UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 16
Prof. Steven Errede
LECTURE NOTES 16
THE MAGNETIC VECTOR POTENTIAL A ( r )
We saw in electrostatics that ∇ × E = 0 {always} (due to intrinsic / microscopic nature of the
electrostatic field) permitted us to introduce a scalar potential V ( r ) such that:
E ( r ) ≡ −∇V ( r ) {n.b. V ( r ) is uniquely defined, up to an (arbitrary) constant.}
Analogously, in magnetostatics, the ∇i B ( r ) = 0 (always) { ⇒ ∃ no magnetic charges / no
magnetic monopoles} permits us to introduce a magnetic vector potential A ( r ) such that:
B (r ) ≡ ∇ × A(r )
Teslas
1
⇒ S.I. units of the magnetic vector potential A ( r ) = Tesla-meters
TeslaMeters
m
(
)
Then: ∇i B ( r ) = ∇i ∇ × A ( r ) = 0 {always}
The divergence of a curl of a vector field F ( r ) is always zero
Ampere’s Law:
In differential form:
(
)
(
)
∇ × B ( r ) = ∇ × ∇ × A ( r ) = ∇ ∇i A ( r ) − ∇ 2 A ( r ) = μo J free ( r )
Now, just as in the case of electrostatics, where V ( r ) was uniquely defined up to an arbitrary
constant (Vo ) , then let: V ′ ( r ) ≡ V ( r ) + Vo
=0
then:
E ( r ) = −∇V ′ ( r ) = −∇ (V ( r ) + Vo ) = −∇V ( r ) − ∇Vo = −∇V ( r )
i.e.
E ( r ) = −∇V ′ ( r ) = −∇V ( r )
An analogous thing occurs in magnetostatics - we can add / we have the freedom to add to the
magnetic vector potential A ( r ) the gradient of any scalar function A ( r ) ≡ ∇Φ m ( r ) where
Φ m ( r ) ≡ magnetic scalar potential SI Units of magnetic scalar potential Φ m ( r ) = Tesla-m2
Then: A′ ( r ) ≡ A ( r ) + A ( r ) = A ( r ) + ∇Φ m ( r ) ⇐ Formally known as a Gauge Transformation
The curl of the gradient of a scalar field ( Φ m ( r ) here) automatically/always vanishes, i.e.:
≡ 0 Always !!!
(
)
B ( r ) = ∇ × A′ ( r ) = ∇ × A ( r ) + A ( r ) = ∇ × A ( r ) + ∇ × A ( r ) = ∇ × A ( r ) + ∇ × ∇Φ m ( r )
= ∇ × A(r )
© 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 16
Prof. Steven Errede
Note that the magnetic scalar potential Φ m ( r ) has same physical units as magnetic flux Φ m :
Tesla-m2 = Weber
(Magnetic flux, Φ m = ∫ B ( r )idA !!)
S
eeek!!!!
⇒ Please do NOT confuse the magnetic scalar potential Φ m ( r ) (= a scalar point function,
whose value can change at each/every point in space, r ) with the magnetic flux Φ m (which is a
constant scalar quantity (i.e. a pure number), independent of position) Φ m ( r ) ≠ Φ m !!!
Thus, like the scalar potential V ( r ) , the magnetic vector potential A ( r ) is (also) uniquely
defined, but only up to an (arbitrary) vector function A ( r ) = ∇Φ m ( r ) .
A′ ( r ) ≡ A ( r ) + A ( r ) = A ( r ) + ∇Φ m ( r )
The definition B ( r ) ≡ ∇ × A ( r ) specifies the curl of A ( r ) , but in order to fully specify the
vector field A ( r ) , we additionally need to specify the divergence of A ( r ) , ∇i A ( r ) .
We can exploit the freedom of the definition of A ( r ) to eliminate the divergence of A ( r )
- i.e. a specific choice of A ( r ) will make A ( r ) divergenceless: ∇i A ( r ) = 0 ⇐ Coulomb Gauge
If:
A′ ( r ) ≡ A ( r ) + A ( r ) = A ( r ) + ∇Φ m ( r )
Then: ∇i A′ ( r ) = ∇i A ( r ) + ∇iA ( r ) = ∇i A ( r ) + ∇i∇Φ m ( r ) = ∇i A ( r ) + ∇ 2 Φ m ( r )
While the original magnetic vector potential, A ( r ) is not/may not be divergenceless, we can
make A′ ( r ) = A ( r ) + A ( r ) = A ( r ) + ∇Φ m ( r ) divergenceless, i.e. ∇i A′ ( r ) = 0 if we chose
A ( r ) = ∇Φ m ( r ) such that ∇iA ( r ) = ∇ 2 Φ m ( r ) = −∇i A ( r ) ⇐ Coulomb Gauge
A Simple Illustrative Example:
Suppose ∃ a region of space that has a uniform/constant magnetic field, e.g. B ( r ) = Bo zˆ .
⎛ ∂A ( r ) ∂Ax ( r ) ⎞
Then: B ( r ) = Bo zˆ = ∇ × A ( r ) = ⎜ y
−
⎟ zˆ .
∂y ⎠
⎝ ∂x
Thus (here): A ( r ) = Ax ( r ) xˆ + Ay ( r ) yˆ + Az ( r ) zˆ = Ax ( r ) xˆ + Ay ( r ) yˆ
=0
If Ax ( r ) = − Bo y and Ay ( r ) = Bo x , then A ( r ) = − 12 Bo yxˆ + 12 Bo xyˆ , and thus we see that this
1
2
1
2
choice of magnetic vector potential indeed gives us the correct B -field:
⎛ ∂A ( r ) ∂Ax ( r ) ⎞
1
1
−
B (r ) = ∇ × A(r ) = ⎜ y
⎟ zˆ = 2 Bo + 2 Bo = Bo zˆ
∂
∂
x
y
⎝
⎠
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 16
Prof. Steven Errede
⎛ ∂A ∂A ∂A ⎞ ⎛ ∂ ( − 12 Bo y ) ∂ ( 12 Bo x ) ∂ ( 0 ) ⎞
Is ∇i A = 0 satisfied? ∇i A = ⎜ x + y + z ⎟ = ⎜
+
+
⎟ = 0 Yes!!!
∂z ⎟⎠
∂y
∂y
∂z ⎠ ⎜⎝
∂x
⎝ ∂x
Note that we could also have instead chosen/used a different magnetic vector potential:
A′ ( r ) = A ( r ) + A ( r ) = A ( r ) + ∇Φ m ( r ) where e.g. A ( r ) = ∇Φ m ( r ) = Ao , i.e. where Ao is any
(arbitrary) constant vector, Ao = Aox xˆ + Aoy yˆ + Aoz zˆ . Since (here) A ( r ) = ∇Φ m ( r ) = Ao , then
A ( r ) = ∇Φ m ( r ) = Ao = Aox xˆ + Aoy yˆ + Aoz zˆ means that the gradient of the magnetic scalar
∂ ( Aoy y )
∂ ( Aox x )
∂ ( Aoz z )
xˆ +
yˆ +
zˆ = Aox xˆ + Aoy yˆ + Aoz zˆ = Ao = A ( r )
∂x
∂y
∂z
and thus the magnetic scalar potential itself (here) is: Φ m ( r ) = Aox xxˆ + Aoy yyˆ + Aoz zzˆ .
potential (here) is: ∇Φ m ( r ) =
Thus, here for the case of a constant/uniform magnetic field B ( r ) = Bo zˆ we see that there is
in fact a continuum of allowed magnetic vector potentials A′ ( r ) = A ( r ) + Ao = A ( r ) + ∇Φ m ( r )
that simultaneously satisfy B ( r ) = Bo zˆ = ∇ × A′ ( r ) and ∇i A′ ( r ) = 0 with the addition of an
(arbitrary) constant magnetic vector potential Ao = Aox xˆ + Aoy yˆ + Aoz zˆ contribution with
corresponding magnetic scalar potential Φ m ( r ) = Aox xxˆ + Aoy yyˆ + Aoz zzˆ . Note that this is exactly
analogous to the situation in electrostatics where the scalar electric potential V ( r ) is unique, up
to an arbitrary constant, Vo because there exists no absolute voltage reference in our universe –
i.e. absolute measurements of the scalar electric potential are meaningless - only potential
differences have physical significance!!!
