Introduction to Electromagnetism Static Magnetic Fields
(6-13)
Yoonchan Jeong
School of Electrical Engineering, Seoul National University Tel: +82 (0)2 880 1623, Fax: +82 (0)2 873 9953
Email: [email protected]
2
Magnetic Forces (1)
Hall effect:
← Hall field Magnetic force:
) N ( B u Fm = q ×
B0
az
B =
u a
J = yJ0 = Nq
B u
F = ×
→ net,steady q = 0
B u E = − ×
→ h → Hall effect
0 0B
xu
= a
0 0)
(−ayu ×azB
−
=
d B u dx E
Vh d h 0 0
0 =
=
→
∫
← Hall voltagee q = −
←
Nq B
J
Ex / y z =1/
→ ← Hall coefficient
For an n-type semiconductor:
What if for a p-type semiconductor?
D. K. Cheng, Field and Wave Electromagnetics, 2nd ed., Addison-Wesley, 1989.
→ Vh will be reversed.
→ To determine N qEh
+
3
Magnetic Forces (2)
Magnetic force on the differential length element dl:
B u F = NqS dl × d m
Newton’s third law of action and reaction
) N ( B l F = ×
→ d m Id
) N
∫
× (=
→ Fm I Cdl B
For two circuits carrying current I1 and I2:
∫
×=
1
21 1
1
21 I C dl B
F ← =
∫
2 ×2 2121 2 2
0
21 4 C
R
R I dl a
B
π
µ
) N ) (
(
4 1 2
21
2 21
2 1
2 1
0
∫ ∫
× ×= C C
R
R d I d
I l l a
π µ
∫
×=
2
12 2
2
12 I C dl B
F
) N ) (
(
4 2 1
12
2 12
1 2
1 2
0
∫ ∫
× ×= C C
R
R d I d
I l l a
π µ
12
21 F
F =? −
→ Ampère’s law of force
→ dl along the direction of I
“Biot-Savart law”
4
Magnetic Forces (3)
) N ) (
(
4 1 2
21
2 21
2 1
2 1 0
21 =
∫ ∫
C C × ×R
R d I d
I l l a
F
π
µ
For two circuits carrying current I1 and I2:
2 21
2
1 ( )
21
R d
dl × l ×aR
→ 2
21 2 1
2 21 1
2( ) ( )
21 21
R d d
R d
dl l aR aR l ⋅ l
⋅ −
=
∫ ∫
⋅→
1 2
21
2 21 1
2( )
C C
R
R d dl l a
∫ ∫
⋅=
2 1
21
2 21 1 2
) (
C C
R
R d dl a
l
1 ] [
2 1
21 1
1
∫
2∫
⋅ −∇ = C dl C dl R = 0
∫ ∫
⋅−
=
→
1 2
12
2 12
2 1
2 1 0 12
) (
4 C C
R
R d I d
I a l l
F
π
µ
F21
−
= Newton’s third law
of action and reaction verified!
0 ) (∇ =
×
∇
← V
Recall:
Recall: “back-cab” rule
∫ ∫
⋅−
=
→
1 2
21
2 21
2 1
2 1 0 21
) (
4 C C
R
R d I d
I a l l
F
π
µ
) N ) (
(
4 2 1
12
2 12
1 2
1 2 0
12 =
∫ ∫
C C × ×R
R d I d
I l l a
F
π
µ
5
Example 6-21
For two parallel current-carrying wires:
) ( 12
2 12
2
12 F a B
F = ′ = I z × l
d I
x π
µ 2
1 0
12 a
B = −
→ ← Ampère’s circuital law
) N/m 2 (
2 1 0
12 d
I I
y π
a µ F′ = −
→
→ Attraction for currents in the same direction
→ Repulsion for currents in opposite directions
D. K. Cheng, Field and Wave Electromagnetics, 2nd ed., Addison-Wesley, 1989.
6
Magnetic Forces and Torques
Magnetic flux density:
Forces on the loop:
B//
B B = ⊥ +
ar
FB ∝
→ ⊥ → Radial symmetry )
// (
,
1 upward
dF B ∝ −az
→
)
// (
,
2 downward
dF B ∝ +az
→ → Rotational motion
→ Torque Torque on the loop:
φ
sin )(
2 dF b dT = ax
φ φ
) sin sin(
2 IdlB// b ax
=
φ φ
d Bx Ib
2 //
2 sin
2
= a
φ
π
φ
d B
Ib
d x
∫
∫
==
→ 0
2 //
2 sin
2 a T
T = axI(
π
b2)B//) m N ( ⋅
×
= m B
D. K. Cheng, Field and Wave Electromagnetics, 2nd ed., Addison-Wesley, 1989.
) (B⊥ = azB⊥
→ No net contribution
7
Forces and Torques - Stored Magnetic Energy (1)
System of circuits with constant flux linkage:
Work done by the source
m
s dW dW
dW = + Mechanical work
done by the system Increase in the stored magnetic energy
Principle of virtual displacement:
Virtual displacement dl with no changes in flux linkages
→ No emf → No energy from to the source to the system
m
s dW dW
dW = +
→
l F d dW = Φ ⋅
) N
m (
−∇W
=
→ FΦ Magnetic forces:
Magnetic torques:
) m N ( )
( ⋅
∂
− ∂
=
→ Φ
φm
z
T W
(
∇Wm)
⋅dl−
m =
−dW
=
= 0
8
Forces and Torques - Stored Magnetic Energy (2)
System of circuits with constant currents:
Virtual displacement dl with constant currents
dt V I
dW k
k k s =
∑
→
→ Emf induced → Energy from the source to the system
) N
m (
I = ∇W
→ F Magnetic forces:
Magnetic torques:
) m N ( )
( ⋅
∂
= ∂
→ I z φm T W
l F d dW = I ⋅
→ = ∇
( )
Wm ⋅dlm
s dW dW
dW = +
→
k k
k
m I d
dW = Φ
→ 21
∑
dWm
=
→ Flux linkage changed
dWm
dW =
k →
k
kd I Φ
=
∑
9
Force and Torque in Terms of Mutual Inductance
For two circuits with currents I1 and I2:
2 2 2 2
1 12 2
1
1 2
1 2
1 L I L I I L I
Wm = + +
→
∑∑
= == N
j N
k
k j jkI I L
1 1
2 1
) ( 12
2
1I L
I = I ∇
→ F
Magnetic force between two circuits:
Magnetic torque between two circuits:
) m N ( )
( 1 2 12 ⋅
∂
= ∂
→ φ
I L I TI z
∑
=Φ
= N
k
k k
m I
W
2 1
1
10
Example 6-23
System of circuits with constant currents:
System of circuits with constant flux linkage:
gap air m
m d W
dW = ( )
→
= B Sdy
0 2
2 2
µ y Wm
y ∂
− ∂
=
→ FΦ a
0 2
µ S B ay
−
=
g
c R
R Vm
+ 2
= Φ
→
∑
Φ=
→
k
k k
m I
W 2
1 = INΦ
2 1
y Wm
y
I ∂
= ∂
→ F a
−Φ
= y 0S
2 2
2 1
a µ (N)
0 2
y µ S
− Φ
= a
D. K. Cheng, Field and Wave Electromagnetics, 2nd ed., Addison-Wesley, 1989.
Magnetic flux
S y
NI
c + 2 / µ0
= R
Magnetic flux Magnetic flux linkgage
) N (
0 2
y µ S
− Φ
= a
11
Example 6-24
Force between two coaxial circular coils:
µ θ
φ sin
4 2
2 0
R a Ib A =
Recall: →
D. K. Cheng, Field and Wave Electromagnetics, 2nd ed., Addison-Wesley, 1989.
µ θ
φ sin
4 2
2 1 1 1 0
12 R
b I a N
A =
→
=
R b R
b I
N 2
2 2 1 1 1 0
4 µ
aφ 2 3/2
2 2
2 2 1 1 1 0
) (
4 z b b b I N
= µ + aφ
∫
⋅= Φ
→
2
2 12
12 C
L A dl µ φ
d b b
z
b b I N
∫
C= +
2
2 2 / 3 2 2 2
2 2 1 1 1 0
) (
4
2 / 3 2 2 2
2 2 2 1 1 2 1 0
) (
2 z b
b b I N N
= µ +π
1 12
12 I
L
ΦL
=
→ 2 3/2
2 2
2 2 2 1 2 1 0
) (
2 z b b b N N
= µ + π
d z
z dz
I dL I
=
=
→ F12 a 1 2 12 2 5/2
2 2
2 2 2 1 2 1 0 2
1 2( )
3
b d
d b b N I N
zI +
−
= µ π
a ) N 2 (
3
4 2 1 0
d m m
z π
a µ
−
≅ ← m1 = N1I1πb12, m2 = N2I2πb22
D. K. Cheng, Field and Wave Electromagnetics, 2nd ed., Addison-Wesley, 1989.
(6-43)
