
⎝ o ⎠⎝
⎠
⎝ o ⎠⎝ m
⎠
Again, note that m = 43 π R 3Μ is parallel (anti-parallel) to the applied external magnetic field
Bext ( r ) = Bo zˆ for χ m > 0 ( χ m < 0 ) {i.e. K m = (1 + χ m ) > 1 ( K m = (1 + χ m ) < 1 ) } respectively.
Let us now investigate / explicitly check out the boundary conditions that we didn’t actually use:
BC 6): ( H outside − H inside )
BC 7): ( Boutside − Binside )
r=R
r =R
⊥
⊥
− H inside
BC 8): ( H outside
)
= K free × nˆ
= μo KTot × nˆ
r=R
surface
surface
⊥
⊥
= − ( Μ outside
− Μ inside
)
r=R
n.b. Because we had many more BC relations than # of unknown coefficients that needed to be
determined in this problem, we see / realize that this problem is in fact over-determined!!!
10
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
BC 6): ( H outside − H inside )
⇒ ( Hθ − Hθ
out
in
)
r=R
Lecture Notes 20.5
Prof. Steven Errede
(Tangential H @ r = R )
= K free × nˆ
= rˆ
=0
r =R
Fall Semester, 2007
zˆ = rˆ cos θ − θ sin θ
= 0 ??
Normal component @ r = R
tangential component @ r = R
⎧⎪ 1
⎫⎪
⎛ K −1 ⎞ ⎛ R ⎞ 1
⎛ 1 ⎞
3
Bo θ sin θ ⎬ −
= ⎨ − Bo θ sin θ + ⎜ m
⎜ ⎟ Bo θ sin θ
⎟⎜ ⎟
⎝ K m + 2 ⎠ ⎝ R ⎠ μo
⎪⎩ μo
⎪⎭ ( K m + 2 ) ⎝ μo ⎠
⎧⎪
⎫⎪ ⎛ 1 ⎞
⎛ K −1 ⎞
3
= ⎨ −1 + ⎜ m
⎬ ⎜ ⎟ Bo θ sin θ
⎟−
K
2
K
2
+
+
(
)
⎪⎩
⎪⎭ ⎝ μo ⎠
m
⎝ m
⎠
⎧ K − 2 + K m − 1 − 3 ⎫⎛ 1 ⎞
= ⎨− m
⎬⎜ ⎟ Bo θ sin θ = 0 !!! Yes!!!
Km + 2
⎩
⎭⎝ μo ⎠
3
i.e.
H outside ( r = R ) = H inside ( r = R ) Tangential-H is continuous across
Hθoutside ( r = R ) = Hθinside ( r = R ) this interface / boundary at r = R!!
r̂
=0
BC 7): ( Boutside − Binside )
⇒ ( Bθ
outside
− Bθ
r =R
inside
)
= μo KTot × nˆ
r =R
r =R
= μo K free × rˆ
= μo K Bound × rˆ
r =R
+ μo K Bound × rˆ
r =R
r=R
3
⎛ χm ⎞ ⎛ R ⎞
⎪⎧
⎪⎫ ⎛ 3 (1 + χ m ) ⎞
B
θ
θ
= ⎨ − Bo θ sin θ + ⎜
sin
⎬+⎜
⎟ Bo θ sin θ
⎟⎜ ⎟ o
⎝ χm + 3 ⎠ ⎝ R ⎠
⎪⎩
⎪⎭ ⎝ χ m + 3 ⎠
⎛ χ m ⎞ ⎛ 3 + 3χ m ⎞ ⎪⎫
⎪⎧
= ⎨−1 + ⎜
⎟+⎜
⎟ ⎬ Bo θ sin θ
⎝ χ m + 3 ⎠ ⎝ χ m + 3 ⎠ ⎭⎪
⎩⎪
⎧⎪ − χ m − 3 + χ m + 3 + 3χ m ⎫⎪
⎛ 3χ m ⎞
=⎨
⎬ Bo θ sin θ = ⎜
⎟ Bo θ sin θ
χm + 3
⎝ χm + 3 ⎠
⎪⎩
⎪⎭
⎛ χm ⎞
=⎜
⎟ Bo θ sin θ ≠ 0
⎝ 1 + χm 3 ⎠
Now what is K Bound ? K Bound ≡ Μ × nˆ
n̂ = outward normal from surface.
surface
⎡ ⎛ 1 ⎞ ⎛ χm ⎞ ⎤
⎛ 1 ⎞ ⎛ Km −1 ⎞
Now: Μ ( r ≤ R ) = Μ o zˆ = ⎢ ⎜ ⎟ ⎜
⎟ Bo ⎥ zˆ = 3 ⎜ ⎟ ⎜
⎟ Bo zˆ
⎝ μo ⎠ ⎝ K m + 2 ⎠
⎣ ⎝ μo ⎠ ⎝ 1 + χ m 3 ⎠ ⎦
And: zˆ × rˆ = ⎡ rˆ cos θ − θ sin θ ⎤ × rˆ = − sin θ θ × rˆ = ϕˆ sin θ { r̂ × θˆ = ϕˆ and θˆ × r̂ = −ϕˆ }
⎣
⎦
⎡ ⎛ 1 ⎞ ⎛ K −1 ⎞ ⎤
Thus: K Bound ( r = R ) = Μ ( r = R ) × rˆ r = R = Μ o zˆ × rˆ = M oϕˆ sin θ = ⎢3 ⎜ ⎟ ⎜ m
⎟ Bo ⎥ sin θϕˆ
2
K
μ
+
⎠ ⎦
⎣ ⎝ o ⎠⎝ m
(
)
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
11
UIUC Physics 435 EM Fields & Sources I
Then: K Bound × rˆ
= Μ ( r = R ) × rˆ
r=R
Fall Semester, 2007
r=R
Lecture Notes 20.5
Prof. Steven Errede
= Μ o zˆ × rˆ = Μ oϕˆ sin θ
⎛ 1 ⎞ ⎛ χm ⎞
⎛ 1 ⎞ ⎛ Km −1 ⎞
Μo = ⎜ ⎟ ⎜
⎟ Bo = 3 ⎜ ⎟ ⎜
⎟ Bo
⎝ μo ⎠ ⎝ 1 + χ m 3 ⎠
⎝ μo ⎠ ⎝ K m + 2 ⎠
Then: K Bound × rˆ r = R = Μ o sin θ ϕˆ × rˆ r = R
{ r̂ × θˆ = ϕˆ , θˆ × ϕˆ = r̂ and ϕˆ × r̂ = θˆ }
=+θ
K Bound × rˆ
Then:
(B
outside
r =R
⎡ ⎛ 1 ⎞ ⎛ K −1 ⎞ ⎤
= Μ o θ sin θ = ⎢3 ⎜ ⎟ ⎜ m
⎟ Bo ⎥ θ sin θ
⎣ ⎝ μo ⎠ ⎝ K m + 2 ⎠ ⎦
− Binside )
r =R
= ( Bθoutside − Bθinside )
r =R
= μo K Bound × rˆ
r=R
⎡ ⎛ 1 ⎞ ⎛ K −1 ⎞ ⎤
⎛ K −1 ⎞
⎛ K −1 ⎞
=3 ⎜ m
=
θ
θ
μ
B
sin
Bo ⎥ θ sin θ = 3 ⎜ m
⎟⎜ m
⎟
⎟ Bo θ sin θ
⎟ o
o ⎢3 ⎜
⎢⎣ ⎝⎜ μo ⎠⎟ ⎝ K m + 2 ⎠ ⎦⎥
⎝ Km + 2 ⎠
⎝ Km + 2 ⎠
Thus we see that BC 7) is indeed satisfied!
