Download 7-5 Magnetic Potentials

11/21/2004
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7-5 Magnetic Potentials
Reading Assignment: pp. 227-236
Recall that the definition of electric scalar potential:
E ( r ) = −∇V ( r
)
led to the integral relationship:
∫C E (r ) ⋅ d A =V (ra ) −V (rb )
Q:
A:
HO: The Integral Definition of Magnetic Vector
Potential
There are numerous analogies between electrostatics
and magnetostatics!
Jim Stiles
The Univ. of Kansas
Dept. of EECS
11/21/2004
section 7_5 Magnetic Potentials blank
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Q:
A:
HO: The Magnetic Dipole
Jim Stiles
The Univ. of Kansas
Dept. of EECS
11/21/2004
The Integral Definition of Magnetic Vector Potential
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The Integral Definition of
Magnetic Vector Potential
Recall for electrostatics, we began with the definition of
electric scalar potential:
E ( r ) = −∇V ( r
)
And then taking a contour integral of each side we discovered:
∫C E (r ) ⋅ d A = −C∫ ∇V (r ) ⋅ d A
∫C E (r ) ⋅ d A =V (ra ) −V (rb )
We can perform an analogous procedure for magnetic vector
potential! Recall magnetic flux density B ( r ) can be written in
terms of the magnetic vector potential A ( r ) :
B ( r ) = ∇ xA ( r
)
Say we integrate both sides over some surface S:
∫∫S B (r ) ⋅ ds = ∫∫S ∇xA (r ) ⋅ ds
Jim Stiles
The Univ. of Kansas
Dept. of EECS
11/21/2004
The Integral Definition of Magnetic Vector Potential
2/4
We can apply Stoke’s theorem to write the right side as:
∫∫S ∇xA (r ) ⋅ ds = vC∫ A (r ) ⋅ d A
Therefore, we find that we can also define magnetic vector
potential in an integral form as:
∫∫S B (r ) ⋅ ds = vC∫ A (r ) ⋅ d A
where contour C defines the border of surface S.
C
S
Consider now the meaning of the integral:
∫∫S B (r ) ⋅ ds
This integral is remarkably similar to:
where:
∫∫S J (r ) ⋅ ds
B ( r ) magnetic flux density
Jim Stiles
The Univ. of Kansas
⎡Webers ⎤
⎢⎣ meters 2 ⎥⎦
Dept. of EECS
11/21/2004
The Integral Definition of Magnetic Vector Potential
and:
J ( r ) current density
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⎡ Amperes ⎤
⎢⎣ meters 2 ⎥⎦
Recall that integrating the current density (in amps/m2) over
some surface S (in m2), provided us the total current I flowing
through surface S:
∫∫S J (r ) ⋅ ds = I
Similarly, integrating the magnetic flux density (in webers/m2)
over some surface S (in m2), provided us the total magnetic
flux Φ flowing through surface S:
∫∫S B (r ) ⋅ ds = Φ
where Φ is defined as:
Φ magnetic flux
Jim Stiles
The Univ. of Kansas
[Webers ]
Dept. of EECS
11/21/2004
The Integral Definition of Magnetic Vector Potential
4/4
Using the equations derived previously, we can directly relate
magnetic vector potential A ( r ) to magnetic flux as:
Φ = v∫ A ( r ) ⋅ d A
C
where we recall that the units for magnetic vector potential are
Webers/m.
Note the similarities of the above expression to the integral
form of Ampere’s Law!
I =
Jim Stiles
1
µ0
vC∫ B (r ) ⋅ d A
The Univ. of Kansas
Dept. of EECS
11/21/2004
The Magnetic Dipole
1/8
The Magnetic Dipole
Consider a very small, circular current loop of radius a,
carrying current I.
Q: What magnetic
flux density is
produced by this loop,
in regions far from the
current (i.e., r>>a)?
z
I
y
A: First, find the
magnetic vector
potential A ( r ) :
x
A (r ) =
µ0
4π
vC∫
I d A′
r −r′
Since the contour C is a circle around the z –axis, with radius a,
we use the differential line vector:
d A′ = ρ ′d φ ′ ˆ
aφ
= a d φ′ ˆ
aφ
(
)
ax + a sinφ ′ ˆ
ay d φ ′
= a cosφ ′ ˆ
Jim Stiles
The Univ. of Kansas
Dept. of EECS
11/21/2004
The Magnetic Dipole
2/8
The location of the current is specified by position vector r ′ .
Since for every point on the current loop we find z ′ = 0 and
ρ ′ = a , we find:
r′ = x′ˆ
ax + y ′ ˆ
ay + z ′ ˆ
az
= ρ ′ cosφ ′ ˆ
ax + ρ ′ sinφ ′ ˆ
ay + z ′ ˆ
az
= a cosφ ′ ˆ
ax + a sinφ ′ ˆ
ay
And finally,
r =x ˆ
ax + y ˆ
ay + z ˆ
az
= r sinθ cosφ ˆ
ax + r sinθ sinφ ˆ
ay + r cosθ ˆ
az
With a little algebra and trigonometry, we find also that:
−1
1
= ⎡⎣r 2 − a (2r sin θ cos (φ − φ ′ ) ) + a 2 ⎤⎦ 2
r − r′
Since the radius of the circle is very small (i.e., a << r), we can
use a Taylor Series to approximate the above expression (see
page 231 of text):
1
1 a sin θ cos (φ − φ ′ )
≈ +
r − r′ r
r2
The magnetic vector potential can now be evaluated !