We used this simple example of the constant/uniform magnetic field B ( r ) = Bo zˆ to elucidate
this particular aspect of the magnetic vector potential A ( r ) . Here in this particular example, we
found that the addition of an arbitrary constant vector
A ( r ) = Ao = Aox xˆ + Aoy yˆ + Aoz zˆ = ∇Φ m ( r ) to the magnetic vector potential A ( r ) was allowed,
i.e. A′ ( r ) = A ( r ) + A ( r ) = A ( r ) + Ao , which leaves the magnetic field B ( r ) unchanged. In
general there are many instances involving more complicated physics situations,
where B ( r ) ≠ constant vector field, where indeed B ( r ) = ∇ × A′ ( r ) and ∇i A′ ( r ) = 0 are
simultaneously satisfied for A′ ( r ) = A ( r ) + A ( r ) , because it is possible to determine/find a
corresponding magnetic scalar potential Φ m ( r ) for the problem satisfying ∇ 2 Φ m ( r ) = −∇i A ( r ) ,
but it is (very) important to understand that, in general, the allowed A ( r ) = ∇Φ m ( r ) (very
likely) may not be simply a constant vector field, but indeed one which varies in space (i.e. with
position vector, r )! Here again, however, the new A′ ( r ) = A ( r ) + A ( r ) will also be such that
B ( r ) = ∇ × A′ ( r ) will be unchanged, exactly analogous to V ′ ( r ) = V ( r ) + Vo leaving
E ( r ) = −∇V ′ ( r ) unchanged.
© 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
Fall Semester, 2007
Lecture Notes 16
Prof. Steven Errede
So we see that if ∇ 2 Φ m ( r ) = −∇i A ( r ) then yes, ∇i A′ ( r ) = 0 .
It is always possible to find an A ( r ) = ∇Φ m ( r ) in order to make ∇i A′ ( r ) = 0 .
Note however that this situation is then formally mathematically identical to Poisson’s Equation,
for the magnetic scalar potential Φ m ( r ) because:
∇ 2 Φ m ( r ) = − ρ m ( r ) ⇐ Analogous to ∇ 2V ( r ) = −
Equivalent magnetic
volume charge density
ρTot ( r )
in electrostatics!!!
ε0
Physically, ρ m ( r ) could e.g. be due to bound effective
⇐
magnetic charges associated with a magnetic material…
If we assume that the equivalent magnetic volume charge density, ρ m ( r ) ≠ 0 and we want
∇i A′ ( r ) = 0
Then: ∇i A ( r ) + ∇iA ( r ) = ∇i A ( r ) + ∇ 2 Φ m ( r ) = 0
Or:
∇i A ( r ) − ρ m ( r ) = 0 ⇒ ∇i A ( r ) = ρ m ( r )
Then, the solution to Poisson’s equation for the magnetic scalar potential Φ m ( r ) is of the form:
Φm ( r ) =
1
4π
∫
v′
ρm ( r′)
r
dτ ′ ⇐ Analogous to V ( r ) =
1
4πε 0
∫
v′
ρTot ( r ′ )
r
dτ ′ in electrostatics
with r = r − r ′
(n.b. these two relations are both valid assuming that ρ m ( r ′ ) and ρToT ( r ′ ) vanish when r ′ → ∞ !)
So then if ρ m ( r ′ ) = ∇i A ( r ′ ) , and ρ m ( r ′ ) = ∇i A ( r ′ ) vanishes as r ′ → ∞ , then the magnetic
scalar potential Φ m ( r ) is given by:
Φm ( r ) =
1
4π
∫
v′
ρm ( r′)
r
dτ ′ =
1
4π
∇i A ( r ′ )
dτ ′
v′
r
∫
{Note that if ∇i A ( r ′ ) = ρ m ( r ′ ) does not go to zero at infinity, then we’ll have to use some other
means in order to obtain an appropriate Φ m ( r ) , e.g. in an analogous manner to that which we’ve
had to do for the (electric) scalar potential V ( r ) associated with problems that have electric
charge distributions extending out to infinity.}
Thus, this choice of Φ m ( r ) ensures that indeed we can always make the magnetic vector
potential A′ ( r ) divergenceless, i.e. the condition ∇i A′ ( r ) = 0 (Coulomb Gauge) can always be
met, for the case of magnetostatics. Note that if ρ m ( r ′ ) = 0 then ρ m ( r ′ ) = ∇i A ( r ′ ) = 0 .
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 16
Prof. Steven Errede
With the choice of the magnetic scalar potential:
1
4π
Φm ( r ) =
∫
ρm ( r′)
v′
r
dτ ′ and A′ ( r ) = A ( r ) + Ao = A ( r ) + ∇Φ m ( r ) , and ∇i A′ ( r ) = 0
Then Ampere’s Law (in differential form) becomes:
(
)
(
)
∇ × B ( r ) = ∇ × ∇ × A ′ ( r ) = ∇ ∇ i A ′ ( r ) − ∇ 2 A′ ( r ) = μ 0 J free ( r )
=0
Which gives: ∇ A′ ( r ) = − μ0 J free ( r ) ⇐ Vector form of Poisson’s equation for magnetostatics.
2
i.e.:
⎧∇ 2 Ax′ ( r ) = − μo J x free ( r ) ⎫
The three separate/independent scalar forms
⎪⎪ 2
⎪⎪
⎨∇ A′y ( r ) = − μo J y free ( r ) ⎬ ⇐ of Poisson's equation are connected by:
⎪ 2
⎪
J free ( r ) = J x free ( r ) xˆ + J y free ( r ) yˆ + J z free ( r ) zˆ
⎩⎪∇ Az′ ( r ) = − μo J z free ( r ) ⎭⎪
n.b. in Cartesian coordinates: ∇ 2 A′ ( r ) = ( ∇ 2 Ax′ ( r ) ) xˆ + ( ∇ 2 A′y ( r ) ) yˆ + ( ∇ 2 Az′ ( r ) ) zˆ
However, in curvilinear coordinates (i.e. spherical-polar or cylindrical coordinates)
⎧rˆ = sin θ cos ϕ xˆ + sin θ sin ϕ yˆ + cos θ zˆ ⎫
⎪
⎪
e.g. spherical-polar coordinates: ⎨θˆ = cos θ cos ϕ xˆ + cos θ sin ϕ yˆ − sin θ zˆ ⎬
⎪ϕˆ = − sin ϕ xˆ + cos ϕ yˆ
⎪
⎩
⎭
Note that the unit vectors rˆ,θˆ + ϕˆ for spherical-polar coordinates are in fact explicit functions of
the vector position, r i.e. rˆ = rˆ ( r ) , θˆ = θˆ ( r ) and ϕˆ = ϕˆ ( r ) and therefore rˆ, θˆ + ϕˆ must also be
explicitly differentiated in calculating the Laplacian ∇ 2 of a vector function (here, A′ ( r ) ) in
curvilinear (i.e. either spherical-polar and/or cylindrical) coordinates!!! This is extremely
important to keep in mind, for the future…
⇒ The safest way to calculate the Laplacian of a vector function ∇ 2 A ( r ) in terms of curvilinear
coordinates, is to NOT use curvilinear coordinates!!! Failing that, then one should use:
(
)
(
)
(
∇ 2 A ( r ) = ∇ ∇i A ( r ) − ∇ × ∇ × A ( r ) = −∇ × ∇ × A ( r )
)
= 0 in the
Coulomb Gauge
If ρ m ( r ′ ) = 0 then (automatically) ρ m ( r ′ ) = ∇i A ( r ′ ) = 0 and we can use A ( r ) directly.
Hence, if ∇ 2 A ( r ) = − μ0 J free ( r ) (vector Poisson equation for magnetostatics),
then if J free ( r ) → 0 as r → ∞ , then A ( r ) =
μ0
4π
∫
v′
J free ( r )
r
dτ ′ where r = r = r − 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 16
Prof. Steven Errede
Generalizing this for a moving point charge as well as for line, surface and volume current
densities (with B ( r ) = ∇ × A ( r ) ), we summarize these results in the following table:
v ( r ′)
μo
q free
4π
r
I free ( r ′ )
μ
A(r ) = o ∫
d ′
4π C ′
r
μ
d ′
= o I free ∫
C′ r
4π
A(r ) =
A(r ) =
μ0
4π
∫
A(r ) =
μo
4π
∫
K free ( r ′ )
S′
r
J free ( r ′ )
v′
r
B (r ) =
v ( r ′ ) × rˆ
μo
q free
4π
r2
B (r ) =
μo
4π
=
∫
(I
free
( r′) d
(
μo
I free ∫
′
C
4π
B (r ) =
μo
4π
∫
dτ ′
B (r ) =
μo
4π
∫
)
2
C′
da′
′ × rˆ
(
r
d ′ ( r ′ ) × rˆ
r
K free ( r ′ ) × rˆ
2
S′
(
v′
)
2
r
J free ( r ′ ) × rˆ
r
2
) da′
) dτ ′
Note that: A v , I , d , K , J i.e. A is always parallel to the direction of motion of current, with
(
)
relative velocity v , whereas B = ∇ × A ⊥ v , I , d , K , J .
Note also that B and A both vanish when v → 0
(e.g. in the rest frame of a current (e.g. a proton or an electron beam)).