(H
Finally, BC 8):
⊥
outside
⊥
− H inside
)
r =R
⊥
⊥
= − ( Μ outside
− Μ inside
)
r=R
zˆ = rˆ cos θ − θ sin θ
normal
component
@ r =R
⎡⎛ 1
= ⎢⎜
⎢⎣⎝ μo
3
⎞
⎛ Km −1 ⎞ ⎛ R ⎞ ⎛ 1
⎟ Bo rˆ cos θ + ⎜
⎟⎜ ⎟ ⎜
⎠
⎝ K m + 2 ⎠ ⎝ R ⎠ ⎝ μo
⎞
⎛ 1
3
⎟ Bo ( 2rˆ cos θ ) −
( K m + 2 ) ⎝⎜ μo
⎠
tangential
component
@ r =R
⎤
⎞
ˆ
⎥
B
r
θ
cos
⎟ o
⎥⎦
⎠
⎡ ⎛ 1 ⎞ ⎛ K −1 ⎞ ⎤
= − [ 0 − Μ o cos θ ] = +Μ o cos θ = ⎢3 ⎜ ⎟ ⎜ m
⎟ Bo ⎥ cos θ
⎣ ⎝ μo ⎠ ⎝ K m + 2 ⎠ ⎦
⎡
⎤⎛ 1 ⎞
⎡ ⎛ 1 ⎞ ⎛ Km −1 ⎞ ⎤
⎛ K −1 ⎞
3
= ⎢1 + 2 ⎜ m
⎥ ⎜ ⎟ Bo cos θ = ⎢3 ⎜ ⎟ ⎜
⎟−
⎟ Bo ⎥ cos θ
⎝ K m + 2 ⎠ ( K m + 2 ) ⎦ ⎝ μo ⎠
⎣
⎣ ⎝ μo ⎠ ⎝ K m + 2 ⎠ ⎦
⎡ K + 2 + 2Km − 2 − 3 ⎤ ⎛ 1 ⎞
⎡ 3K m − 3 ⎤ ⎛ 1 ⎞
=⎢ m
⎥ ⎜ ⎟ Bo cos θ = ⎢
⎥ ⎜ ⎟ Bo cos θ
Km + 2
⎣ K m + 2 ⎦ ⎝ μo ⎠
⎣
⎦ ⎝ μo ⎠
⎡ K −1 ⎤ ⎛ 1 ⎞
⎛ μ ⎞
with: K m = ⎜ ⎟ = 1 + χ m or: K m − 1 = χ m
= 3⎢ m
⎥ ⎜ ⎟ Bo cos θ
⎣ K m + 2 ⎦ ⎝ μo ⎠
⎝ μo ⎠
⎡ 3χ m ⎤ ⎛ 1 ⎞
⎡ χm ⎤ ⎛ 1 ⎞
=⎢
⎥ ⎜ ⎟ Bo cos θ = ⎢
⎥ ⎜ ⎟ Bo cos θ
⎣ χ m + 3 ⎦ ⎝ μo ⎠
⎣1 + χ m 3 ⎦ ⎝ μ o ⎠
Thus, we see that BC 8) is also indeed satisfied:
(H
12
⊥
outside
⊥
− H inside
)
r=R
⊥
⊥
= − ( Μ outside
− Μ inside
)
r=R
⎡ 3χ m ⎤ ⎛ 1 ⎞
⎡ χm ⎤ ⎛ 1 ⎞
=⎢
⎥ ⎜ ⎟ Bo cos θ = ⎢
⎥ ⎜ ⎟ Bo cos θ
⎣ χ m + 3 ⎦ ⎝ μo ⎠
⎣1 + χ m 3 ⎦ ⎝ μo ⎠
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 20.5
Prof. Steven Errede
Now let us examine and discuss these results in more detail:
•
The magnetization (magnetic dipole moment per unit volume) Μ inside the magnetic sphere
is uniform / constant in the ẑ -direction (n.b. same as the externally applied magnetic field
Bext = Bo zˆ ):
⎛ 1 ⎞ ⎛ χm ⎞
⎛ 1 ⎞ ⎛ Km −1 ⎞
Μ ( r ≤ R ) = Μ o zˆ = ⎜ ⎟ ⎜
⎟ Bo zˆ = 3 ⎜ ⎟ ⎜
⎟ Bo zˆ SI Units = Amps/meter
⎝ μo ⎠ ⎝ 1 + χ m 3 ⎠
⎝ μo ⎠ ⎝ K m + 2 ⎠
•
The corresponding magnetic dipole moment of the magnetized sphere (of radius R) is:
m = Μ ⋅Vsphere = ( 43 π R 3 ) Μ SI Units = Ampere-meters2 {recall “ m = Ia ”}
⎛ 1 ⎞⎛ χ m ⎞
3 ⎛ 1 ⎞ ⎛ Km −1 ⎞
m = ( 43 π R 3 ) Μ = 43 π R 3Μ o zˆ = 43 π R 3 ⎜ ⎟⎜
⎟ Bo zˆ = 4π R ⎜ ⎟ ⎜
⎟ Bo zˆ
+
+
μ
1
χ
3
μ
K
2
m
⎝ o ⎠⎝
⎠
⎝ o ⎠⎝ m
⎠
•
The corresponding magnetic field inside the magnetized sphere (with Bext = Bo zˆ ) is:
3 (1 + χ m )
⎛ 3K m ⎞
(1 + χ m ) B
B inside ( r ≤ R ) = ⎜
Bo zˆ =
SI units = Teslas
⎟ Bo zˆ =
(1 + χ m 3) ext
(3 + χm )
⎝ Km + 2 ⎠
We can rearrange / manipulate this relation to further illuminate the physics of what is going on
here, as follows:
⎛ 1 + χm ⎞
⎛ 1 + 13 χ m + 23 χ m ⎞
ˆ
=
B inside ( r ≤ R ) = ⎜
B
z
⎟ o
⎜
⎟ Bo zˆ
⎝ 1 + χm 3 ⎠
⎝ 1 + χm 3 ⎠
⎛ 1+ 1 χ ⎞
⎞
⎛ 2 ⎞⎛ χ
B inside ( r ≤ R ) = ⎜ 13 m ⎟ Bo zˆ = ⎜ ⎟ ⎜ 1m ⎟ Bo zˆ
⎝ 3 ⎠ ⎝ 1 + 3 χm ⎠
⎝ 1 + 3 χm ⎠
=1
=Μ=Μ o zˆ
⎤
⎞
⎞
⎛ 2 ⎞⎛ χ
⎛ 2 ⎞ ⎡⎛ 1 ⎞ ⎛ χ
B inside ( r ≤ R ) = Bo zˆ + ⎜ ⎟ ⎜ 1m ⎟ Bo zˆ = Bext + ⎜ ⎟ μo ⎢⎜ ⎟ ⎜ 1m ⎟ Bo zˆ ⎥
⎝ 3 ⎠ ⎝ 1 + 3 χm ⎠
⎝ 3 ⎠ ⎢⎣⎝ μo ⎠ ⎝ 1 + 3 χ m ⎠
⎥⎦
This is identical with the result in Griffiths Example 6.1 pp. 266-67 (it better be!!!):
⎞
⎛ 1 ⎞⎛ χ
⎛2⎞
B inside ( r ≤ R ) = Bext + ⎜ ⎟ μo Μ
with Μ = Μ o zˆ = ⎜ ⎟ ⎜ 1m ⎟ Bext
⎝3⎠
⎝ μo ⎠⎝ 1 + 3 χ m ⎠
Thus we see that the magnetic field inside the sphere is the linear superposition of the externally
applied magnetic field Bext = Bo zˆ plus the internal B -field of the magnetized sphere (alone):
inside
Bsphere
( r ≤ R ) = 23 μo Μ !!!
Outside the magnetized sphere, the magnetic field is:
3
⎛ 1 ⎞ ⎛ χm ⎞ ⎛ R ⎞
outside
B
( r ≥ R ) = Bo zˆ + ⎜ ⎟ ⎜ 1 ⎟ ⎜ ⎟ Bo ⎡⎣ 2rˆ cos θ + θ sin θ ⎤⎦ SI Units = Teslas
⎝ 3 ⎠ ⎝ 1 + 3 χm ⎠ ⎝ r ⎠
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
13
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 20.5
Prof. Steven Errede
Again, we can rearrange / manipulate this relation further to elucidate the underlying physics:
⎛ μ ⎞ ⎡⎛ 4π 3 ⎞ ⎛ 1 ⎞ ⎛ χ m ⎞ ⎤ ⎛ 1 ⎞ ⎡
B outside ( r ≥ R ) = Bext + ⎜ o ⎟ ⎢⎜
R ⎟⎜ ⎟⎜ 1
⎟ Bo ⎥ ⎜ 3 ⎟ 2rˆ cos θ + θ sin θ ⎤⎦
⎠ ⎝ μo ⎠ ⎝ 1 + 3 χ m ⎠ ⎥⎦ ⎝ r ⎠ ⎣
⎝ 4π ⎠ ⎢⎣⎝ 3
4π 3 ⎛ 1 ⎞ ⎛ χ m ⎞
⎛ 4π 3 ⎞ ⎛ 1 ⎞ ⎛ χ m ⎞
R ⎜ ⎟⎜ 1
R ⎟⎜ ⎟⎜ 1
But: m = ( 43 π R 3 ) Μ = 43 π R 3Μ o zˆ =
⎟ Bo zˆ = ⎜
⎟ Bext
3
⎝ 3
⎠ ⎝ μo ⎠ ⎝ 1 + 3 χ m ⎠
⎝ μo ⎠ ⎝ 1 + 3 χ m ⎠
⎛ 4π 3 ⎞ ⎛ 1 ⎞ ⎛ χ m ⎞
R ⎟⎜ ⎟⎜ 1
Thus: m = m = 43 π R 3Μ o = ⎜
⎟ Bo
⎝ 3
⎠ ⎝ μo ⎠ ⎝ 1 + 3 χ m ⎠
⎛ μ ⎞⎛ m ⎞
B outside ( r ≥ R ) = Bext + ⎜ o ⎟ ⎜ 3 ⎟ ⎡ 2rˆ cos θ + θ sin θ ⎤
∴
⎦
⎝ 4π ⎠ ⎝ r ⎠ ⎣
Which again can be seen as the linear superposition of the external magnetic field and the
(external) magnetic field of a physical magnetic dipole, with magnetic dipole moment m :
⎛μ
B dipole ( r ≥ R ) = ⎜ o
⎝ 4π
∴
⎞m
⎟ 3 ⎣⎡ 2rˆ cos θ + θ sin θ ⎦⎤
⎠r
See / compare to Griffiths 5.86 p. 246
& also P435 Lecture Notes 16, p. 14
B outside ( r ≥ R ) = Bext + B dipole ( r ≥ R )
We have also shown that B dipole ( r ≥ R ) can be written in coordinate-free form as:
⎛μ ⎞1
B dipole ( r ≥ R ) = ⎜ o ⎟ 3 ⎡⎣3 ( mirˆ ) rˆ − m ⎤⎦
SI Units = Teslas
⎝ 4π ⎠ r
⎛ 4π 3 ⎞ ⎛ 1 ⎞ ⎛ χ m ⎞
2
R ⎟⎜ ⎟⎜ 1
With: m = ( 43 π R 3 ) Μ = ⎜
⎟ Bext SI Units = Ampere-meters (“ m = Ia ” )
⎝ 3
⎠ ⎝ μo ⎠ ⎝ 1 + 3 χ m ⎠
•
Comments on the relative strengths of internal & external magnetic fields vs. the applied
external field - we can gain some additional physics insight on the nature of this problem by
taking ratios of B inside ( r ≤ R ) and B outside ( r ≥ R ) to Bext = Bo :
B
⎞
⎛ 2 ⎞⎛ χ
Bo + ⎜ ⎟ ⎜ 1m ⎟ Bo
(r ≤ R)
⎞
⎝ 3 ⎠ ⎝ 1 + 3 χm ⎠
⎛ 2 ⎞⎛ χ
=
= 1 + ⎜ ⎟ ⎜ 1m ⎟ n.b. = constant
Bo
Bext
⎝ 3 ⎠ ⎝ 1 + 3 χm ⎠
inside
For the outside ratio, since this is polar angle dependent, let’s simply do it for θ = 0 :
3
⎛ 2 ⎞ ⎛ χm ⎞ ⎛ R ⎞
Bo + ⎜ ⎟ ⎜ 1
⎟ ⎜ ⎟ Bo
3
B outside ( r ≥ R, θ = 0 )
⎝ 3 ⎠ ⎝ 1 + 3 χm ⎠ ⎝ r ⎠
⎛ 2 ⎞ ⎛ χm ⎞ ⎛ R ⎞
=
= 1+ ⎜ ⎟⎜ 1
⎟⎜ ⎟
Bo
Bext
⎝ 3 ⎠ ⎝ 1 + 3 χm ⎠ ⎝ r ⎠
Note that at r = R , the inside ratio = outside ratio (i.e. normal component of B is continuous
across an interface/boundary).
14
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 20.5
Prof. Steven Errede
For either linear diamagnetic ( χ m < 0 ) or linear paramagnetic ( χ m > 0 ) materials, the typical
values of magnetic susceptibilities associated with these materials are χ m ~ 10−3 − 10−6 . Thus
for diamagnetic & paramagnetic materials with χ m
B inside ( r ≤ R )
Bext
i.e.
