Lecture 7
Applications of Magnetostatics
Today’s topics
1. Inductance
2. Magnetic materials
3. Boundary conditions
4. Shielding
Inductance
1. Recall how to calculate capacitance from electrostatics
1
F0E 2
2
CV = ¨
d ra
2
2
CV = Q
(more fundamental)
2. The definitions are the same if the plates are located in a vacuum region
3. For two parallel plates C = F0A / d
4. As a simple magnetostatic analog let’s calculate the inductance of a finite
diameter wire carrying a current I surrounded by a perfectly conducting shell
carrying a return current I .
5. First we calculate the magnetic field inside and outside the wire.
1
6. Inside the wire ‹ × B = N0 J while outside the wire ‹× B = 0 .
7. By symmetry we see that the non-trivial field components are B = BR eR and
J = J z ez .
8. If the current in the wire is uniform then J z = I / Qa 2
9. Apply Ampere’s law in the wire
1 d
(rBR ) = N0J z
r dr
NI r
NI r
c
BR = 0 2 + = 0 2
r
2Q a
2Q a
10. Outside the wire
1 d
(rBR ) = 0
r dr
c
BR =
r
11. Match across r = a : aBR ba = 0
l
c = N0 I / 2 Q
BR =
N0I
2Qr
12. Therefore
13. Below is a plot of BR due to the wire
14. Now let’s calculate the field due to the return current
Inside the shell:
‹ × B = 0 l BR = c / r = 0
Outside the shell:
‹×B = 0
l BR = c / r = N0I / 2Qr
2
15. The total field is found by superposition and is shown below.
Calculating the inductance
1. In general there is both internal and external inductance.
2. This makes it a little more complicated than calculating capacitance between
thin conducting plates.
3. Here is a simple widely used definition of external inductance
Le =
Ze
I
Ze = external flux
4. Evaluate the external flux
Ze =
¨ B ¸ n dS = ¨
L
0
b
dz ¨ BRdr =
a
N0IL
2Q
¨
b
a
dr N0IL b
ln
=
2Q
r
a
5. The inductance is thus given by
L=
Ze
NL b
= 0 ln
2Q a
I
6. We get the same result from the more basic definition
1 2
LI =
2
¨
B2
d ra
2N0
3
7. The details are as follows
2
¨
B2
1  N0I ¬­
ž
d ra =
­
2N0
2N0 žŸ 2Q ®­
NL b
Le = 0 ln
2Q a
¨
a
b
L
rdr 2 Q
N0I 2  b ¬­
žln ­ (2QL )
d R ¨ dz =
0
r 2 ¨0
8Q 2 Ÿž a ®­
Equivalence of the two definitions
1. The equivalence is shown as follows
1
Le I 2 =
2
=
=
=
=
=
¨
B2
1
d ra =
2N0
2N0
1
2N0
1
2N0
1
2N0
1
2N0
1
2N0
2
¨ (‹ × A) d ra
¨ [‹ ¸ (A × ‹ × A) A ¸ ‹ × ‹ × A ]d ra
1
¨ ‹ ¸ (A × ‹ × A)d ra = 2N ¨ A × B ¸ n dS
0
¨ A × B ¸ d l ×d l
&
?
=
1
2N0
¨ A ¸ B ×d l ×d l
&
?
¯
¨ ¡¢(B ¸ d l )(A ¸ d l ) + (B ¸ d l )(A ¸ d l )°±
?
&
&
¯
¨ ¡¢(B ¸ d l )(A ¸ d l )°± = 2N ¨ A ¸d l
?
N0I
&
?
?
0
2. The last term is further simplified by noting that
Ze =
¨ B ¸dS = ¨ ‹×A ¸dS = ¨ A ¸dl
?
4
3. Therefore,
1
1
Le I 2 = I Ze
2
2
Ze = Le I
l
4. Both definitions are equivalent for vacuum fields (i.e. when ‹×‹× A = 0 )
Internal inductance
1. For the internal inductance Li we always must use the general definition
2
 N0I r ¬­
B2
2QL a
N0I 2L
a
ž
r
d
rdr
=
=
­
¨ 2N0
žŸ 2Q a 2 ®­
2N0 ¨0
16Q
NL
Li = 0
8Q
2. What does the definition Zi = Li I give?
1
Li I 2 =
2
Zi =
BRdrdz =
¨
Li =
N0L
4Q
N0IL
2Q
¨
a
0
r
N IL
dr = 0
2
a
4Q
3. This is an incorrect answer.
4. For distributed currents we must use the general energy definition to calculate
inductance.
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Magnetic materials
1. Recall that in electrostatics an applied electric field induced a small electric