A Tale of Two Reference Frames:
For a pure point electric charge/point electric monopole moment, q we know that if it is moving
in the lab frame with speed v << c {c = speed of light in vacuum} that the magnetic field Bq ( r )
observed in the lab frame is:
Bq ( r ) = ∇ × Aq ( r ) =
q ⎛
rˆ ⎞ μ ⎛
rˆ ⎞
1
v × Eq ( r ) =
v × 2 ⎟ = o ⎜ qv × 2 ⎟
2
2 ⎜
c
r ⎠ 4π ⎝
r ⎠
4πε 0 c ⎝
(
)
Thus in the lab frame where this charged particle is moving, the magnetic vector potential Aq ( r )
associated with this moving charged particle (as observed in the lab frame) has a non-zero curl.
Contrast this with the situation in the rest frame of this pure point electric charge particle, where
the magnetic field vanishes, i.e. the magnetic vector potential Aq ( r ) associated with this charged
particle has no curl!!!
1 ∂V ( r , t )
in electrodynamics.
c2
∂t
⇒ ∃ connection between the A ( r , t ) − field and electric scalar potential V ( r , t ) - they are in fact
We will find out (next semester, in P436) that: ∇i A′ ( r , t ) = −
the 3 spatial & 1 temporal components of the relativistic 4-potential in electrodynamics !!!
6
© 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 16
Prof. Steven Errede
Uses of the Magnetic Scalar Potential Φ m ( r ) :
In certain (limited) circumstances for magnetostatics, it is actually possible to have the
magnetic field B ( r ) directly related to the (negative) gradient of a magnetic scalar potential
Φ m ( r ) , i.e. B ( r ) = − μo∇Φ m ( r ) , in direct analogy to that for electrostatics E ( r ) = −∇V ( r ) .
However, while ∇i B ( r ) = − μo∇i∇Φ m ( r ) = − μo ∇ 2 Φ m ( r ) = 0 is satisfied, i.e. ∇ 2 Φ m ( r ) = 0
is Laplace’s equation for the magnetic scalar potential, Φ m ( r ) (n.b. implying that ρ m ( r ) ≡ 0 ),
Ampere’s law ∇ × B ( r ) = − μo ∇ × ∇Φ m ( r ) = μo J ( r ) is not satisfied/is violated (!!!) unless
≡ 0 Always !!!
J ( r ) ≡ 0 everywhere in the region(s) of interest. These current-free regions must also be simplyconnected. {A region D (e.g. in a plane) is connected if any two points in the region can be
connected by a piecewise smooth curve lying entirely within D. A region D is a simply
connected region if every closed curve in D encloses only points that are in D.}
The use of B ( r ) = −∇Φ m ( r ) is in fact helpful for determining the magnetic fields associated
with e.g. current-carrying filamentary wires, current loops/magnetic dipoles, and e.g. the
magnetic fields associated with magnetized materials/magnetized objects.
The SI Units of the magnetic vector potential A are Tesla-meters (= magnetic field strength
per unit length), which is also equal to Newtons/Ampere (force per unit current) =
kg-meter/Ampere-sec2 = kg-meter/Coulomb-sec = (kg-meter/sec)/Coulomb = momentum per
unit charge, since (kg-meter/sec) are the physical units associated with momentum p = " mv " .
Thus, for the A -field:
1 Tesla-meter = 1 unit of
force
momentum
= 1 unit of
Ampere of current
Coulomb of charge
and for the B -field, from B ( r ) = ∇ × A ( r ) :
1 Tesla
= 1 unit of
force
N
momentum
meter =
=
meter
Ampere of current
A − m Coulomb of charge
Physically, the A -field has units of force per Ampere of current (or momentum per Coulomb of
electric charge), and thus physically, the magnetic field B ( r ) = ∇ × A ( r ) is the curl of the force
per unit current (or momentum per unit charge) field. Note also that force, F = dP dt and
current, I = dQ dt such that the magnetic vector potential A physically also has units of
⎛ Force ⎞ F dp / dt dp / dt Δp
=
=
and thus B is the curl of this physical quantity.
⎜
⎟= =
⎝ Current ⎠ I dq / dt dq / dt Δq
© 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 16
Prof. Steven Errede
The Magnetic Vector Potential of a Long Straight Wire Carrying a Steady Current
For a filamentary wire carrying steady current I, the magnetic vector potential and magnetic field
d ′ × rˆ
d ′
⎛μ ⎞
⎛μ ⎞
B (r ) = ∇ × A(r ) = ⎜ o ⎟ I ∫
and
are: A ( r ) = ⎜ o ⎟ I ∫ ′
r2
⎝ 4π ⎠ C r
⎝ 4π ⎠ C ′
Let the length of wire = 2L, and I = Izˆ and thus d ′ = d ′zˆ = dzzˆ
(
r = r = r − r′ = R2 +
)
2
The infinitesimal magnetic vector potential,
dA ( r = Ryˆ ) due to the current-carrying
line segment d ′ carrying current I is:
⎛μ
dA ( r = Ryˆ ) = ⎜ o
⎝ 4π
⎞ d ′ ( r ′ ) ⎛ μo
=⎜
⎟I
r
⎠
⎝ 4π
⎞ d ′ ( r′)
⎟I
⎠
R2 + 2
The corresponding infinitesimal magnetic field increment is:
d ′ ( r ′ ) ⎛ μo ⎞ ⎛
dzzˆ ⎞
⎛μ ⎞
=⎜
dB ( r = Ryˆ ) = ∇ × dA ( r = Ryˆ ) = ⎜ o ⎟ I ∇ ×
⎟
⎟ I ⎜∇×
r
⎝ 4π ⎠
⎝ 4π ⎠ ⎝
R2 + 2 ⎠
d
⎛ μ ⎞ z =+ L dzzˆ
⎛μ ⎞ L
= 2⎜ o ⎟ ∫
A ( r = Ryˆ ) = ∫ dA ( r = Ryˆ ) = ⎜ o ⎟ I ∫
2
2
=−
0
C′
z
L
⎝ 4π ⎠
⎝ 4π ⎠
R +
R2 + 2
Now:
L
⎛μ ⎞
⎛μ ⎞
= ⎜ o ⎟ I ln ⎡ + R 2 + 2 ⎤ zˆ = ⎜ o ⎟ I ln ⎡ L + R 2 + L2 ⎤ − ln R
⎣
⎦0
⎣
⎦
⎝ 4π ⎠
⎝ 4π ⎠
{
}
More generally, for a ⊥ distance R away from a long straight wire of length 2L carrying a steady
current I:
2 ⎫⎤
⎡⎛ L ⎞ ⎧
⎛μ ⎞
A ( r = Ryˆ ) = ⎜ o ⎟ I ln ⎢⎜ ⎟ ⎨1 + 1 + R
⎬⎥ zˆ
L
R
⎝
⎠
⎝ 4π ⎠
⎩
⎭⎦
⎣
( )
⎛μ
If L >> R, then: A ( r = Ryˆ ) ≈ ⎜ o
⎝ 4π
⎞
⎛ 2L ⎞
⎟ zˆ {Since 1 + ε ≈ 1 for ε
⎟ I ln ⎜
⎝ R ⎠
⎠
1 .}
Note that if L → ∞ (or R → 0), then A ( r = Ryˆ ) diverges (logarithmically)!
This is OK (unphysical anyway!) because even if A → ∞, B = ∇ × A ≠ ∞ !!! (necessarily!!!)