~1
B outside ( r = R, θ = 0 )
and
B inside ( r ≤ R ) ~ Bext
1 we see that:
Bext
~1
and B outside ( r = R, θ = 0 ) ~ Bext simply because: χ m
1
For ferromagnetic materials, where formally / technically speaking, the magnetization Μ is
history-dependent, if we imagine that we have an initially unmagnetized sphere of ferromagnetic
material and place it in our experimental apparatus and then slowly turn on the external magnetic
field, from Bext = 0 (initially) to Bext = Bo zˆ (finally) then we trace out a curve along the Μ vs.
Bext relation as shown below:
Let’s suppose that at point a on this curve, the magnetization, Μ corresponds to a magnetic
susceptibility χ m = 1000 (i.e. χ m 1 ). Then for this ferromagnetic material we see that for
χm 1 :
B inside ( r ≤ R )
Bext
Likewise:
⎞
⎛ 2 ⎞⎛ χ
= 1 + ⎜ ⎟ ⎜ 1m ⎟ ≈ 1 + 2 = 3 for χ m
⎝ 3 ⎠ ⎝ 1 + 3 χm ⎠
B outside ( r = R, θ = 0 )
Bext
i.e. B inside ( r ≤ R )
1
⎞
⎛ 2 ⎞⎛ χ
= 1 + ⎜ ⎟ ⎜ 1m ⎟ ≈ 1 + 2 = 3 for χ m
⎝ 3 ⎠ ⎝ 1 + 3 χm ⎠
3 Bext
B outside ( r = R, θ = 0 )
for χ m
3 Bext
1
1
(at surface) for χ m
1
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
15
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 20.5
Prof. Steven Errede
For the uniformly-magnetized sphere of radius R, we have also seen that the magnetization
⎛ 1 ⎞ ⎛ χm ⎞
⎛ 1 ⎞ ⎛ Km −1 ⎞
Μ ( r ≤ R ) = Μ o zˆ = ⎜ ⎟ ⎜
⎟ Bo zˆ = 3 ⎜ ⎟ ⎜
⎟ Bo zˆ can be replicated by an
⎝ μo ⎠ ⎝ 1 + χ m 3 ⎠
⎝ μo ⎠ ⎝ K m + 2 ⎠
equivalent, bound surface current density, K Bound ( r = R ) = Μ × nˆ
r =R
= Μ × rˆ
r =R
= Μ o sin θϕˆ
circulating in the +ϕ̂ direction on the surface of the sphere with magnitude:
⎛ 1 ⎞ ⎛ χm ⎞
K Bound ( r = R, θ ) = Μ o sin θ = ⎜ ⎟ ⎜
SI Units = Amperes/meter
⎟ Bo sin θ
⎝ μo ⎠ ⎝ 1 + χ m 3 ⎠
This corresponds to an equivalent bound current of: I bound ( r = R, θ ) = ∫ K Bound ( r = R, θ ) d ⊥
C⊥
north
pole
south
pole
I bound ( r = R, θ ) = ∫
K Bound ( r = R,θ ) d
⊥
I bound ( r = R,θ ) = π RK Bound ( r = R, θ )
⎛ 1 ⎞ ⎛ χm ⎞
I bound ( r = R, θ ) = π R ⎜ ⎟ ⎜
⎟ Bo sin θϕˆ
⎝ μo ⎠ ⎝ 1 + χ m 3 ⎠
SI Units = Amperes
Now, recall for the charged, spinning hollow conducting sphere of radius R (n.b. Bext = 0 there)
with surface electric charge density σ (Coulombs per meter2) and angular velocity of rotation ω
(radians/sec)
The corresponding free surface current density K free = K ofreeϕˆ = [σω R ] sin θϕˆ (See Griffiths
Example 5.11 pp. 236-37; P435 Lecture Notes 19 pp. 12-13; See also Griffiths Example 6.1 p.
264). This spinning free surface current density produced internal ( r ≤ R ) and external ( r ≥ R )
2
⎛μ ⎞m
μo [σω R ] zˆ and Boutside ( r ≥ R ) = ⎜ o ⎟ 3 ⎡⎣ 2rˆ cos θ + θ sin θ ⎤⎦
3
⎝ 4π ⎠ r
3
with: m = 43 π R [σω R ] which are identical to those of a permanently magnetized sphere, of
magnetic fields: Binside ( r ≤ R ) =
uniform magnetization Μ = M o zˆ (i.e. Bext = 0 here) provided Μ o ≡ [σω R ] :
inside
Bspinning
(r ≤ R) =
sphere
2
2
2
μo Μ = μo Μ o zˆ = μo [σω R ] zˆ
3
3
3
μ
outside
Bspinning
( r ≥ R ) = ⎛⎜ o
sphere
⎝ 4π
16
⎞m⎡
⎟ 3 ⎣ 2rˆ cos θ + θ sin θ ⎦⎤
⎠r
n.b. Bext = 0 here!!
4
4
with m = π R 3Μ o = π R 3 [σω R ]
3
3
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 20.5
Prof. Steven Errede
Using the principle of linear superposition the magnetic field associated with a charged, spinning
hollow conducting sphere of radius R, surface electric charge density σ and angular velocity of
rotation ω that is additionally immersed in an external magnetic field Bext = Bo zˆ is:
2
inside
Bspinning
( r ≤ R ) = Bext + μo Μ where Μ = Μ o zˆ = [σω R ] zˆ
3
sphere
μ
outside
Bspinning
( r ≥ R ) = Bext + ⎜⎛ o
sphere
⎝ 4π
4 3
4 3
⎞m
⎟ 3 ⎡⎣ 2rˆ cos θ + θ sin θ ⎤⎦ and m = π R M o = π R [σω R ]
3
3
⎠r
We can investigate one more aspect of the uniformly-magnetized sphere in a uniform external
magnetic field Bˆext = Bo zˆ . In P435 Lecture Notes 20 p. 8, we introduced the concepts(s) of
effective bound surface and volume densities of magnetic pole strength (i.e. magnetic charge):
σ mBound ( r = R ) ≡ Μ inˆ
surface
ρ mBound ( r ) ≡ −∇iΜ ( r )
SI Units = Amperes/meter
SI Units = Amperes/meter2
Recall here that the SI Units of magnetic charge g m are Ampere-meters
( g m = " qv " = Coulombs * meters/sec = Ampere-meters)
These relations for σ mBound and ρ mBound were / are defined in complete analogy to the bound surface
and volume electric charge densities for electrostatic dielectric materials:
σ eBound ( r = R ) ≡ Ρinˆ
surface
ρeBound ( r ) ≡ −∇iΡ ( r )
SI Units = Coulombs/meter2
SI Units = Coulombs/meter3
Since the magnetization of the sphere is uniform/constant:
⎛ 1 ⎞ ⎛ χm ⎞
⎛ 1 ⎞ ⎛ Km −1 ⎞
Μ ( r ≤ R ) = Μ o zˆ = ⎜ ⎟ ⎜
⎟ Bo zˆ = 3 ⎜ ⎟ ⎜
⎟ Bo zˆ
⎝ μo ⎠ ⎝ 1 + χ m 3 ⎠
⎝ μo ⎠ ⎝ K m + 2 ⎠
we see that the effective volume density of magnetic pole strength ρ mBound ( r ) ≡ −∇iΜ ( r ) = 0.
On the other hand, the effective surface density of magnetic pole strength is non-zero:
⎛ 1 ⎞ ⎛ χm ⎞
σ mBound ( r = R ) ≡ Μ inˆ surface = Μ o zˆ irˆ = Μ o cos θ = ⎜ ⎟ ⎜
⎟ Bo cos θ
⎝ μo ⎠ ⎝ 1 + χ m 3 ⎠
Thus we see that at the north pole on the surface of the magnetized sphere ( r = R, θ = 0 ) :
⎛ 1 ⎞ ⎛ χm ⎞
⎟⎜
⎟ Bo
⎝ μo ⎠ ⎝ 1 + χ m 3 ⎠
And at the south pole on the surface of the magnetized sphere ( r = R, θ = π ) :
σ mBound ( r = R, θ = 0 ) = +Μ o = + ⎜
⎛ 1 ⎞ ⎛ χm ⎞
⎟⎜
⎟ Bo
⎝ μo ⎠ ⎝ 1 + χ m 3 ⎠
σ mBound ( r = R,θ = π ) = −Μ o = − ⎜
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
17
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 20.5
Prof. Steven Errede
We can now also understand BC 8 in a new light. We can rewrite it, noting that nˆ = rˆ here, as:
⎛ ⎞⎡
⎤
⎛ ⎞⎡
⎤
⊥
⊥
⊥
⊥
− H inside
− M inside
( H outside
) r =R = − ( M outside
) r =R = ⎜ μ1 ⎟ ⎢ χ3χ+m 3 ⎥ Bo cos θ = ⎜ μ1 ⎟ ⎢1 + χχm 3 ⎥ Bo cos θ
m
⎝ o ⎠⎣ m
⎦
⎝ o ⎠⎣
⎦
(
)
H outside ( r ) − H inside ( r ) inˆ
as:
surface
⎛
⎞
= Μ inside ( r )inˆ
= σ mBound
= − ⎜ Μ outside ( r ) − Μ inside ( r ) ⎟inˆ
surface
⎜
⎟
⎝ ≡0 here
⎠ surface
⎛ 1 ⎞ ⎡ χm ⎤
= σ mBound = ⎜ ⎟ ⎢
⎥ Bo cos θ
⎝ μ o ⎠ ⎣1 + χ m 3 ⎦
i.e. the point here is that this boundary condition is actually:
⊥
⊥
⊥
⊥
− H inside
− M inside
( H outside
) r = R = − ( M outside
) r = R = σ mBound !!!
More explicitly, this boundary condition actually is:
(H
outside
( r ) − H inside ( r ) )inˆ surface = − ( Μ outside ( r ) − Μ inside ( r ) )inˆ surface = σ mBound
We also know that the net effective bound magnetic charge on surface of the magnetized sphere
must be = 0, i.e.