dipole whose electric field direction was opposite to the applied field.
2. There is a magnetic analog when a material is placed in a DC magnetic field.
3. What happens when such atoms are placed in an applied magnetic field Ba ? A
torque develops that tends to flip the electron current so that the current flows in
a plane perpendicular to Ba .
4. The flips to a new position where the torque is zero.
5. The direction of the flipping is such as to enhance the field (paramagnetic).
6. As in electrostatics we can introduce the property of permeability. This is
analogous to polarizability which allowed us to introduce a relative dielectric
constant.
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7. The comparisons are as follows:
Electrostatics
Magnetostatics
Eind = Dp E
S
S
‹ ¸ E = free + ind
F0
F0
Sind
‹ ¸ Eind =
F0
S
‹ ¸ (E Eind ) = free
F0
S
‹ ¸ ¡(1 + Dp ) E ¯° = free
¢
±
F0
Bind = +Dm B
D = (1 + Dp ) F0E = FE = Fr F0 E
H = (1 Dm ) B / N0 = B / N = B / Nr N0
‹¸D = S
‹×H = J
‹ × B = N0 Jfree + N0 Jind
‹ × Bind = N0 Jind
‹ × (B Bind ) = N0 J free
‹ × ¢(1 Dm ) B¯± = N0 J free
8. Note that for typical materials E is reduced by the dielectric effect while B is
increased by the paramagnetic effect.
Boundary conditions
1. Let’s determine the boundary conditions across the interface between a magnetic
material and a vacuum region.
2. There are two conditions in analogy with electrostatics. The first is given by
Electrostatics
‹×E = 0
l
Magnetostatics ‹ ¸ B = 0 l
¨v E ¸ d l = 0 l a n × E b = 0
¨ B ¸ n dS = 0 l a n ¸ Bb = 0
3. The second condition, assuming no surface charge or surface current is given by
Electrostatics
‹¸D = S
Magnetostatics
‹×H = J
¨ D ¸ n dS = 0 l
l ¨
v H ¸dl = 0 l
l
a Fn ¸ E b = 0
an × B / Nb = 0
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Statics in a resistive medium
1. Consider a piece of metal with finite thickness.
2. It has a resistivity I because of the friction felt by electrons as they try to flow
through the fixed lattice of ions.
3. Therefore a piece of metal carrying a current density J generates a resistive
electric field given by the familiar ohm’s law E = I J .
4. Static problems of this type are solved in the following order
5. First, in a static problem ‹ × E = 0 so that as before E = ‹G . When no free
charges flow in the conductor G satisfies
‹2G = 0
6. We solve this equation first.
7. Assuming that G is known we next turn to Ampere’s law. For a magnetostatic
problem we can again write B = ‹ × A with ‹ ¸ A = 0 . The vector potential
then satisfies
‹2 A =
N0
‹G
I
8. Since we know G we can now solve for A
DC Shielding
1. We next examine the important practical question of DC shielding.
2. The goals are to see whether (1) dielectrics can shield electric fields and (2)
magnetic materials can shield magnetic fields.
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3. Start with the dielectric problem as illustrated below.
4. With no dielectric the interior electric field is obviously E = E 0ey .
5. With a shell of dielectric we have to solve the Laplace’s equation in three regions
and match across the interfaces.
6. In each region E = ‹G and ‹2G = 0 .
7. The solutions are given by
I.
II.
III.
c1
c ¬