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 16
Prof. Steven Errede
So, for a ⊥ distance R away from a long straight wire of length 2L carrying a steady current I:
⎛μ
A ( r = Ryˆ ) = ⎜ o
⎝ 4π
2 ⎫⎤
⎡⎛ L ⎞ ⎧
⎞
R
ln
1
1
I
zˆ = Az ( r = Ryˆ ) zˆ
+
+
⎨
⎢⎜ ⎟
⎟
L ⎬⎥
⎠
⎭⎦
⎣⎝ R ⎠ ⎩
( )
Then B ( r ) = ∇ × A ( r ) Let’s do this in cylindrical coordinates: (note: ρ = x 2 + y 2 = R here):
=0
=0
⎛
⎜ 1 ∂Az ∂Aϕ
−
B ( ρ = R) = ⎜
∂z
ρ
ϕ
∂
⎜
⎝
ρˆ
B ( ρ = R) =
Or:
=0
⎞
⎛ =0
⎞
⎛
=0
∂Aρ
1⎜ ∂
⎟ˆ
⎜ ∂Aρ ∂Az ⎟ ˆ
+⎜
−
+ ⎜
( ρ Aϕ ) − ∂ϕ
⎟ρ
⎟ϕ
∂
∂
∂
z
ρ
ρ
ρ
⎟
⎜
⎟
⎜
⎠ ρ =R ⎝
⎠ ρ =R
⎝
ϕˆ
1 ∂
R ∂ρ
0
zˆ
∂
∂z
Az
0
0
In Cylindrical Coordinates:
=−
∂Az
ϕˆ
∂ρ ρ = R
ẑ
ρ =R
ϑ
Bρ = 0
Bz = 0
x̂
Then: B ( ρ = R ) = −
∂Az
ϕˆ
∂ρ
ρ =R
(
Now if: U ( R ) ≡ L + L2 + R 2
dU ( R )
=
dR
ϕ̂
ρ̂
)
(
⎛μ ⎞ ∂ ⎡
= −⎜ o ⎟ I
ln L + L2 + R 2 − ln R ⎤ ϕˆ
⎢
⎥⎦
⎝ 4π ⎠ ∂R ⎣
)
R
1 ⎡ dU ( R ) ⎤
∂
ln (U ( R ) ) =
⎢
⎥=
∂R
U ( R ) ⎣ dR ⎦
L + L2 + R 2
Since:
ŷ
ϕ
1
∂Az
∂Az
=−
ϕˆ
ϕˆ
Thus: Bϕ = i R
∂ρ ρ = R
∂ρ ρ = R
R
Then:
⎞
∂Az
⎟ˆ
=−
ϕˆ
⎟z
∂
ρ
ρ =R
⎟
⎠ ρ =R
(
R
L2 + R 2
and:
)(
L2 + R 2
)
∂
1
ln ( R ) ) =
(
∂R
R
⎧
⎫
B-field associated
⎪
⎪
with filamentary
μ
R
1
⎪
⎛
⎞ ⎪
− ⎬ ϕˆ ⇐ wire of length 2L
Then finally: B ( ρ = R ) = − ⎜ o ⎟ I ⎨
2 ⎛
2 ⎞
R⎪
⎝ 2π ⎠ ⎪ 2
carrying steady
L 1+ R
1+ 1+ R
⎜
⎟
L
L
current I.
⎪⎩
⎪
⎝
⎠
⎭
⎛μ ⎞I
Note that as L → ∞, then B ( ρ = R ) = − ⎜ o ⎟ ϕˆ ⇐ i.e. exactly the same as we obtained for
⎝ 2π ⎠ R
∞-long straight filamentary wire carrying steady current I (see previous P435 Lecture Notes)!!!
( )
( )
© 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 16
Prof. Steven Errede
The Magnetic Vector Potential A ( r ) and Magnetic Field B ( r ) = ∇ × A ( r ) Associated with
a Pair of Long, Parallel Wires Carrying Steady Currents I1 = + Izˆ and I 2 = − Izˆ ,
Separated by a Perpendicular Separation Distance, d
Plane ⊥ to
wires and
containing ⊥
separation
distance d
d
1
= + d zˆ = + dzzˆ
r 1 = r1 − r1′ = r − r1′
d
2
= − d zˆ = − dzzˆ
r 2 = r2 − r2′ = r − r2′
( r1 = r2 = r )
For simplicity’s sake, assume L >> R1, R2.
⎛μ
Then: A1 ( R1 ) ≈ + ⎜ o
⎝ 2π
⎛ 2L ⎞
⎞
⎟ zˆ and
⎟ I ln ⎜
⎠
⎝ R1 ⎠
⎛μ
A2 ( R2 ) ≈ − ⎜ o
⎝ 2π
⎛ 2L ⎞
⎞
⎟ zˆ
⎟ I ln ⎜
⎠
⎝ R1 ⎠
Then, using the principle of linear superposition:
⎛ 2 L ⎞ ⎛ μo
⎛μ ⎞
ATOT ( r ) = A1 ( r ) + A2 ( r ) ≈ + ⎜ o ⎟ I ln ⎜
⎟ zˆ − ⎜
⎝ 2π ⎠
⎝ R1 ⎠ ⎝ 2π
⎛ 2L ⎞
⎞
⎟ zˆ
⎟ I ln ⎜
⎠
⎝ R2 ⎠
2
Or:
10
⎛R ⎞
⎛R ⎞
⎛μ ⎞
⎛μ ⎞
ATOT ( r ) ≈ ⎜ o ⎟ I ln ⎜ 2 ⎟ zˆ = ⎜ o ⎟ I ln ⎜ 2 ⎟ zˆ for L >> R1, R2.
⎝ 2π ⎠
⎝ 4π ⎠
⎝ R1 ⎠
⎝ R1 ⎠
© 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 16
Prof. Steven Errede
Now let us re-locate the local origin to be at the LHS wire, where it intersects the ⊥ -plane:
Top View
I1 (out of page) d
I 2 (into page)
R12 = x 2 + y 2
d-y
y
R1
x
Then: ATOT ( r )
R22 = x 2 + ( d 2 − y 2 )
R2
⎛ μo
⎜ 4π
⎝
2
⎛ R2 ⎞
⎞
⎛ μo
⎟ I ln ⎜ R ⎟ zˆ = ⎜ 4π
⎠
⎝
⎝ 1⎠
⎡ x 2 + ( d − y )2 ⎤
⎞
⎟ I ln ⎢ x 2 + y 2 ⎥ zˆ for L >> R1, R2.
⎠
⎢⎣
⎥⎦
ATOT ( y = 0 ) diverges (x = 0)
(
ATOT ( r )
(for x = 0
i.e. observation
point on ŷ -axis) y = 0
ATOT y = d
2
) = 0 for I
y = d/2
1
= − I 2 = Izˆ
y=d
y
ATOT ( y = d ) diverges (x = 0)
Note that ATOT = Az zˆ i.e. AxTOT = ATOT
= 0 (since currents only in ± ẑ -direction) !!!
y
Note also that ATOT = Az zˆ changes sign – its direction is parallel to the closest current!!!
Then: BTOT ( r ) = ∇ × ATOT ( r ) , in Cartesian ( xˆ − yˆ − zˆ ) coordinates:
TOT
x
B
∂AzTOT
⎛ μ ⎞ ⎡(d − y) y ⎤
=+
= −⎜ o ⎟ I ⎢
+ 2⎥
2
R1 ⎦
∂y
⎝ 2π ⎠ ⎣ R2
BTOT
=+
y
∂AzTOT
⎛μ
= −⎜ o
∂x
⎝ 2π
x ⎤
⎞ ⎡ x
⎟I ⎢ 2 − 2 ⎥
⎠ ⎣ R2 R1 ⎦
R12 = x 2 + y 2
R22 = x 2 + ( d 2 − y 2 )
BzTOT = 0
© 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
Fall Semester, 2007
Lecture Notes 16
Prof. Steven Errede
At the point x = 0 and y = d/2 (i.e. at the center point, midway between the two conductors):
{where R1 = R2 = R = d/2}:
TOT
x
B
⎛μ
= −⎜ o
⎝ 2π
⎡ d
d ⎤
⎞ ⎢ 2
2 ⎥ = − ⎛ μo
+
⎟I ⎢
⎜
2
2⎥
⎠
⎝ 2π
d
d
⎢⎣ 2
2 ⎥⎦
( ) ( )
2
⎡
⎛ μo ⎞ 4
1 ⎤
⎛ 2 μo ⎞ I
⎞ ⎢ 1
⎟
⎟ I = −⎜
⎟ I d + d ⎥ = −⎜
2π ⎠ d
⎥
⎝ π ⎠d
⎠ ⎢
⎝
⎣ 2
2⎦
BTOT
=0
y
BzTOT = 0
Top View:
2μ ⎛ I ⎞
BTOT x = 0, y = d , z = 0 = − o ⎜ ⎟ xˆ
2
π ⎝d ⎠
(
)
The Magnetic Vector Potential A ( r ) and Magnetic Field B ( r ) = ∇ × A ( r )
Associated with a Magnetic Dipole Loop
(For Large Source-Observer Separation Distances)
For simplicity’s sake, let us choose the observation / field point P ( r ) to lie in the x-z plane:
Observation / field
Point P ( r ) = P ( x, 0, z )
Magnetic Dipole
Loop has radius R
R = r′
I = Iϕˆ
12
© 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
d ′ = Rdϕϕˆ
(d
Lecture Notes 16
Prof. Steven Errede
cos Ψ = rˆirˆ′ = opening angle between r̂ and r̂ ′
r i r ′ r ir ′ r ir ′
=
=
cos Ψ =
where: r ′ = r ′ = R
r i r ′ r ir ′ rR
′ = Rdϕ )
{from the arc length formula “ S = Rθ ”}
Id ′ ( r ′ ) ⎛ μo
⎛μ ⎞
=⎜
Now: A ( r ) = ⎜ o ⎟ ∫
′
r
⎝ 4π ⎠ C
⎝ 4π
d ′ ( r ′)
⎞
⎟ I ∫C′
r
⎠
⇒ A ( r ) (here) is a function of xˆ and yˆ only (actually only ϕ̂ ) and not ẑ since Id ′ = Id ′ lies
in the x-y plane and I = Iϕˆ (here).