Bound
∫
QmNET =
S
σ mbound da =
∫ ( Μ inˆ ) da = ∫
S
S
Μ i da = 0
Explicit check:
Bound
QmNET =
∫
S
∫
Μ i da =
= M o R2 ∫
ϕ = 2π
ϕ =0
S
ˆ =
Μ o zˆ inda
dϕ ∫
θ =π
θ =0
∫
S
Μ o ⎡ rˆ cos θ − θ sin θ ⎤ irˆ da = Μ o ∫ cos θ da
⎣
⎦
S
cos θ sin θ dθ = 2π R 2 M o ∫
θ =π
θ =0
+1
2 +1
−1
−1
= 2π R 2 M o ∫ u du = π R 2 M o u
Bound
=0
cos θ d cos θ
=u
= du
θ = π : u = cos π = −1
θ = 0 : u = cos 0 = +1
∴ QmNET = 0
We can also compare the equivalent bound surface magnetic charge vs. the bound surface
electric current σ mbound ( r = R, θ ) vs. K Bound ( r = R,θ )
σ mBound ( r = R ) ≡ Μ inˆ
surface
K Bound ( r = R ) = Μ × nˆ
surface
⎛ 1 ⎞ ⎛ χm ⎞
= Μ o cos θ = ⎜ ⎟ ⎜
⎟ Bo cos θ
⎝ μo ⎠ ⎝ 1 + χ m 3 ⎠
= Μ o zˆ irˆ
= Μ × rˆ
r=R
⎛ 1 ⎞ ⎛ χm ⎞
= Μ o sin θϕˆ = ⎜ ⎟ ⎜
⎟ Bo sin θϕˆ
⎝ μo ⎠ ⎝ 1 + χ m 3 ⎠
Both of these produce the exact same magnetization Μ and associated / corresponding magnetic
fields (internal and external)! They are simply two equivalent, but different ways / methods of
viewing the same physics problem.
18
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 20.5
Prof. Steven Errede
Example #2: Magnetic Fields Associated with a Uniformly Magnetized Sphere
Consider a permanently magnetized sphere of radius R that has uniform magnetization:
Μ ( r ) = Μ o zˆ ( r ≤ R )
Since the magnetization Μ ( r ) = Μ o zˆ is constant, then ∇iΜ ( r ) = 0
(
)
But B ( r ) = μo H ( r ) + Μ ( r ) and ∇i B ( r ) = 0 ⇒ ∇i H ( r ) = −∇iΜ ( r ) , ∴ ∇i H ( r ) = 0
And since ∇ × H ( r ) = J free ( r ) = 0 (here), since ∇i H ( r ) = 0 and ∇ × H ( r ) = 0
Then we may write H ( r ) = −∇Vm ( r )
And thus ∇i H ( r ) = ∇ 2Vm = 0 (i.e. Laplace’s Equation – Magnetic Scalar Potential Vm ( r ) )
Note that this problem has azimuthal symmetry (i.e. no ϕ -dependence):
∞
B ⎞
⎛
Thus ∇ 2Vm ( r , θ ) = 0 has a general solution of the form: Vm ( r , θ ) = ∑ ⎜ A r + +1 ⎟P ( cos θ )
r ⎠
=0 ⎝
Where P ( cos θ ) = the ordinary Legendré Polynomial of order .
Sufficient relations to
determine all
coefficients A and B
inside and outside the
sphere
θ -direction
Problem is actually
over-determined /
over-constrained
Boundary Conditions:
0) Vm ( r ) = finite everywhere
1) VmOUT ( r = R ) = VmIN ( r = R )
⊥
2) Bout
( r = R ) = Bin⊥ ( r = R ) ⇒ Brout ( r = R ) = Brin ( r = R )
Normal component of B is continuous at the surface of sphere
( ⊥ = rˆ direction at r = R interface / boundary)
3) H out ( r = R ) = H in ( r = R ) = K free × nˆ surface = K free × rˆ surface = 0
( K free = 0 here ⇒ KTOT = K free + K bound = K bound )
4) Bout ( r = R ) − Bin = μo KTOT × nˆ
5)
surface
= μo KTOT × rˆ
surface
= μo K bound × rˆ
( H ( r = R ) − H ( r = R )) = − ( M ( r = R ) − M ( r = R ))
( H ( r = R ) − H ( r = R )) = − ( M ( r = R ) − H ( r = R ))
⊥
out
⊥
in
⊥
out
⊥
in
out
r
in
r
out
r
in
r
surface
( ⊥ = rˆ direction at r = R interface / boundary)
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
19
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 20.5
Prof. Steven Errede
General solutions for the magnetic scalar potential inside/outside of the sphere are of the form:
∞
B ⎞
⎛
Vmin ( r , θ ) = ∑ ⎜ A r + +1 ⎟P ( cos θ )
r ⎠
=0 ⎝
∞
B′ ⎞
⎛
Vmout ( r , θ ) = ∑ ⎜ A′r l + +1 ⎟P ( cos θ )
r ⎠
=0 ⎝
(r ≤ R)
(r ≥ R)
Impose BC 0): Vm ( r ) must be finite everywhere:
→ for Vmin ( r , θ ) : B = 0
∀
→ for Vmout ( r , θ ) : A′ = 0
∀
(r ≤ R)
∞
⇒ Vmin ( r ,θ ) = ∑ A r P ( cos θ )
(r ≤ R)
=0
(r ≥ R)
B′
P ( cos θ )
+1
=0 r
∞
⇒ Vmout ( r ,θ ) = ∑
(r ≥ R)
Impose BC 1): Vmout ( r = R ) = Vmin ( r = R )
(i.e. Vm ( r = R ) is continuous across the interface / boundary of sphere at r = R )
⇒ At r = R we must have for each :
AR =
Impose BC 2): Brout ( r = R ) = Brin ( r = R )
B′
R +1
or:
B′ = A R 2
+1
(Normal component of B is continuous across
interface / boundary of sphere at r = R )
Now: H ( r ) ≡ −∇Vm ( r )
⇒ radial component of H: H r ( r ) = − ∂Vm ( r ) ∂r (in spherical polar coordinates)
1
But: H in ( r ≤ R ) =
⇒
μo
H in ( r ≤ R ) =
1
μo
B in ( r ≤ R ) − Μ ( r ≤ R )
B in ( r ≤ R ) − Μ ( r ≤ R )
where Μ ( r ) = Μ o zˆ = Μ o ⎡⎣cosθ rˆ − sin θθˆ ⎤⎦
(r ≤ R)
since zˆ = rˆ cos θ − θ sin θ
∴ radial component of H in ( r ≤ R ) :
H rin ( r ≤ R ) =
=
1
μo
1
μo
Brin ( r ≤ R ) − Μ r ( r ≤ R )
Brin ( r ≤ R ) − Μ o cos θ
∴ radial component of B in ( r ≤ R ) :
Brin ( r ≤ R ) = μo H rin ( r ≤ R ) + μo Μ o cos θ , but H r ( r ) = − ∂Vm ( r ) ∂r
∂Vmin ( r ≤ R )
+ μo Μ o cos θ
∴ B ( r ≤ R ) = − μo
∂r
in
r
20
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
Outside the sphere ( r ≥ R ) :
∴ H out ( r ≥ R ) =
1
μo
Fall Semester, 2007
Lecture Notes 20.5
Prof. Steven Errede
Μ ( r ≥ R ) = 0 for r > R
B out ( r ≥ R )
∴ radial component of H out ( r ≥ R ) : H rout ( r ≥ R ) =
1
μo
Brout ( r ≥ R )
∂Vmout ( r ≥ R )
∂Vmout ( r ≥ R ) 1 out
out
B ( r ≥ R ) or: Br ( r ≥ R ) = − μo
=
⇒ −
∂r
∂r
μo r
Thus BC2 is: Brout ( r = R ) = Brin ( r = R )
∂Vmin ( r ,θ )
∂Vmout ( r ,θ )
out
+ μo Μ o cos θ and Br ( r , θ ) = − μo
with: B ( r ,θ ) = − μo
∂r
∂r
out
in
∂V ( r ,θ )
∂Vm ( r ,θ )
gives: − μo m
r = R = − μo
r = R + μ o Μ o cos θ
∂r
∂r
∂V out ( r ,θ )
∂V in ( r ,θ )
− m
=− m
+ Μ o cos θ
or:
r
∂r
∂
r =R
r=R
in
r
(n.b. Μ o cos θ only contains the P ( cos θ ) = P1 ( cos θ ) = cos θ term)
B′
P ( cos θ )
+1
=0 r
∞
Now: Vmout ( r , θ ) = ∑
(r ≥ R)
∞
and Vmin ( r ,θ ) = ∑ A r P ( cos θ )
(r ≤ R)
=0
Carrying out the radial differentiation on both sides of BC 2 relation, we obtain:
∞
+∑
=0
(
∞
+ 1) B′
=
−
P
cos
A R −1 P ( cos θ ) + Μ o cos θ
θ
(
)
∑
+2
R
=1
This relation can only be satisfied term-by-term, i.e. for each -value in the infinite series, thus:
For = 0:
For = 1:
For
≥ 2:
B0′ = 0 (and therefore, from BC 1: B0′ = A0 R 2 +1 ⇒ A0 = 0 )
2 B′
+ 31 = − A1 + Μ o
R
( + 1) Bl′ = − A R −1 or: B′ = − ⎛
⎞
2 +1
⎜
⎟AR
+2
1
+
R
⎝
⎠
But from BC 1: B′ = A R 2
+1
∴ A R2
+1
⎛
⎞
2
= −⎜
⎟AR
⎝ +1 ⎠
+1
⎛
⎞
The relation A = − ⎜
⎟ A can only be satisfied for each
⎝ +1 ⎠
⎛
⎞
⇒ A = −⎜
⎟A
⎝ +1 ⎠
(
≥ 2 ) if A = 0 ∴ B′ = 0
∴ The only surviving term in the series expansion(s) for Vm ( r , θ ) is the
= 1 term!