Jim Stiles
The Univ. of Kansas
Dept. of EECS
11/21/2004
A (r ) =
The Magnetic Dipole
µ0
4π
vC∫
µ I
= 0
4π
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I d A′
r − r′
⎛ 1 a sin θ cos (φ − φ ′ ) ⎞
ax + a sinφ ′ ˆ
ay d φ ′
⎟ a cosφ ′ ˆ
2
∫0 ⎜⎝ r +
r
⎠
2π
(
)
π a 2I
ax + cosφ ˆ
ay )
=
sinθ ( -sinφ ˆ
r2
π a 2I
=
sinθ âφ
r2
Note that π a 2 equals the surface area S of the circular loop.
Therefore, we can write that magnetic vector potential
produced by a very small current loop is:
A (r ) =
µ0 S I
aφ
sinθ ˆ
2
4π r
(a << r )
We can now determine magnetic flux density B ( r
) by taking
the curl:
B ( r ) = ∇xΑ ( r
=
Jim Stiles
)
µ 0 SI
(2cos θ ˆar + sin θ ˆaθ )
4π r 3
The Univ. of Kansas
Dept. of EECS
11/21/2004
The Magnetic Dipole
4/8
Q: Hey! Something about this result looks very familiar !
A: Compare this result to that of an electric dipole:
E (r ) =
Qd
2 cos θ ˆ
ar + sin θ ˆ
aθ )
3 (
4πε r
Both results have exactly the same form!:
⎛ 2 cos θ ˆ
ar + sin θ ˆ
aθ ⎞
⎟
3
4
π
r
⎝
⎠
c⎜
z
x
where c is a constant.
Jim Stiles
The Univ. of Kansas
Dept. of EECS
11/21/2004
The Magnetic Dipole
5/8
Because of this similarity, we can refer to a small current loop
of are S and current I as a Magnetic Dipole.
Note that the only difference between the mathematical
description of an electric field produced by an electric dipole
and the magnetic flux density produced by a magnetic dipole is
a constant c:
electric dipole
→
magnetic dipole
→
c=
Qd
ε
c = µ 0S I
Recall that we defined a dipole moment for electric dipoles,
where:
p = Qd
Clearly, the analogous product to Qd for a magnetic dipole is
SI. We can, in fact, define a magnetic dipole moment m:
m Magnetic Dipole Moment
⎡⎣Amps ⋅ m2 ⎤⎦
Analogous to the electric dipole, the magnetic dipole moment
has magnitude:
m = SI
Jim Stiles
The Univ. of Kansas
Dept. of EECS
11/21/2004
The Magnetic Dipole
6/8
Q: We now know the magnitude of the magnetic dipole moment,
but what is its direction ??
A: The magnetic dipole m points in the direction orthogonal to
the circular surface S, e.g.:
m
m
m
Note the direction is defined using the right-hand rule with
respect to the direction of current I.
Instead of plus (+) and minus (-), the poles of a magnetic dipole
are defined as north (N) and south(S):
m
N
S
Jim Stiles
The Univ. of Kansas
Dept. of EECS
11/21/2004
The Magnetic Dipole
7/8
Thus, for the example provide on this handout, the magnetic
dipole moment is:
az
m = SI ˆ
aφ = SI ˆ
az x ˆ
ar = mx ˆ
ar , therefore we can
We note that SI sinθ ˆ
write:
µ0 S I
sinθ ˆ
aφ
4π r 2
µ mxâr
= 0
4π r 2
A (r ) =
The above equation is in fact valid for any magnetic dipole m
located at the origin, regardless of its direction! In other
words, we can also use the above expression if m is pointed in
some direction other than âz ,e.g.:
m
m
Q: What if the magnetic dipole is not located at the origin?
A: Just like we have many times before, we make the
substitutions:
r → r −r′
Jim Stiles
ˆar = ˆaR =
The Univ. of Kansas
r −r′
r −r′
Dept. of EECS
11/21/2004
The Magnetic Dipole
Therefore, we find the magnetic flux density A ( r
8/8
) produced
by an arbitrary magnetic dipole m, located at an arbitrary
position r ′ , is:
A (r ) =
µ0 m x ( r - r ′ )
3
4π
r - r′
To determine the magnetic flux density B ( r ) , we simply take
the curl of the above expression.
Note this is analogous to the expression of the electric scalar
potential generated by an electric dipole with moment p:
V (r ) =
1 p ⋅ (r - r ′)
4πε r - r ′ 3
and then taking the gradient of this function to determine the
electric field E ( r ) .
Jim Stiles
The Univ. of Kansas
Dept. of EECS