Since we evaluate A in the x-z plane, the only component of d ′ that will contribute to A (there)
will be in the ŷ direction (n.b. A is parallel to the closest current from the observation point P(r)).
⇒ We only want the component of d ′ ( r ′ ) along the ŷ -axis, d
( r ′) cos ϕ
(Note: If we wanted
to evaluate A e.g. in the y-z plane, then we would want only the component of d ′ ( r ′ ) along the
x̂ -axis, d
( r ′) sin ϕ )
⎛ μ ⎞ 2π ( Rdϕ ) cos ϕ
ϕˆ ← ϕˆ = yˆ in the x-z plane for r = ( x, 0, z ) .
Then: A ( r ) = A ( x, 0, z ) = ⎜ o ⎟ I ∫
r
⎝ 4π ⎠ 0
Now: r 2 = r 2 + R 2 − 2rR cos Ψ (from the Law of Cosines) and r = r − r ′
1
And:
And:
2
⎛ R 2 2rR
⎞ 2
1⎛ R⎞ ⎛R⎞
= ⎜ 1 − 2 + 2 cos Ψ ⎟ ≈ 1 − ⎜ ⎟ + ⎜ ⎟ cos Ψ if R
r ⎝ r
r
2⎝ r ⎠ ⎝ r ⎠
⎠
r ir ′ = rr ′ cos Ψ = ( xxˆ + zzˆ )i( R cos ϕ xˆ + R sin ϕ yˆ ) = xR cos ϕ
r
r, r
and r ≈ r .
(if the observation / field point P ( r ) = P ( x, 0, z ) lies in the x-z plane)
2
1 ⎛ R ⎞ xR cos ϕ
1− ⎜ ⎟ +
Thus:
for r >> R and r r = r − r ′
2⎝ r ⎠
r2
r
⎛ 1 ⎛ R ⎞ 2 xR
⎞
⎛ μ ⎞ I ϕ = 2π
R cos ϕ ⎜1 − ⎜ ⎟ + 2 cos ϕ ⎟ dϕ yˆ
Then: A ( r ) = A ( x, 0, z ) ⎜ o ⎟ ∫
⎜ 2⎝ r ⎠
⎟
r
⎝ 4π ⎠ r ϕ =0
⎝
⎠
2
⎛ μ ⎞ Iπ R
A ( r ) = A ( x, 0, z ) ⎜ o ⎟ 3 xyˆ for r >> R and r r = r − r ′
Or:
⎝ 4π ⎠ r
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 16
But notice that in the x-z plane:
P ( r ) = P ( x, 0, z )
Prof. Steven Errede
ẑ
⎛x⎞
sin θ = ⎜ ⎟
⎝r⎠
θ
r
z
ŷ out of page
x
x̂
⎛ μ ⎞ Iπ R
A ( r ) = A ( x, 0, z ) ⎜ o ⎟ 2 sin θ yˆ for r >> R and r r = r − r ′
⎝ 4π ⎠ r
But in the x-z plane, ϕˆ = ŷ , and since the direction of A ( r ) is always parallel to the current:
2
∴
∴
A(r )
2
⎛ μo ⎞ I π R
⎜
⎟ 2 sin θϕˆ for r >> R and r
⎝ 4π ⎠ r
r = r − r ′ {n.b. A ( r ) I ϕˆ }
The magnetic dipole moment associated with this current-carrying loop is m = mzˆ
m = Ia = Iazˆ (for this planar loop) where (here): a = π R 2 zˆ (by the right-hand rule)
m = I π R 2 zˆ and: m = m = Ia = I π R 2 = π R 2 I
(
)
(
)
Note that: zˆ × rˆ = cos θ rˆ − sin θθˆ × rˆ = − sin θ θˆ × rˆ = − sin θ ( −ϕˆ ) = + sin θϕˆ
Thus, the quantity: m × rˆ = Iazˆ × rˆ = I π R 2 ( zˆ × rˆ ) = I π R 2 sin θϕˆ
∴
⎛ μ ⎞ m × rˆ ⎛ μ ⎞ m × r
A ( r ) ⎜ o ⎟ 2 = ⎜ o ⎟ 3 for r >> R and r
⎝ 4π ⎠ r
⎝ 4π ⎠ r
r = r − r′
Now B ( r ) = ∇ × A ( r ) in spherical coordinates for the magnetic dipole (with magnetic dipole
moment m = Ia = I π R 2 zˆ ) is:
⎛ μ ⎞ 2m
Br ( r ) = ⎜ o ⎟ 3 cos θ
⎝ 4π ⎠ r
⎛μ ⎞m
Bθ ( r ) = ⎜ o ⎟ 3 sin θ
⎝ 4π ⎠ r
valid for r >> R and r
r = r − r′
Bϕ ( r ) = 0
B ( r ) = Br ( r ) rˆ + Bθ ( r ) θ + Bϕ ( r ) ϕˆ
⎛μ
Thus: B ( r ) = ⎜ o
⎝ 4π
(
⎞m
ˆ
⎟ 3 2 cos θ rˆ + sin θθ
⎠r
)
for r >> R and r
r = r − r ′ and m = Ia = I π R 2 zˆ .
cf w/ the E -field associated with a physical electric dipole with dipole moment p = qd :
⎛ 1
E (r ) = ⎜
⎝ 4πε o
14
⎞ p
⎟ 3 2 cos θ rˆ + sin θθˆ for r >> d and r
⎠r
(
)
r = r − r ′ and p = qd .
© 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 16
Prof. Steven Errede
We already know what this B ( r ) looks like it – it is “solenoidal” around the current loop:
Cross-Sectional View of a Magnetic Dipole Loop:
⎛μ
Thus if m = Ia = I π R 2 zˆ = a constant vector, A ( r ) ⎜ o
⎝ 4π
m× r
⎛μ ⎞
B ( r ) = ∇ × A ( r ) = ⎜ o ⎟ ∇ × 3 for r >> R and r r
r
⎝ 4π ⎠
⎡
⎢
⎛
r
⎛ μo ⎞ ⎢
⎜ ∇i r 3
−
∇
+
i
B (r ) = ⎜
m
m
⎟⎢
3
r
⎜
⎝ 4π ⎠
r
⎝
⎢
=0
⎣
(
e.g. mx
and: ∇i
r
r
∴
3
=
)
⎞ m× r
⎟ 3 for r >> R and r
⎠ r
r = r − r′
= r − r′
⎤
⎞⎥
⎛
⎞
⎟ ⎥ ⇐ ⎜ ∇i r 3 = 0 ⎟
⎜ r
⎟
⎟⎥
⎝
⎠
⎠⎥
⎦
3 ( mi r ) r
3 ( mirˆ ) rˆ − m
r
m
∂ ⎛ r ⎞ mx xˆ
r
=−
⎜ 3 ⎟ = 3 − 3mx x 5 ∴ mi∇ 3 = 3 −
5
3
∂x ⎜⎝ r ⎟⎠ r
r
r
r
r
r
(
3
r
3
− ri
3r
r
5
=
⎛μ
B (r ) = ∇ × A(r ) = ⎜ o
⎝ 4π
3
r
3
−
3r ir
r
5
=
3
r
3
m × r ⎛ μo
⎞
⎟∇× 3 = ⎜
⎠
⎝ 4π
r
for r >> R (far away) and r
−
)
3r
r
2
5
=
3
r
3
−
3
r
3
⎞ ⎡ 3 ( mirˆ ) rˆ − m ⎤
⎥ ⇐
⎟⎢
3
⎠ ⎢⎣
r
⎥⎦
=0
Magnetic Field
of a Magnetic
Dipole Loop
r = r − r′
⇒ The magnetic field of a distant circuit (r >> R) does not depend on its detailed geometry,
but only its magnetic dipole moment, m !!! ⇐ Important (conceptual) result!!!
© 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 16
Prof. Steven Errede
Compare this result for B ( r ) for the magnetic dipole loop, with magnetic dipole moment
m = Ia , with result for the electric dipole field E ( r ) associated with a physical electric dipole
moment p = qd :
⎛μ
B (r ) = ⎜ o
⎝ 4π
⎞ ⎡ 3 ( mirˆ ) rˆ − m ⎤
⎥
⎟⎢
3
⎠ ⎢⎣
r
⎥⎦
m = Ia , for r
R (far away) and r
r = r − r′
⎛ 1 ⎞ ⎡ 3 ( p irˆ ) rˆ − p ⎤
E (r ) = ⎜
⎥ p = qd , for r
⎟⎢
3
r
⎝ 4πε o ⎠ ⎢⎣
⎦⎥
d (far away) and r
r = r − r′
+q
When r ≈ R (or less) for magnetic dipole loop
Or
r ≈ d (or less) for electric dipole, then
will be able to “see” / observe / detect higherorder moments - e.g. quadrupole, octupole,
sextupole, etc. . .moments of the B E fields.
p = qd
d
( )
−q
The statement that the magnetic field of a distant circuit (r >> R) does not depend on its detailed
geometry, but only its magnetic dipole moment, m = Ia (n.b. this is also true for the electrostatic
case, with p = qd ) are very useful!!!