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
21
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 20.5
Prof. Steven Errede
Thus, the solutions for the magnetic scalar potential inside/outside the sphere, for this particular
physics problem are:
B′
Vmin ( r , θ ) = A1r cos θ and Vmout ( r ,θ ) = 21 cos θ
r
3
B1′ = A1 R
(from BC 1)
With: 1)
2 B1′
= − A1 + Μ o
(from BC 2) ⇒
2’) 2 B1′ = − A1 R 3 + Μ o R 3
And: 2)
R3
Simultaneously solve 1) and 2) above for A1 and B1′ :
Add 1) and 2’): (i.e. eliminate A1):
B1′ + 2 B1′ = A1 R 3 − A1 R 3 + Μ o R 3
1
⇒ 3B1′ = M o R 3 or: B1′ = Μ o R 3
3
1
1
B1′ = Μ o R 3 = A1 R 3 ⇒ A1 = Μ o
Plug this back into eq. 1) above:
3
3
Thus, the specific solutions for the magnetic scalar potential inside/outside the sphere, unique for
this particular physics problem are:
1
Vmin ( r ,θ ) = Μ o r cos θ
3
(r ≤ R)
2
out
m
V
1
R
( r ,θ ) = Μ o R ⎛⎜ ⎞⎟ cos θ
3
⎝r⎠
(r ≥ R)
∂
1 ∂
1
∂
rˆ +
θ+
ϕˆ (in spherical-polar coordinates)
r ∂θ
r sin θ ∂ϕ
∂r
1
1
1
Thus: H in ( r ,θ ) = −∇Vmin ( r , θ ) = − Μ o cos θ rˆ + Μ o sin θθ = − Μ o ⎡ cos θ rˆ − sin θθ ⎤
⎣
⎦
3
3
3
1
1
1
i.e. H in ( r , θ ) = − Μ o ⎡⎣ rˆ cos θ − θ sin θ ⎤⎦ = − Μ o zˆ = − Μ
3
3
3
Now: H ( r ) = −∇Vm ( r ) where ∇ =
≡ zˆ
Notice that here, in this problem, that H in ( r , θ ) points in the − ẑ direction, opposite to the
direction of the magnetization Μ ( r ) = Μ o zˆ
(r ≤ R)
!!!
3
3
2
1
⎛R⎞
⎛R⎞
Outside the magnetic sphere, H out ( r ,θ ) = −∇Vmout ( r ,θ ) = + Μ o ⎜ ⎟ cos θ rˆ + Μ o ⎜ ⎟ sin θθ
3
3
⎝r⎠
⎝r⎠
3
i.e. H
22
out
1
R
( r ,θ ) = + Μ o ⎛⎜ ⎞⎟ ⎡⎣ 2rˆ cos θ + θ sin θ ⎤⎦ ⇐ H -field associated with a magnetic dipole!
3
⎝r⎠
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 20.5
Prof. Steven Errede
Thus:
1
1
1
H in ( r , θ ) = − Μ o ⎡ rˆ cos θ − θ sin θ ⎤ = − Μ o zˆ = − Μ where Μ ( r ) = Μ o zˆ
⎣
⎦
3
3
3
(r ≤ R)
≡ zˆ
3
1
⎛R⎞
H out ( r ,θ ) = + Μ o ⎜ ⎟ ⎡ 2rˆ cos θ + θ sin θ ⎤
⎦
3
⎝r⎠ ⎣
(
Now: B in ( r , θ ) = μo H in ( r , θ ) + Μ
∴
and B out ( r ,θ ) = μo H out ( r , θ ) (outside (r > R) M = 0)
(
1
1
2
2
B in ( r ,θ ) = − Μ o zˆ + Μ o zˆ = + Μ o zˆ = + Μ o rˆ cos θ − θ sin θ
3
3
3
μo
1
μo
Or:
)
)
3
B
out
1
R
( r ,θ ) = Μ o ⎛⎜ ⎞⎟ ⎡⎣ 2rˆ cos θ + θ sin θ ⎤⎦
3
⎝r⎠
(
)
2
2
2
B in ( r , θ ) = + μo Μ o zˆ = μo Μ o rˆ cos θ − θ sin θ = μo Μ where Μ ( r ) = Μ o zˆ
3
3
3
(r ≤ R)
3
B
out
1
R
( r ,θ ) = μo Μ o ⎛⎜ ⎞⎟ ⎡⎣ 2r cos θ + θ sin θ ⎤⎦
3
⎝r⎠
Déjà vu! We have seen before that the magnetic field associated with a magnetic dipole moment
m (see Griffiths Equation 5.86 and/or P435 Lecture Notes 16, page14) is:
⎛μ ⎞m
B dipole ( r ≥ R ) = ⎜ o ⎟ 3 2rˆ cos θ + θ sin θ
⎝ 4π ⎠ r
1 dipole
1 m
H dipole ( r ≥ R ) =
B
( r ≥ R ) = ⎛⎜ ⎞⎟ 3 2rˆ cos θ + θ sin θ
μo
⎝ 4π ⎠ r
(
)
(
)
Thus, we see here that for a uniformly, permanently magnetized sphere of radius R that:
Μ
Μo
3 Μo
m= m =
=
=
3
4
volume
π R 3 4π R
3
4π R 3
m
or: Μ o =
3
→ Compare these results with those of the previous magnetostatic boundary value problem
example above, pages 9-18.
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
23
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 20.5
Prof. Steven Errede
Note that the lines of B are continuous across the boundary / interface at r = R (i.e. closed)
whereas the lines of H are discontinuous across the boundary / interface at r = R. This is
because the lines of H originate / terminate from / on the effective bound magnetic charges gm
on the surface of the magnetized sphere:
ẑ
+ gm N
σ mbound ( R,θ ) = Μ irˆ
r =R
= Μ o zˆ irˆ = Μ o cos θ
R
(See the previous BVP example, pages 17-18 above)
North Magnetic Poles (magnetic charges) have +gm.
− gm S
South Magnetic Poles (magnetic charges) have −gm.
Lines of H and B Produced by Uniformly Magnetized Sphere ( Μ = Μ o zˆ ):
Μ = Μ o zˆ
The lines of H and B produced by a uniformly magnetized sphere.
Again, we wish to emphasize the fact that H in ( r ) = − 13 Μ o zˆ points in the direction opposite
to the magnetization Μ = Μ o zˆ and also B in ( r ) = + 23 Μ o zˆ . The lines of H emanate/terminate
from the effective bound magnetic charge on the surface of the magnetized sphere. Note that the
lines of B close on themselves – they do not terminate/emanate from the effective bound
magnetic charges on the surface of the magnetized sphere.
Since the H in ( r ) “bucks” the magnetization Μ , it results in a demagnetizing effect, which
occurs over over a long period of time – e.g. centuries, for AlNiCo materials at room
temperature, T 300 K. How fast depends on the nature of the magnetic material, and on the
geometry of the magnetic material!
24
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 20.5
Prof. Steven Errede
The demagnetization effect of having H antiparallel to Μ can be quantified / characterized by
defining a quantity known as the Demagnetization Factor γ, defined as follows:
Vmin ( r , θ ) ≡ γ
Μo
r cos θ
4π
For the uniformly magnetized sphere, we found:
M
1
1
Vmin ( r ,θ ) = Μ o r cos θ hence we see that γ sphere o = M o
3
4π 3
∴ The demagnetization factor for a uniformly magnetized sphere is γ sphere =
4π
≈ 4.19 ~ 4.2
3
Different geometries of uniformly magnetized objects will have different values of
H in ( r ,θ ) = − [ _____ ] Μ o zˆ and hence different values of demagnetization factors γ.
e.g. For a large flat thin sheet lying in the x-y plane with uniform magnetization Μ = Μ o zˆ ,
γ sheet 4π 12.57 (very unstable magnetization has γ → ∞!!!)
For a very long, thin rod of radius R << length L with uniform magnetization Μ = Μ o zˆ ║ to the
long axis of rod, γ rod 0 !! (i.e. very stable magnetization has γ → 0)
Let us now also explicitly verify / show that the remaining boundary conditions (i.e. the ones
we didn’t use for determining the A and B′ coefficients are indeed satisfied, i.e. that this
particular physics problem is actually over-determined:
Bout ( r = R ) − Bin ( r = R ) = μo K bound × rˆ
BC 4):
i.e.
Then:
or:
= Bθ
out
r =R
( r = R ) − Bθ ( r = R ) = μo Kbound × rˆ r = R
in
1
2
μo Μ o sin θθ + μo Μ o sin θθ = μo K bound × rˆ
3
3
Μ o sin θθ = K bound × rˆ r = R
But we know that: K bound ( r = R, θ ) ≡ Μ × rˆ
r=R
= Μ o zˆ × rˆ
r=R
r =R
(
= −Μ o sin θ θ × rˆ
)
Since: zˆ = rˆ cos θ − θ sin θ and r̂ × θ = ϕˆ ⇒ θ × r̂ = −ϕˆ
Thus: K bound ( r = R,θ ) = +Μ o sin θϕˆ
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
25
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
∴ K bound × rˆ = Μ o sin θ (ϕˆ × rˆ ) = Μ o sin θθ
Lecture Notes 20.5
Prof. Steven Errede
with r × θ = ϕ and θ × ϕ = r and ϕ × r = θ
=θ
∴ Bθ
out
BC 5):
( r = R ) − Bθ ( r = R ) = μo Kbound × rˆ r = R
in
Yes!!!
( H ( r = R ) − H ( r = R )) = − ( Μ ( r = R ) − Μ ( r = R ))
out
r
in
r
out
r
{n.b. originally derived from:
1
μo
∫
S
in
r
Bida = 0 =
( ∫ H ida + ∫ Μida ) }
S
Then:
2
1
Μ o cos θ + Μ o cos θ = − ( 0 − Μ o cos θ )
3
3
But:
Μ = Μ o zˆ = Μ o rˆ cos θ − θ sin θ for r < R
∴
Μ o cos θ = +Μ o cos θ Yes!!!
(
S
)
This boundary condition can be rewritten (with n̂ = outward unit normal here!) as:
(H
out
inˆ − H in inˆ
)
r =R
(
= − Μ out inˆ − Μ in inˆ
)
r =R
= −σ mbound ( r = R )
Bound effective surface magnetic charge / magnetic pole density:
σ mbound ( r = R,θ ) = +Μ o cos θ ≡ Μ inˆ (for n̂ =radial outward normal unit vector here)
26
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 20.5
Prof. Steven Errede
Example 3 – Magnetic Field of Uniformly Magnetized Bar Magnet
Consider a rectangular bar magnet of dimensions x, y, z = a, b, c with uniform magnetization
M = Mo z :
Since problem has
Since J free = 0 everywhere in space,
manifest rectangular
symmetry → use
rectangular coordinates
∇ × H = 0 ; and since ∇ × M = 0 here,
then ∇ × B = ∇ × H = ∇ × M = 0
∴We may write
∇i H = −∇ 2Vm = 0
i.e. H = −∇Vm
∇ 2Vm ( x, y, z ) = 0 everywhere
So:
Then we need to solve ∇ 2Vm = 0 (Laplace’s Equation) for the magnetic scalar potential
()
Vm r = Vm ( x, y, z ) . In rectangular coordinates as in electrostatics case, try product solution of
the form:
()
Vm r = Vm ( x, y, z ) = X ( x ) Y ( y ) Z ( z ) i.e. use separation of variables technique)
⎛ ∂2
∂2
∂2 ⎞
∇ Vm ( x, y, z ) = ⎜ 2 + 2 + 2 ⎟ Vm ( x, y, z ) = 0
⎝ ∂x ∂y ∂z ⎠
⎛ ∂2
∂2
∂2 ⎞
X ( x)Y ( y ) Z ( z ) = 0
+
+
⎜ 2
2
2 ⎟
⎝ ∂x ∂y ∂z ⎠
Give three separated equations:
2
1 d X ( x)
= −α 2 → general solution X ( x ) ~ cos α x + sin α x
X ( x ) dx 2
2
2
1 d Y ( y)
= − β 2 → general solution Y ( y ) ~ cos β x + sin β x
2
Y ( y ) dy
2
1 d Z (z)
= γ 2 = α 2 + β 2 → general solution Z ( z ) ~ eγ z + e −γ z or: ~ cos γ x + sin γ x
Z ( z ) dz 2
Or:
1 iu − iu
(e + e )
2
1
sin u = ( eiu − e − iu )
2i
i ≡ −1
cos u =
1 u −u
(e + e )
2
1
sinh u ≡ ( eu − e −u )
2
cosh u ≡
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
27
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 20.5
Prof. Steven Errede
What are the boundary conditions for this problem?