sin R = žžE 0r + 1 ­­ sin R
Ÿ
r
r®
c ¬
c ¬


E = žžE 0 + 12 ­­ sin R er + žžE 0 12 ­­ cos R eR
Ÿ
Ÿ
r ®
r ®
c ¬

G = žžc2r + 3 ­­ sin R
Ÿ
r®
c ¬
c ¬


E = žžc2 + 32 ­­ sin R er + žžc2 32 ­­ cos R eR
Ÿ
Ÿ
r ®
r ®
G = c4r sin R
G = E 0y +
E = c4 sin R er + c4 cos R eR
8. The goal is to calculate c 4 = E inside and see how it compares in magnitude to the
applied field E 0 .
9. The matching conditions across r = b yield two relations
aE R b = 0
a Fr Er b = 0
c1
c
= c2 32
2
b
b
c
c ¬

E 0 + 12 = Fr žžc2 + 32 ­­
Ÿ
b
b ®
E0 9
10. There are two similar conditions across r = a
c3
a2
aE R b = 0
c 4 = c2 a Fr Er b = 0
c ¬

c4 = Fr žžc2 + 32 ­­
Ÿ
a ®
11. These are 4 equations with four unknowns. We can easily solve them and
evaluate c 4 .
Einside
4Fr
=
2
2
2
E0
(Fr + 1) (Fr 1) (a / b )
12. In the limit of a large dielectric constant Fr 1 we have
Einside
4
=
1
E0
Fr (1 a 2 / b 2 )
13. The conclusion is that a material with a high dielectric constant does a good job
shielding DC electric fields.
Magnetic shielding
1. We can now ask a similar question with respect to magnetostatics. Does a highly
permeable material shield out magnetic fields?
2. The calculation is very similar to the electrostatic case. Consider the problem
illustrated below.
10
3. Clearly with no permeable material the field on the inside is just B0 .
4. When a permeable material is present we again have to solve a three region
problem. The solutions in each region satisfy B = ‹ × (Aez ) with ‹2A = 0 .
5. The solutions are given by
I.
II.
III.
c1
c ¬

cos R = žžB0r + 1 ­­ cos R
Ÿ
r
r®
c ¬
c ¬


B = žžB0 12 ­­ sin R er + žžB0 + 12 ­­ cos R eR
Ÿ
Ÿ
r ®
r ®
c ¬

A = žžc2r + 3 ­­ cos R
Ÿ
r®
c ¬
c ¬


B = žžc2 32 ­­ sin R er + žžc2 + 32 ­­ cos R eR
Ÿ
®
Ÿ
r
r ®
A = c4r cos R
A = B0x +
B = c4 sin R er + c4 cos R eR
6. We again have matching conditions across the two interfaces.
c1
c
= c2 + 32
2
b
b
c
c ¬
1
aBR / N bb = 0 B0 12 = žŸžc2 32 ®­­
b
b
Nr
c
c4 = c2 + 32
aBr ba = 0
a
c ¬
1 
c4 = žžc2 32 ­­
aBR / N ba
a ®
Nr Ÿ
aBr bb = 0
B0 +
7. Solve for c 4 to find the interior field.
Binside
4Nr
4
1
=
2
2
2 x
B0
Nr (1 a 2 / b 2 )
(Nr + 1) (Nr 1) (a / b )
8. A material with a high permeability (Nr 1) does a good job shielding out DC
magnetic fields.
9. What happens if I try to shield electric or magnetic fields with a conductor which
has Fr = Nr = 1 ?
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Conducting materials
1. To learn a little about conductors consider the single turn solenoidal magnetic
illustrated below.
2. Our goal is to calculate the electric field, magnetic field, and current density. We
start with the electric field in the magnet which by symmetry is in the R
direction.
3. The electric field equation reduces to
‹×E = 0
1 drE R
=0
r dr
l
l
ER =
c1
r
4. The constant c1 is found by noting that around the circumference of the coil
V = ¨
2Q
0
E Rrdr
l
c1 = V
2Q
l
ER = V
2Qr
5. The current density in the magnet is then given by
JR =
ER
V
=
I
2QIr
6. The magnetic field in the magnet satisfies Ampere’s law
dBz
NV
= N0J R = 0
2QIr
dr
7. The solution is
Bz =
N0V
ln r + c2
2QI
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8. The integration constant c2 is found by noting that in a long solenoid the field is
zero outside the coil. Therefore,
Bz =
N0V r
ln
2QI b
9. The magnetic field in the coil is uniform. Its value is given by
Bz (r )inside =
N0V a
ln
2QI b
10. The fields are sketched below.
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