If one can compute m = Ia then one can obtain A ( r ) and hence B ( r ) = ∇ × A ( r )
(or if have p = qd then can obtain E ( r ) ) for r >> R (or d) and r
r = r − r ′ . EASY!!!
Magnetic Flux Conservation
(
)
If B ( r ) = ∇ × A ( r ) then ∇i B ( r ) = ∇i ∇ × A ( r ) = 0 is automatically satisfied everywhere ( ∀ r )
If ∇i B ( r ) ≡ 0 for each/every point, r in a volume v bounded by its surface S
ˆ
(by the Divergence Theorem)
∫ ∇i B ( r ) dτ = ∫ B ( r )indA
ˆ ?? ∫ B ( r )indA
ˆ ≡0
What is ∫ B ( r )indA
Then:
v
S
S
S
Recall Gauss’ Law for E (and/or D ) were:
ˆ = Q enclosed
Φ D ≡ ∫ DindA
= net electric displacement flux through closed surface S .
free
S
ΦE ≡
ˆ =Q
∫ E ( r )indA
Thus: Φ m ≡
ˆ =0
∫ B ( r )indA
S
S
enclosed
Tot
ε o = net electric flux through closed surface S .
= net magnetic flux through closed surface S ≡ 0 !!!
⇒ Magnetic flux is conserved → magnetic field lines have no beginning / no end points
(because ∃ no magnetic charge(s)!)
The SI units of magnetic flux Φ m are Tesla-m2 = Webers
16
© 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 16
Prof. Steven Errede
Again: Do not confuse magnetic flux, Φ m with the magnetic scalar potential Φ m ( r )
(they even have the same units!!! Webers / Tesla-m2) WAA-HEE !!!
They are not the same thing!!!
A′ ( r ) = A ( r ) + ∇Φ m ( r )
B ( r ) = ∇ × A′ ( r ) = ∇ × A ( r ) + ∇ × ∇Φ m ( r )
ˆ
Φ m = ∫ B ( r )indA
S
Magnetic Flux
Magnetic Vector Potential
Area element
Magnetic Scalar Potential
Don’t confuse
these either!!!
Φm ( r ) =
∇ × A ( r ) = ∇ 2Φ m ( r ) = − ρm ( r )
1
4π
∫
∇i A ( r )
r
v
dτ
gives ∇i A′ ( r ) = 0
The magnetic flux through a surface S (not necessarily closed!!!):
ˆ = ∫ ∇ × A ( r ) indS
ˆ = ∫ A ( r )id by Stoke’s Theorem
Φ m = ∫ B ( r )indS
S
S
(
)
C
n.b. not a closed surface!
Φm =
Magnetic flux enclosed by contour C :
n.b. This d
S
(
)
C
is NOT a line segment associated with a line current I !!!
ˆ = ∫ ∇ × A ( r ) indS
ˆ =
Φ m = ∫ B ( r )indS
S
∫ A ( r )i d
∫ A ( r )i d
C
=
∫
C′
∇Φ m ( r )id = Φ m
© 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 16
Prof. Steven Errede
Griffiths Example 5.11:
A spherical shell of radius R carries a uniform surface charge density σ and rotates with
constant angular velocity ω , Determine the magnetic vector potential it produces at point r .
A rotating surface charge density σ produces a surface/sheet current density K ( r ′ ) = σ v ( r ′ )
The magnetic vector potential is thus: A ( r ) =
μo
4π
∫
S′
K ( r′)
r
da′
For ease of integration, choose the observation/field point P ( r ) = P ( zzˆ ) (i.e. r = zzˆ ) along the
+ ẑ -axis and ω to lie in the x-z plane. Choose the origin ϑ to be at the center of the sphere, as
shown in the figure below:
r = r − r ′ = R 2 + r 2 − 2 Rr cos θ ′ from the law of cosines
K ( r ′) = σ v ( r′)
v ( r ′) = ω × r ′
ω = ω sinψ xˆ + ω cosψ zˆ (here)
r ′ = R sin θ ′ cos ϕ ′xˆ + R sin θ ′ sin ϕ ′ yˆ + R cos θ ′zˆ
and:
v ( r ′) = ω × r ′
xˆ
v ( r ′) =
yˆ
zˆ
ω sinψ
ω cosψ
0
R sin θ ′ cos ϕ ′ R sin θ ′ sin ϕ ′ R cos θ ′
v ( r ′ ) = Rω ⎡⎣ − ( cosψ sin θ ′ sin ϕ ′ ) xˆ + ( cosψ sin θ ′ cos ϕ ′ − sinψ cos θ ′ ) yˆ + ( sinψ sin θ ′ sin ϕ ′ ) zˆ ⎤⎦
Now since
∫
2π
0
2π
sin ϕ ′dϕ ′ = ∫ cos ϕ ′dϕ ′ = 0
0
Then terms involving only sin ϕ ′ or cos ϕ ′ in the integral for A ( r ) contribute nothing.
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
Lecture Notes 16
Prof. Steven Errede
σ (ω × r ′ )
da′
r
r
with K ( r ′ ) = σ (ω × r ′ ) and da′ = R 2 sin θ ′dθ ′dϕ ′ and r = R 2 + r 2 − 2 Rr cos ϕ ′
A(r ) =
μo
4π
∫
S′
Then: A ( r ) =
Let:
K ( r ′)
Fall Semester, 2007
da′ =
μo
4π
μo R 3σω sinψ ⎛
u ≡ cos θ ′
du = sin θ ′dθ ′
π
⎞
dθ ′ ⎟ yˆ
R + r − 2 Rr cos θ ′
⎠
cos θ ′ sin θ ′
2
cos θ ′ sin θ ′
R 2 + r 2 − 2 Rr cos θ ′
(R
=−
2
θ ′ = 0 ⇒ u = +1
θ ′ = π ⇒ u = −1
π
0
S′
⎜ ∫0
⎝
2
∫
∫
2
+ r 2 + Rru )
3R 2 r 2
dθ ′ = ∫
+1
−1
udu
R 2 + r 2 − 2 Rru
u =+1
R 2 + r 2 − 2 Rru
u =−1
1
⎡ R 2 + r 2 + Rr ) R − r − ( R 2 + r 2 − Rr ) ( R + r ) ⎤
2 2 ⎣(
⎦
3R r
⎛ 2r ⎞
If r < R (i.e. inside sphere) then this integral = ⎜ 2 ⎟
⎝ 3R ⎠
⎛ 2R ⎞
If r > R (i.e. outside sphere) then this integral = ⎜ 2 ⎟
⎝ 3r ⎠
=−
Now: ω × r = −ω r sinψ yˆ
Then: A ( r ) =
μo Rσ
(ω × r ) for r < R (inside sphere)
3
μo R 4σ
A(r ) =
(ω × r ) for r > R (outside sphere)
3r 3
If we now rotate the problem so that ω = ω ẑ and r = ( r , ϑ , ϕ )
then ω × r = −ω r sinψ yˆ ⇒ ω r sin θϕˆ , thus with ω rotated to ω = ω ẑ and field point now located
at r = ( r , ϑ , ϕ ) , the magnetic vector potential A ( r ) inside/outside the rotating sphere becomes:
A ( r ,ϑ , ϕ ) =
μo Rωσ
r sin θϕˆ ( r < R , inside sphere)
3
μo R 4ωσ sin θ
ϕˆ ( r > R , outside sphere)
A ( r ,ϑ , ϕ ) =
3
r2
A(r )
Amax ( r = R )
~r
= μo R 2ωσ sin θ
1
3
~ 1 r2
r=R
Then: B ( r ) = ∇ × A ( r ) =
= zˆ
r
2μo Rωσ
2
2
cos θ rˆ − sin θθ = μo Rω zˆ = μo Rω !!! ( ω = ω ẑ )
3
3
3
(
)
© 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 16
Prof. Steven Errede
Griffiths Example 5.12:
Determine the magnetic vector potential A ( r ) of an infinitely long solenoid with n turns / unit
length, radius R and steady current I
⇒ n.b. The current extends to infinity, so we cannot use A ( r ) =
μo
4π
∫
C′
But we do know that:
(
)
I
r
d ′ because it diverges!