Because M = M o z and from intrinsic geometrical symmetries associated with this problem, and
from the fact that we know that we can replace the magnetization M = M o z with effective
bound magnetic pole strength (magnetic charge) surface charge densities σ m = − M in ( n =
outward unit normal here) on the top and bottom surfaces. Thus, this problem has many
similarities to the electrostatics problem of a six-sided hollow, rectangular conducting box with 5
of its 6 sides at ground, and the top surface at potential Vm ( x, y, z = c ) = +Vo . Recall that only
potential differences have physical significance, thus ΔVm = Vm ( x, y, z = c ) + Vm ( x, y, z = 0 ) will
ultimately need to be tied in with the magnetization M = M o z .
Thus, on each of the six sides of the rectangular bar magnet, each side (i.e. face) is a magnetic
equipotential, and from symmetry of this problem:
Dirichlet
1) VmLHS ( x, 0, z ) = VmRHS ( x, b, z ) = 0
Boundary
2) Vmback ( 0, y, z ) = Vmfront ( a, y, z ) = 0
Conditions
3) Vmbottom ( x, y, 0 ) = 0
3’) Vmtop ( x, y, c ) = +Vo
BC 0) Of course, Vm ( x, y, z ) must be finite everywhere.
⊥
= Bin⊥ at each surface
BC 4) Bout
BC 5) H out − H in = K free × n
@ each
surface
= 0 (because K free = 0 here)
i.e. 5) H out − H in at each surface
⊥
⊥
sides
− H in⊥ ⎤⎦ = − ⎡⎣ M out
− M in⊥ ⎤⎦ =0±σonmonfourtop(-)
BC 6) ⎡⎣ H out
and bottom(+), respectively
BC 7) Bout − Bin = μo K TOT × n
@ each
surface
= μo K bound × n
@ each
surface
(because K free = 0 here)
Inside the rectangular bar magnet:
- The Dirichlet Boundary Conditions 1): VmLHS ( x, 0, z ) = VmRHS ( x, b, z ) = 0 on y require sin β y
solutions, with: sin β b = 0 or: β b = nπ , n = 1, 2, 3, . . . (i.e. β n =
nπ
, n = 1, 2, 3, . .)
b
n.b. n = 0 and m = 0 solutions not allowed because then Vm ( x, y, z ) = 0 everywhere.
28
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 20.5
Prof. Steven Errede
- The Dirichlet Boundary conditions 2): Vmback ( 0, y, z ) = Vmfront ( a, y, z ) = 0 on x require sin α x
solutions, with: sin α a = 0 or: α a = mπ , m = 1, 2, 3, . . . (i.e. α m =
mπ
, m = 1, 2, 3, . . .)
a
- The Dirichlet Boundary conditions 3): Vmbottom ( x, y, 0 ) = 0 and 4): Vmtop ( x, y, c ) = +Vo require
2
⎛ mπ ⎞ ⎛ nπ ⎞
γ m,n ≡ α + β = ⎜
⎟ +⎜
⎟
⎝ a ⎠ ⎝ b ⎠
m = 1, 2, 3, . . . and n = 1, 2, 3, . . .
sinh γ z solutions, with:
2
m
2
2
n
∴ Inside the rectangular bar magnet, the general solution for the magnetic scalar potential is of
the form:
⎛ ⎛ mπ ⎞2 ⎛ nπ ⎞2 ⎞
∞
∞
⎛ mπ ⎞ ⎛ nπ ⎞
in
in
Vm ( x, y, z ) = ∑ ∑ Am ,n sin ⎜
x ⎟ sin ⎜
y ⎟ sinh ⎜ ⎜
⎟ +⎜
⎟ z ⎟⎟
⎜
a
b
a
b
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠ ⎠
m =1 n =1
⎝
At z = c we must have (BC 4)):
∞
in
m
V
∞
( x, y, c ) = +Vo = ∑ ∑ A
m =1
n =1
in
m ,n
⎛ ⎛ mπ ⎞2 ⎛ nπ ⎞2 ⎞
⎛ mπ x ⎞ ⎛ nπ y ⎞
sin ⎜
⎟ sin ⎜
⎟ sinh ⎜⎜ ⎜
⎟ +⎜
⎟ c⎟
a ⎠ ⎝ b ⎠ ⎟
⎝ a ⎠ ⎝ b ⎠
⎝
⎝
⎠
- Now, take inner products (i.e. use orthogonality properties of sin α px and sin β qy ) to “project
out” the p,qth term (i.e. coefficient Apq where p,q = 1, 2, 3, . . . )
⎛ pπ x ⎞ ⎛ qπ y ⎞
→ Multiply both sides of above expression (BC 4) and z = c) by sin ⎜
⎟ sin ⎜
⎟ and then
⎝ a ⎠ ⎝ b ⎠
integrate over
∫
x =a
x =0
dx ∫
y =b
y =0
dy :
x=a
y =b
⎛ pπ x ⎞ ⎛ qπ y ⎞
⎛ pπ x ⎞ ⎛ qπ y ⎞
in
=
+
,
,
sin
sin
sin ⎜
V
x
y
c
dxdy
V
(
)
m
o
⎜
⎟
⎜
⎟
⎟ sin ⎜
⎟ dxdy
∫x=0 ∫y =0
∫
∫
x
y
0
0
=
=
⎝ a ⎠ ⎝ b ⎠
⎝ a ⎠ ⎝ b ⎠
⎛ ⎛ mπ ⎞2 ⎛ nπ ⎞2 ⎞ x = a y =b ⎛ mπ x ⎞ ⎛ nπ y ⎞ ⎛ pπ x ⎞ ⎛ qπ x ⎞
∞
∞
in
sin ⎜
= ∑ ∑ Am, n sinh ⎜ ⎜
+
c⎟
⎟ sin ⎜
⎟ sin ⎜
⎟ sin ⎜
⎟ dxdy
⎜ ⎝ a ⎟⎠ ⎜⎝ b ⎟⎠ ⎟ ∫x =0 ∫y =0
a ⎠ ⎝ b ⎠ ⎝ a ⎠ ⎝ a ⎠
⎝
m =1 n =1
⎝
⎠
x =a
y =b
Now define:
When:
⎛π ⎞
u ≡⎜ ⎟x
⎝a⎠
⎛π ⎞
du = ⎜ ⎟ dx
⎝a⎠
x=0→u=0
x=a→u=π
⎛π ⎞
v≡⎜ ⎟y
⎝b⎠
⎛π ⎞
dv = ⎜ ⎟ dy
⎝b⎠
when:
→
→
⎛a⎞
x = ⎜ ⎟u
⎝π ⎠
⎛a⎞
dx = ⎜ ⎟ du
⎝π ⎠
⎛b⎞
y = ⎜ ⎟v
⎝π ⎠
⎛b⎞
dy = ⎜ ⎟ dv
⎝π ⎠
y=0→v=0
y=b→v=π
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
29
UIUC Physics 435 EM Fields & Sources I
Then:
Fall Semester, 2007
Lecture Notes 20.5
Prof. Steven Errede
a ⎛π ⎞
a
⎜ ⎟ δ m, p = δ m, p
2
π ⎝ 2⎠
y =b
b v =π
b ⎛π ⎞
b
⎛ nπ y ⎞ ⎛ qπ y ⎞
∫y =0 sin ⎜⎝ b ⎟⎠ sin ⎜⎝ b ⎟⎠ dy ⇒ π ∫v=0 sin ( nv ) sin ( qv ) dv = π ⎜⎝ 2 ⎟⎠ δ n,q = 2 δ n,q
a
⎛ mπ x ⎞ ⎛ pπ x ⎞
∫x=0 sin ⎜⎝ a ⎟⎠ sin ⎜⎝ a ⎟⎠ dx ⇒ π
x =a
∫
u =π
u =0
sin ( mu ) sin ( pu ) du =
δ i , j = Kroenecker δ-function =
Where:
0 for i ≠ j
i, j = 1, 2, 3, . . .