Magnetic flux Φ m = ∫ A ( r )id = ∫ ∇ × A ( r ) ida = ∫ Bida
C
S′
S′
Flux-enclosing loop / contour
Φ m = ∫ B ( r )ida
S
But we know from Ampere’s Circuital Law that:
Binside ( r ≤ R ) = μo nIzˆ = uniform & constant
= ( μo nI ) ∗ (π R 2 ) = μo nI π R 2
∴ Φ inside
m
= ∫ A ( r = R )i d
But: Φ inside
m
C
where d = Rdϕϕˆ
= A ( r = R ) 2π R
∴ Φ inside
m
Now Asolenoid must be parallel to I = Iϕˆ
for the “ideal solenoid” (i.e. no pitch angle)
⇒ A ( r ) = A ( r ) ϕˆ
Then: A ( r = R ) =
Φ inside
μ nI π R 2
1
m
ϕˆ = μo nIRϕˆ
= o
2
2π R
2π R
If r > R, then more generally, we have:
For r < R, then:
Aoutside ( r > R ) =
Ainside
⎛ R2 ⎞
1
μo nI ⎜ ⎟ ϕˆ
2
⎝ r ⎠
1
( r < R ) = μo nIrϕˆ
2
A(r )
Amax ( r = R )
= 12 μo nIR
~r
~1 r
r=R
Note that: A ( r ) = Aϕ ( r ) ϕˆ (only) for the infinitely long ideal solenoid.
20
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
r
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 16
Prof. Steven Errede
Does B ( r ) = ∇ × A ( r ) ?
B (r ) = ∇ × A(r ) =
1 ∂
( rAϕ ( r ) ) zˆ in cylindrical coordinates
r ∂r
=0
2
1
1
1 ∂
⎛1⎞ ∂ ⎛ R ⎞
Boutside ( r > R ) = μo nI ⎜ ⎟ ⎜ r
R 2 ) zˆ ≡ 0
(
⎟ zˆ = μo nI
2
r ∂r
2
⎝ r ⎠ ∂r ⎝ r ⎠
1
1
⎛1⎞ ∂
⎛1⎞
Binside ( r < R ) = μo nI ⎜ ⎟ ( r 2 ) zˆ = μo nI ⎜ ⎟ ( 2r ) zˆ = μo nIzˆ
2
2
⎝ r ⎠ ∂r
⎝r⎠
A ( r ) = Aϕ ( r ) ϕˆ
Does ∇i A ( r ) = 0 ?? (Coulomb Gauge)
In Cylindrical Coordinates:
1 ∂Aϕ ( r )
∇i A ( r ) =
= 0 because A ( r ) has NO explicit ϕ -dependence!
r ∂ϕ
Aoutside ( r > R ) =
Ainside ( r < R ) =
⎛ R2 ⎞
1
μo nI ⎜ ⎟ ϕˆ
2
⎝ r ⎠
1
μo nIrϕˆ
2
Magnetostatic Boundary Conditions
In the case of electrostatics, we learned (via use of Gauss’ Law enclosed
ˆ = QTot
Φ E = ∫ E ( r )inda
ε o ) that the normal component of E ( r ) suffers a discontinuity
S
whenever there is a surface charge density (free or bound) present on a surface / interface:
E2⊥ above − E1⊥ below =
(Ε
above
2
)
σ Tot (σ free + σ bound ) ∂V2above
=
=
εo
εo
∂n
−
surface
∂V1below
∂n
surface
= Ε1above ( Ε is continuous across interface)
n.b. ⊥ = perpendicular component relative to surface, = parallel component relative to surface:
© 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 16
Prof. Steven Errede
Consider a thin conducting sheet of material carrying a surface current density of
K = − Kxˆ = K ( − xˆ ) Amperes/meter
Now imagine that this current sheet K = − Kxˆ = K ( − xˆ ) is “placed” in an external magnetic field,
e.g. created / emanating from some other current-carrying circuit below this current sheet.
Call this external magnetic field that is below the original current sheet B1below
.
ext
What we discover is that the magnetic field above the current sheet B2above
is not parallel to B1below
ext
ext
- it has been refracted by the current sheet (in the tangential direction - with respect to the
surface)!
The physical origin for this is simple to understand. Below the current sheet, the current sheet
1
below
below
itself adds to the tangential component of Bext
a component Bsheet
= − μo Kyˆ (for K = − Kxˆ ),
2
above
however, above the current sheet, the current sheet adds to the tangential component of Bext
a
1
above
component Bsheet
= + μo Kyˆ (for K = − Kxˆ ).
2
⊥
zˆ
So if: Bext = Bextx xˆ + Bext y yˆ + Bext
z
below
below
below
⊥below
= Bextx xˆ + Bext y yˆ + Bext
zˆ
Then: Bext
z
And:
22
above
above
⊥ above
above
Bext
= Bextx xˆ + Bext y yˆ + Bext
zˆ
z
║ = parallel to surface
⊥ = perpendicular to surface
© 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 16
Prof. Steven Errede
Then by the principle of linear superposition, BTot = Bext + Bsheet .
Hence, below the current sheet ( K = − Kxˆ ):
= B sheet ⎞
⎛
⎜
⎟
1
below
⊥below
below
below
below
ˆ
⎜
⎟ yˆ + Bext
= Bextbelow
+
−
x
B
μ
K
zˆ = BTOT
xˆ + BTOT
yˆ + BTOT
zˆ
ext y
o
x
z
x
y
z
2
⎜
⎟
⎜
⎟
⎝
⎠
below
below
BTOT
And above the current sheet ( K = − Kxˆ ):
= B sheet ⎞
⎛
⎜ above 1
⎟
⊥ above
above
above
above
ˆ
⎜
⎟ yˆ + Bext
= Bextabove
+
+
x
B
μ
K
zˆ = BTOT
xˆ + BTOT
yˆ + BTOT
zˆ
ext
o
x
y
z
x
y
z
2
⎜
⎟
⎜
⎟
⎝
⎠
above
above
BTOT
above
below
Thus, (comparing BTOT
vs. BTOT
component-by-component), we see that:
1)
2)
below
above
BTOT
= BTOT
x
x
Tangential (to sheet / surface) component of BTOT parallel
Bextbelow
= Bextabove
x
x
to sheet current K = − Kxˆ is continuous.
BTOTy ≠ BTOTy
Tangential (to sheet/surface) component of BTOT perpendicular
below
above
above
below
BTOTy − BTOTy = μo K to sheet current K = − Kxˆ is discontinuous by an amount
μo K across sheet / surface.
3)
⊥below
⊥ above
BTOT
= BTOT
z
z
Normal (to sheet/surface) component of BTOT is continuous
⊥below
⊥ above
Bext
= Bext
z
z
across sheet / surface.
Mathematically, these 3 statements can be compactly combined into a single expression:
above
below
BTOT
− BTOT
= μo K × nˆ where the unit normal to the surface, nˆ = zˆ (here, as drawn above).
© 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 16
Prof. Steven Errede
As we found in electrostatics, that the scalar electric potential V ( r ) was continuous across
any boundary Vabove ( r ) = Vbelow ( r ) , likewise, the magnetic vector potential A ( r ) is also
continuous across any boundary, i.e. Aabove ( r ) = Abelow ( r ) provided that: ∇i A ( r ) = 0 , which
(
)
guarantees that A⊥above ( r ) = A⊥below ( r ) and also provided that: ∇ × A ( r ) = B ( r ) , which, in
integral form, i.e.
∫ A ( r )i d
C
=
∫ B ( r )ida = Φ
S
guarantees that Aabove ( r ) = Abelow ( r ) .
m
However, note that the normal derivative of A ( r ) , since A ( r ) K ( r ) then A ( r ) also
“inherits” the discontinuity associated with B ( r ) : BTOTy − BTOTy = μo K (see #2 on previous
above
below
page), and since B ( r ) = ∇ × A ( r ) , thus we have a discontinuity in the (normal) slope(s) of
A ( r ) on either side of the boundary/current sheet.
We can understand the origin of this condition on the normal derivative(s) of A ( r ) taken just
above/below an “interface” e.g. for the specific case of the current sheet K = − Kxˆ . From
above
below
BTOTy − BTOTy = μo K we know that the discontinuity in the B -field is in the ŷ -direction,
whereas since the magnetic vector potential associated with the current sheet A ( r ) is always
parallel to the current, and since K = − Kxˆ we know that the component of ATOT ( r ) that we are
(
concerned with here is in the ŷ -direction. But from: BTOT = ∇ × ATOT , then: BTOTy = ∇ × ATOT
)
y
thus we need to worry only about the ŷ -component of the curl of ATOT ( r ) , which is:
(
BTOTy = ∇ × ATOT
)
y
⎛ ∂ATOTx ∂ATOTz
=⎜
−
⎜ ∂z
∂x
⎝
⎞
⎟⎟
⎠
Then, noting that the ẑ -direction is perpendicular (i.e. normal) to the plane of the current sheet:
above
below
⊥ above
⊥below
⎛ ∂ATOT
⎞
⎛ ∂ATOT
⎞
∂ATOT
∂ATOT
above
below
x
x
z
z
⎟
⎟
−
−⎜
−
BTOTy − BTOTy = μo K = ⎜
⎜ ∂z
⎟
⎜ ∂z
⎟
∂
∂
x
x
⎝
⎠ surface ⎝
⎠ surface
above
below
⊥ above
⊥below
⎛ ∂ATOT
⎛ ∂ATOT
⎞
∂ATOTx ⎞
∂ATOT
x
z
z
⎟
=⎜
−
−⎜
−
⎟
⎜ ∂x
⎜ ∂z
∂z ⎟⎠
∂x ⎟⎠
⎝
⎝
surface
above
below
⎛ ∂ATOT
∂ATOTx ⎞
x
⎟
=⎜
−
⎜ ∂n
⎟
∂
n
⎝
⎠ surface
surface
=0
⊥
ATOT
suffers no discontinuity
z
⊥
nor ATOTy suffer discontinuities in their slopes at the current sheet – only
Neither ATOT
z
above
ATOT
does - in the normal (i.e. ẑ ) direction. Therefore, we can most generally write this
x
condition on the discontinuity in the normal derivative on A ( r ) as:
∂Aabove ( r )
∂Abelow ( r )
−
= − μo K
∂n
∂
n
surface
surface
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 16
Prof. Steven Errede
The Magnetic Vector Potential A ( r ) Associated with a Finite Circular Disk Sheet Current
We wish to delve a bit deeper into the nature of the magnetic vector potential, A ( r ) and also
B ( r ) = ∇ × A ( r ) associated with current sheets. Consider a sheet current K = K o xˆ flowing on the
surface of a finite circular disk of radius R, lying in the x-y plane as shown in the figure below:
ẑ
+z
r
ρ
xˆ , K = K o xˆ
ŷ
−z
To keep it simple, we’ll just calculate A ( r ) at an arbitrary point along the ẑ -axis above and
below the x-y plane. The magnetic vector potential A ( r ) associated with a sheet current is:
A(r ) =
μo
4π
K ( r ′)
μ K xˆ da′
da′ = o o ∫
S′
4π S ′ r
r
∫
We deliberately chose a sheet current flowing on a finite circular disk of radius R so that we
could easily carry out the integration. The area element da′ on the circular disk (in cylindrical
coordinates) is da′ = d ρ ( ρ dϕ ) = ρ d ρ dϕ , and from the figure above, we see that: r = ρ 2 + z 2 .
A( z ) =
Thus:
=
ρ =R
μo K o xˆ ρ = R ϕ = 2π ρ d ρ dϕ 2π μo K o xˆ ρ = R ρ d ρ
1
= μo K o xˆ ⎡ ρ 2 + z 2 ⎤
=
∫
∫
∫
2
2
2
2
=
0
=
0
=
0
ρ
ρ
ϕ
⎣
⎦ ρ =0
2
4π
4π
ρ +z
ρ +z
1
1
μo K o xˆ ⎡ R 2 + z 2 − z 2 ⎤ = μo K o ⎡ R 2 + z 2 − z 2 ⎤ xˆ
⎣
⎦ 2
⎣
⎦
2
Now there is a subtlety here that we need to notice before proceeding further – since we are
interested in knowing A ( z ) at an arbitrary point along the ẑ -axis - above and/or below the x-y
plane, thus z can be either positive or negative. Note that both the
R 2 + z 2 and z 2 terms are
always ≥ 0 for both positive and/or negative z (in particular: z 2 = z ≠ z !). Thus, in order to
preserve this fact, we explicitly keep expression for the magnetic vector potential A ( z ) as:
A( z ) =
1
μo K o
2
(
)
R 2 + z 2 − z 2 xˆ
© 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
Lecture Notes 16
Prof. Steven Errede
A plot of the magnetic vector potential A ( z ) vs. z is shown in the figure below for a circular
disk of radius R = 10 m and sheet current K = K o xˆ = 1.0 xˆ Amperes/meter.
Note that A ( z ) is a maximum when z = 0, right on the sheet current. Note also the discontinuity
in the slope(s) of A ( z ) on either side of z = 0, which arises due to the presence of the sheet
current in the x-y plane, since:
∂Aabove ( r )
−
∂n
surface
∂Abelow ( r )
= − μo K
∂n
surface
∂Aabove ( z ≥ 0 )
∂Abelow ( z ≤ 0 )
−
= − μo K
∂z
∂z
z =0
z =0
or:
Care/thought must also be taken when carrying out the normal derivatives (slopes) above and
below the x-y plane – look carefully at the slopes for z > 0 and z < 0 in the above figure, and
compare this information to what we calculate:
∂A
above
( z ≥ 0) = 1 μ
o
Ko
∂
∂z
(
⎛
z
z
R 2 + z 2 − z 2 xˆ = 12 μo K o ⎜
−
2
2
z2
⎝ R +z
( z ≤ 0) = 1 μ
o
Ko
∂
∂z
(
⎛
⎛
⎞
z
z ⎞
z
1
R 2 + z 2 − z 2 xˆ = 12 μo K o ⎜
−
+ 1⎟ xˆ
⎟ xˆ = 2 μo K o ⎜
2
2
2
2
2
z ⎠
⎝ R +z
⎝ R +z
⎠
∂z
∂A
below
∂z
26
2
2
)
⎞
⎛
⎞
z
1
− 1⎟ xˆ
⎟ xˆ = 2 μo K o ⎜
2
2
⎠
⎝ R +z
⎠
)
© 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 16
Prof. Steven Errede
Thus we see that indeed:
∂Aabove ( z ≥ 0 )
∂Abelow ( z ≤ 0 )
−
= − 12 μo K o xˆ − 12 μo K o xˆ = − μo K o xˆ = − μo K
∂z
∂
z
z =0
z =0
The magnetic field B ( z ) at an arbitrary point along the along the ẑ -axis – either above and/or
below the x-y plane is calculated using B ( z ) = ∇ × A ( z ) in Cartesian coordinates. Since
A ( z ) = Ax ( z ) xˆ (only), then: B ( z ) = ∇ × A ( z ) = ∇ × Ax ( z ) xˆ =
∂Ax ( z )
yˆ
∂z
Thus:
∂Axabove ( z ≥ 0 )
⎛
⎞
z
− 1⎟ yˆ
B
yˆ = 12 μo K o ⎜
( z ≥ 0 ) = ∇ × A ( z ≥ 0 ) = ∇ × A ( z ≥ 0 ) xˆ =
2
2
∂z
⎝ R +z
⎠
below
∂A
( z ≤ 0 ) yˆ = 1 μ K ⎛ z + 1⎞ yˆ
B below ( z ≤ 0 ) = ∇ × Abelow ( z ≤ 0 ) = ∇ × Axbelow ( z ≤ 0 ) xˆ = x
⎟
2 o o⎜
2
2
∂z
⎝ R +z
⎠
above
above
above
x
The figure below shows the magnetic field B ( z ) vs. z along the ẑ -axis with a sheet current
K = K o xˆ flowing on the surface of the finite disk of radius R, lying in the x-y plane:
© 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 16
Prof. Steven Errede
We now investigate what happens in the limit that the radius of the sheet current-carrying
circular disc, R → ∞ , i.e. it becomes an infinite planar sheet current. We discover that the
magnetic vector potential A ( r ) associated with the sheet current K = K o xˆ becomes infinite
(i.e. A ( r ) diverges):
(
)
lim A ( z ) =
R→∞
1
μo K o
2
(
)
R 2 + z 2 − z 2 xˆ →
1
μo K o
2
(
)
∞ 2 + z 2 − z 2 xˆ
whereas the boundary condition on the discontinuity in the normal derivative of A ( r ) across the
sheet current lying in the x-y plane at z = 0 still exists, and is well-behaved (i.e. finite):
∂Aabove ( z ≥ 0 )
∂Abelow ( z ≤ 0 )
−
= − 12 μo K o xˆ − 12 μo K o xˆ = − μo K o xˆ = − μo K
∂z
∂
z
z =0
z =0
We also discover that the magnetic field B ( r ) is also well-behaved (i.e. finite) – and constant –
independent of the height/depth z above/below the x-y plane (!!):
⎛
⎞
z
− 1⎟ yˆ = − 12 μo K o yˆ
lim B above ( z ≥ 0 ) = 12 μo K o ⎜
2
2
R→∞
⎝ ∞ +z
⎠
⎛
⎞
z
+ 1⎟ yˆ = + 12 μo K o yˆ
lim B below ( z ≤ 0 ) = 12 μo K o ⎜
2
2
R→∞
⎝ ∞ +z
⎠
28
(
)
(
)
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.