1 for i = j
Then: +Vo ∫
x=a
x =0
Again, let:
When:
Then:
∫
y =b
y =0
⎛ ⎛ pπ ⎞2 ⎛ qπ ⎞2 ⎞ ⎛ a ⎞ ⎛ b ⎞
⎛ pπ x ⎞ ⎛ qπ y ⎞
in
sin ⎜
⎟ sin ⎜
⎟ dxdy = Ap , q sinh ⎜⎜ ⎜
⎟ +⎜
⎟ c ⎟⎟ ⎜ ⎟ ⎜ ⎟
a
b
a
b
⎝
⎠ ⎝
⎠
⎠ ⎝
⎠ ⎠⎝ 2 ⎠⎝ 2 ⎠
⎝ ⎝
⎛ pπ ⎞
u ≡⎜
⎟x
⎝ a ⎠
⎛ pπ ⎞
du = ⎜
⎟ dx
⎝ a ⎠
⎛ qπ ⎞
v≡⎜
⎟y →
⎝ b ⎠
⎛ qπ ⎞
dv = ⎜
⎟ dy →
⎝ b ⎠
x=0→u=0
x = a → u = pπ
when:
⎛ a ⎞
x=⎜
⎟u
⎝ pπ ⎠
⎛ a ⎞
dx = ⎜
⎟ du
⎝ pπ ⎠
⎛ b ⎞
y =⎜
⎟v
⎝ qπ ⎠
⎛ b ⎞
dy = ⎜
⎟ dv
⎝ qπ ⎠
y=0→v=0
y = b → v = qπ
u = pπ
v = qπ
⎛ b ⎞⎛ a ⎞
⎜
⎟⎜
⎟ Vo ∫u =0 ∫v =0 sin ( u ) sin ( v ) dudv
⎝ qπ ⎠ ⎝ pπ ⎠
⎛ a ⎞⎛ b ⎞
pπ
qπ
= +Vo ⎜
⎟⎜
⎟ [ cos u ] 0 [ cos v ] 0
⎝ pπ ⎠ ⎝ qπ ⎠
⎛ a ⎞⎛ b
= +Vo ⎜
⎟⎜
⎝ pπ ⎠ ⎝ qπ
⎞
⎟ ⎡⎣cos ( pπ ) − 1⎤⎦ ⎡⎣cos ( qπ ) − 1⎤⎦
⎠
with: p = 1, 2, 3, 4, . . .
q = 1, 2, 3, 4, . . .
when p or q = odd integer (1, 3, 5, . . .):
⎛ 2a ⎞ ⎛ 2b ⎞
cos podd π = cos qodd π = −1 ⇒ above expression = +Vo ⎜
⎟⎜
⎟
⎝ podd π ⎠ ⎝ qodd π ⎠
But when p or q = even integer (2, 4, 6, . . .):
cos pevenπ = cos qevenπ = +1 ⇒ above expression vanishes for either p = even integer or q
= even integer!!
∴Only odd integer values of p and q give non-zero values for above expression (due to manifest
symmetry of problem in x and y directions!!)
30
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
⎛
⎛ 2a ⎞ ⎛ 2b ⎞
in
∴+Vo ⎜
⎟⎜
⎟ = Apodd ,qodd sinh ⎜
⎜
⎝ podd π ⎠ ⎝ qodd π ⎠
⎝
⎛ 4 ⎞⎛ 4 ⎞
Ainpodd ,qodd = +Vo ⎜
⎟⎜
⎟
⎛ ⎛
⎝ podd π ⎠ ⎝ qodd π ⎠
sinh ⎜ ⎜
⎜ ⎝
⎝
Or:
Lecture Notes 20.5
Prof. Steven Errede
2
2
⎛ podd π ⎞ ⎛ qodd π ⎞ ⎞⎟ ⎛ a ⎞ ⎛ b ⎞
⎜
⎟ +⎜
⎟ c ⎜ ⎟⎜ ⎟
⎝ a ⎠ ⎝ b ⎠ ⎟⎠ ⎝ 2 ⎠ ⎝ 2 ⎠
1
podd =1,3,5,...
qodd =1,3,5,...
2
2 ⎞
podd π ⎞ ⎛ qodd π ⎞ ⎟
⎟ +⎜
⎟ c
a ⎠ ⎝ b ⎠ ⎟
⎠
Therefore, inside the rectangular bar magnet, the specific solution for the magnetic scalar
potential is of the form:
∞
in
m
V
( x, y , z ) = ∑
m = odd
integers
∞
∑
A
n = odd
integers
in
m,n
⎛ mπ
sin ⎜
⎝ a
⎛ 4 ⎞⎛ 4 ⎞
With: Amin,n ≡ +Vo ⎜
⎟⎜
⎟
⎝ mπ ⎠⎝ nπ ⎠
⎛
sinh ⎜
⎜
⎝
⎛ ⎛ mπ ⎞2 ⎛ nπ ⎞2 ⎞
⎞ ⎛ nπ ⎞
+
x ⎟ sin ⎜
y ⎟ sinh ⎜ ⎜
z⎟
⎜ ⎝ a ⎟⎠ ⎜⎝ b ⎟⎠
⎟
⎠ ⎝ b ⎠
⎝
⎠
1
m =1,3,5,7,...
(
n =1,3,5,7,... )
2
2
⎞
⎛ mπ ⎞ ⎛ nπ ⎞
⎜
⎟ +⎜
⎟ c ⎟⎟
⎝ a ⎠ ⎝ b ⎠
⎠
Physically, these terms represent the 3-D spatial Fourier Harmonic Amplitudes associated with a
3-D rectangular “wave” – i.e. a 3-D rectangular box potential (here) an infinite series of such
terms is required in order to properly mathematically define the abrupt / sharp edges of this
object (in 3-D):
+Vo on top
z
c
V = 0 everywhere else on remaining 5 sides
(LHS, RHS, front, back and bottom)
b
y
a
x
Outside the rectangular bar magnet, we require solutions which either vanish or constant value
(at least) when x → ± ∞, y → ± ∞ and / or when z → ± ∞, i.e. when an observer is infinitely far
away from the bar magnet, because for either Vmout r = constant or ? when r → ∞ , since
()
H
out
( r ) ≡ −∇V ( r ) , then H ( r ) → 0 when r → ∞ (hence B ( r ) = μ H ( r ) → 0 when
out
m
out
out
out
o
r →∞.
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
31
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 20.5
Prof. Steven Errede
However, we also require continuity of the magnetic scalar potential at / on each of the six sides
of the rectangular bar magnet, i.e.:
Back and Front surfaces:
Vmout ( 0, y, z ) = Vmin ( 0, y, z ) = 0
Vmout ( a, y, z ) = Vmin ( a, y, z ) = 0
LHS and RHS surfaces:
Vmout ( x, 0, z ) = Vmin ( x, 0, z ) = 0
Vmout ( x, b, z ) = Vmin ( x, b, z ) = 0
Bottom and Top surfaces:
Vmout ( x, y, 0 ) = Vmin ( x, y, 0 ) = 0
Vmout ( x, y, c ) = Vmin ( x, y, c ) = ?
()
The “natural” choice for general form solutions to Vmout r would be e.g. e ± kx however we
cannot choose such exponential type solutions for all of x and y and z because of the constraint
γ 2 = α 2 + β 2 - i.e. at least one solution in x or y or z must be oscillatory (i.e. sine or cosine),
because of this constraint.
Let us re-examine ∇ 2Vmout ( x, y, z ) = 0 again. We still want product-type solutions of the form
Vmout ( x, y, z ) = X out ( x ) Y out ( y ) Z out ( z ) with:
d 2 X out ( x )
= − A2 X out ( x )
dx 2
d 2Y out ( y )
= − B 2Y out ( y )
2
dy
and with:
d 2 Z out ( z )
= +C 2 Z out ( z )
dz 2
⎛
⎞
⎜ ∂2
∂2
∂ 2 ⎟ out
⎜ 2 + 2 + 2 ⎟ Vm ( x, y, z ) = 0
⎜ ∂x ∂y ∂z ⎟
⎝
⎠
⎛
⎞
⎜ − A2 − B 2 + C 2 = 0 ⎟
C 2 = A2 + B 2
⎜
⎟
⎝
⎠
However, here we will define:
A12 ≡ (α + i β ) = α 2 + 2iαβ − β 2 ⇒ − A2 = ⎡⎣i (α + i β ) ⎤⎦ = − (α + i β )
2
2
B12 ≡ ( γ + iδ ) = γ 2 2iγδ − δ 2 ⇒ − B 2 = ⎡⎣i ( γ + iδ ) ⎤⎦ = − ( γ + iδ )
2
2
2
2
C12 ≡ ( μ + iv ) = μ 2 + 2i μ v − v 2
2
With:
C12 = A12 + B12 ⇒ (α 2 − β 2 ) + ( γ 2 − δ 2 ) = ( μ 2 − v 2 )
And:
αβ + γδ = μ v
Solutions are then of the form:
X ( x ) ~ ei(α +iβ ) x = e( iα − β ) x
Y ( y ) ~ ei(γ +iδ ) y = e( iγ −δ ) y
Z ( z ) ~ e( μ +iv ) z = e( μ +iv ) z
32
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 20.5
Prof. Steven Errede
dX ( x )
= i (α + i β ) e( iα − β ) x
dx
dY ( y )
i γ + iδ y
= i ( γ + iδ ) e ( )
dy
dZ ( z )
= ( μ + iv ) e( μ +iv )
dz
d 2 X ( x)
2
= − ( α + β ) e i (α − i β ) x
2
dx
d 2Y ( y )
2
= − ( γ + iδ ) ei(γ +iδ ) y
2
dy
2
d Y ( y)
= − B12Y ( y )
2
dy
d 2Z ( z )
2
= ( μ + iv ) e( μ +iv )
2
dz
Or:
d 2 X ( x)
= − A12 X ( x )
2
dx
d 2Z ( z )
= +C12 Z ( z )
2
dz
However, most / more generally there are actually four possible acceptable relations for each of
A, B and C (simply changing ± signs):
A12 ≡ (α + i β )
2
A2 ≡ ( −α + i β )
A3 ≡ (α − i β )
2
2
A4 ≡ ( −α − i β )
2
with
B12 ≡ ( γ + iδ )
with
B21 ≡ ( −γ + iδ )
with
B3 ≡ ( γ − iδ )
with
B4 ≡ ( −γ − iδ )
With: C12 = A12 + B12
(α
2
C22 = A22 + B22
− β 2 ) + (γ 2 − δ 2 ) = ( μ 2 − v2 )
2
2
2
2
and with
C1 ≡ ( μ + iv )
and with
C2 ≡ ( − μ + iv )
and with
C3 ≡ ( μ − iv )
and with
C4 ≡ ( − μ − iv )
C32 = A32 + B32
(α
αβ + γδ = μ v
2
2
2
2
2
C42 = A42 + B42
− β 2 ) + (γ 2 − δ 2 ) = ( μ 2 − v2 )
αβ + γδ = μ v
n.b all relations the same for i = 1, 2, 3, 4
Thus, the most general solution for Vmout ( x, y, z ) will be of the form:
Vmout ( x, y, z ) = Ke ± i(α ±iβ ) x e ± i(γ ±iδ ) y e ±( u ±iv ) z
= Ke ±(iα ± β ) x e ±(iγ ±δ ) y e ±( u ±iv ) z
K = constant
For each variable / in each direction x, y and z, we will have to match nine separate solutions, e.g.
for x-z plane, when y ≤ 0 (-∞ < y ≤ 0):
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
33
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 20.5
Prof. Steven Errede
Then this also must be completed / repeated for a ≤ y ≤ b and completely repeated again for y ≥ b
region. This gives a total of 27 – 1 separate solutions for Vmout ( x, y, z ) , one for each of
9 x 3 = 27 – 1 = 26 regions each with unknown coefficients, and in general, we will again require
infinite odd-integer series solutions, once we start matching Vmout ( x, y, z ) = Vmm ( x, y, z ) at
surfaces / boundaries of rectangular bar magnet. Lots of equations / constraints to
simultaneously solve!! Doable, but with much, much work!!
Assuming we succeeded in uniquely and correctly determining the solution(s) Vmout ( x, y, z ) in all
⊥
26 regions exterior to the bar magnet, we would then e.g. apply BC 4) Bout
= Bin⊥ at each surface
and / or BC 5) H out = H in at each surface to then formally connect Vmout ( x, y, z ) solution(s) to
Vmin ( x, y, z ) .
Even though we do not explicitly have solution(s) for Vmout ( x, y, z ) , we can still easily determine
the fields inside the rectangular bar magnet, because H
in
( x, y, z ) ≡ −∇Vmin ( x, y, z ) and
Vmin ( x, y, z ) is explicitly known.
⎞
z ⎟ Vmin ( x, y, z )
⎠
⎛ ⎛ mπ ⎞2 ⎛ nπ ⎞2 ⎞
∞
∞
⎛ mπ ⎞ ⎛ nπ ⎞
With:
+
Vmin ( x, y, z ) = ∑ ∑ Amin,n sin ⎜
x ⎟ sin ⎜
y ⎟ sinh ⎜ ⎜
z⎟
⎜ ⎝ a ⎟⎠ ⎜⎝ b ⎟⎠ ⎟
⎝ a ⎠ ⎝ b ⎠
odd m odd n
⎝
⎠
1
⎛ 4 ⎞⎛ 4 ⎞
And with:
Amin,n = +Vo ⎜
H ≡ Hx x + H y y + Hz z
⎟⎜
⎟
⎛ ⎛ mπ ⎞ 2 ⎛ nπ ⎞ 2 ⎞
⎝ mπ ⎠ ⎝ nπ ⎠
c⎟
sinh ⎜ ⎜
+
⎜ ⎝ a ⎟⎠ ⎜⎝ b ⎟⎠
⎟
⎝
⎠
⎛ ⎛ mπ ⎞2 ⎛ nπ ⎞2 ⎞
∞
∞
⎛ mπ ⎞ in
⎛ mπ ⎞ ⎛ nπ ⎞
⎜ ⎜
∴ H xin ( x, y, z ) = − ∑ ∑ ⎜
cos
sin
sinh
+
A
x
y
z⎟
⎟ m,n
⎜
⎟ ⎜
⎟
⎜ ⎝ a ⎟⎠ ⎜⎝ b ⎟⎠
⎟
⎝ a ⎠ ⎝ b ⎠
odd m odd n ⎝ a ⎠
⎝
⎠
2
2
⎛ ⎛ mπ ⎞ ⎛ nπ ⎞
⎞
∞
∞
⎛ nπ ⎞ in
⎛ mπ ⎞
⎛ nπ ⎞
⎜
⎟
sin
cos
sinh
+
H yin ( x, y, z ) = − ∑ ∑ ⎜
A
x
y
z
⎟ m,n
⎜
⎟
⎜
⎟
⎜ ⎜⎝ a ⎟⎠ ⎜⎝ b ⎟⎠
⎟
⎝ a ⎠
⎝ b ⎠
odd m odd n ⎝ b ⎠
⎝
⎠
2
2
⎛ ⎛ mπ ⎞2 ⎛ nπ ⎞2 ⎞
∞
∞
⎛ mπ ⎞ ⎛ nπ ⎞ in
⎛ mπ ⎞ ⎛ nπ ⎞
⎜
sin
sin
cosh
+
+
H zin ( x, y, z ) = − ∑ ∑ ⎜
A
x
y
z⎟
⎟ ⎜
⎟ m,n
⎜
⎟ ⎜
⎟
⎜ ⎜⎝ a ⎟⎠ ⎜⎝ b ⎟⎠
⎟
⎝ b ⎠
⎝ a ⎠ ⎝ b ⎠
odd m odd n ⎝ a ⎠
⎝
⎠
H
in
⎛ ∂
∂
∂
x+
y+
∂y
∂z
⎝ ∂x
( x, y, z ) ≡ −∇Vmin ( x, y, z ) = − ⎜
** Note that H zin is anti-parallel to magnetization M = M o z
34
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
in
1
in
Fall Semester, 2007
(
in
in
B − M ⇒ B = μo H + M
Lecture Notes 20.5
)
Prof. Steven Errede
Then since:
H =
Then:
=0 ⎞
=0 ⎞
in
⎛
⎛
B = Bxin x + Byin y + Bzin z = μo ⎜ H xin + M x ⎟ x + μo ⎜ H yin + M y ⎟ y + μo ( H zin + M z ) z
⎝
⎠
⎝
⎠
μo
M = Mo z = M z z
= μo H xin x + μo H yin y + μo ( H zin + M o ) z
∴B
in
x
∞
( x, y , z ) = μ o H ( x , y , z ) = − μ o ∑
in
x
odd m
Byin ( x, y, z ) = μo H yin ( x, y, z ) = − μo
∞
⎛ ⎛ mπ ⎞2 ⎛ nπ ⎞2 ⎞
⎛ mπ ⎞ in
⎛ mπ ⎞ ⎛ nπ ⎞
∑ ⎜ ⎟ Am,n cos ⎜⎝ a x ⎟⎠ sin ⎜⎝ b y ⎟⎠ sinh ⎜⎜ ⎜⎝ a ⎟⎠ + ⎜⎝ b ⎟⎠ z ⎟⎟
odd n ⎝ a ⎠
⎝
⎠
∞
⎛ nπ
b
odd n
∞
∑ ∑ ⎜⎝
odd m
⎞ in
⎛ mπ
⎟ Am ,n sin ⎜
⎠
⎝ a
Bzin ( x, y, z ) = μo ( H zin ( x, y, z ) + M o )
= − μo
With:
∞
∞
odd m
odd n
2
2
⎛ mπ ⎞ ⎛ nπ ⎞ in
⎛ mπ
⎜
⎟ +⎜
⎟ Am, n sin ⎜
⎝ a ⎠ ⎝ b ⎠
⎝ a
∑ ∑
⎛ 4 ⎞⎛ 4 ⎞
Amin,n = +Vo ⎜
⎟⎜
⎟
⎝ mπ ⎠ ⎝ nπ ⎠
⎛ ⎛ mπ ⎞2 ⎛ nπ ⎞ 2 ⎞
⎞
⎛ nπ ⎞
x ⎟ cos ⎜
y ⎟ sinh ⎜ ⎜
z⎟
+
⎜ ⎝ a ⎟⎠ ⎜⎝ b ⎟⎠
⎟
⎠
⎝ b ⎠
⎝
⎠
⎛ ⎛ mπ ⎞2 ⎛ nπ ⎞2 ⎞
⎞ ⎛ nπ ⎞
x ⎟ sin ⎜
y ⎟ cosh ⎜ ⎜
⎟ +⎜
⎟ z ⎟⎟ + μo M o
⎜
b
a
b
⎠ ⎝
⎠
⎠ ⎝
⎠
⎝ ⎝
⎠
1
⎛ ⎛ mπ ⎞ 2 ⎛ nπ ⎞ 2 ⎞
c⎟
sinh ⎜ ⎜
+
⎜ ⎝ a ⎟⎠ ⎜⎝ b ⎟⎠
⎟
⎝
⎠
Now we can use
BC 6) to connect
0 on 4 sides
(H
⊥
out
⎛
− H in⊥ ) = − ⎜ M
⎜
⎝
=0
⊥
out
⎞
− M in⊥ ⎟ =
⎟
⎠
+σ m on bottom surface (@ z = 0)
−σ m on top surface (@ z = 0)
to connect Vo to Mo
i.e.
@ z = 0:
@ z = c:
(H
(H
out
z
out
z
− H zin )
− H zin )
z =0
= M o = +σ m
±σ m = bound magnetic surface
z =c
= − M o = −σ m
charge densities (“pole strength”
surface charge densities)
Obviously, we need to explicitly solve H zout ( x, y, z ) first in order to carry this out . . .
However, we can also turn this around, so that:
H zout z =0 = H zin z =0 + M o
H zout
z =c
From symmetry arguments, we also know that:
More generally:
H
out
= H zin
H zout
z =c
z =0
− Mo
= − H zout
z =c
( x , y , z ≤ 0 ) = − H ( x, y , z ≥ c )
out
Δz by same amounts
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
35
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 20.5
Prof. Steven Errede
If we go back to BC 3’):
Vmin ( x, y, z = c ) = +Vo =
With:
⎛ mπ
Amin,n sin ⎜
∑
⎝ a
n = odd
∞
∞
∑
m = odd
⎛ 4 ⎞⎛ 4 ⎞
Amin,n = +Vo ⎜
⎟⎜
⎟
⎝ mπ ⎠ ⎝ nπ ⎠
⎛ ⎛ mπ ⎞2 ⎛ nπ ⎞2 ⎞
⎞ ⎛ nπ ⎞
+
x ⎟ sin ⎜
y ⎟ sinh ⎜ ⎜
c⎟
⎜ ⎝ a ⎟⎠ ⎜⎝ b ⎟⎠
⎟
⎠ ⎝ b ⎠
⎝
⎠
1
⎛ ⎛ mπ ⎞ 2 ⎛ nπ ⎞ 2 ⎞
+
sinh ⎜ ⎜
c⎟
⎜ ⎝ a ⎟⎠ ⎜⎝ b ⎟⎠
⎟
⎝
⎠
Due to orthonormality properties of sine functions → we realize that:
1=
Because:
36
∞
∞
m = odd
n = odd
⎛ 4 ⎞⎛ 4 ⎞
⎛ mπ ⎞ ⎛ nπ ⎞
x ⎟ sin ⎜
y ⎟ = 1 ∗1
a ⎠ ⎝ b ⎠
∑ ∑ ⎜⎝ mπ ⎟⎠ ⎜⎝ nπ ⎟⎠ sin ⎜⎝
1=
∞
⎛ 4
∑
⎜
m = odd ⎝ mπ
⎞ ⎛ mπ
⎟ sin ⎜
⎠ ⎝ a
⎞
x⎟
⎠
and
1=
∞
⎛ 4 ⎞
⎛ nπ ⎞
y⎟
b ⎠
∑ ⎜⎝ nπ ⎟⎠ sin ⎜⎝
n = odd
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.