Download Lecture 9 - web page for staff

ENE 325
Electromagnetic Fields
and Waves
Lecture 9 Magnetic Boundary Conditions,
Inductance and Mutual Inductance
31/01/08
1
Review (1)

Scalar magnetic potential (Vm) is different from the electric potential
in that the scalar magnetic potential is not a function of positions
and there is no physical interpretation.

Vector magnetic potential (A) is useful to find a magnetic field for
antenna and waveguide. We can transform Bio-Savart’s law and can
show that
for the current line
for current sheet
for current volume
31/01/08
0 Id L
A 
4 R
0 KdS
A  S
4 R
0 Jdv
A   vol
.
4 R
2
Review (2)

Magnetic force on a moving charge F  qv  B N.

Magnetic force on a current element F   Id L  B N.



For a straight conductor in a uniform magnetic field F  I L  B or
F = ILBsin N.
Torque on a closed circuit in which current is uniform can be
expressed as T  R  F Nm.
For current loop that has uniform current and magnetic field, torque
can be expressed as T  I S  B.
31/01/08
3
Magnetic boundary conditions (1)
Gauss’s law for magnetostatics
Bn1
 B dS  0
s
S
Bn1S  Bn 2 S  0
1
31/01/08
Bn1  Bn 2
2
Bn2
H n2
1

H n1
2
4
Magnetic boundary conditions (2)
Use Ampere’s circuital law
 H dL  I
L
1
2
Ht1L  Ht 2 L  K L
H t1  H t 2  K
a n12
( H 1  H 2 )  an12  K
or
31/01/08
H 1  H 2  a n12  K .
5
Ex1 The interface between two magnetic materials is
defined by the equation shown. Given H1 = 30
ax
A/m, determine the following
y>
2x-5
0
y<
5
x
2
a)
31/01/08
 r1=3
 r2=3
0
B1
6
b) B n1
c) H t1
d) H 2
31/01/08
7
Ex2 From the interface shown, given B1  2a x 10a y
mT, determine B2 and the angle that it makes with
the interface.
y-z
1
r1  1
2
r2  5000
x
31/01/08
8
Ex3 Let the permeability be 5 H/m in region A
where x < 0, and 20 H/m in region B where x > 0. If
there is a surface current density
K  150a y  200
az
A/m at x = 0, and if
A/m,
H A  300a x  400a y  500a z
find
a)
H tA
b)
H nA
31/01/08
9
c) H tB
d)
31/01/08
H nB
10
Potential energy of magnetic
materials
1
wH   B Hdv
2 vol
2
1
1
B
   H 2 dv  
dv J/m3.
2 vol
2 vol 
31/01/08
11
Duality of magnetostatics and
electrostatics
Electrostatics
E  V
b
Vab    E d L
a
J  E
I   J dS
31/01/08
Magnetostatics
H  Vm
b
Vm,ab   H d L
a
B  H
   B dS
V = IR
d
R
S
Vm = 
d
=
S
 E dL  0
 H d L  Itotal  NI
12
Inductance and mutual inductance


31/01/08
Flux linkage is the total flux passing through the surface
bounded by the contour of the circuit carrying the current.
Inductane L is defined as the ratio of flux linkage to the current
generating the flux,
Ntotal henrys or Wb/A.
L
I
13
A procedure for finding the inductance
1. Assume a current I in the conductor
2. Determine B using the law of Bio-Savart, or Ampere’s circuital
law if there is sufficient symmetry.
3. Calculate the total flux  linking all the loops.
4. Multiply the total flux by the number of loops to get the flux
linkage.
5. Divide the flux linkage by I to get the inductance. The assumed
current will be divided out.
31/01/08
14
Inductance for a coaxial cable
total flux
31/01/08
0
b

Id ln
2
a
15
Inductance for a solenoid with N
turns
31/01/08
16
Inductance for a toroid that has N
turns and current I.
If the wired is tightly wound, the flux linkage will be the same
for the adjacent turns of toroid. If the adjacent turns are
separated by some finite distance, the total flux must be
calculated from the flux from each turn.
31/01/08
( N )total  1  2  .......
17
More about inductance

The definition of the inductance can be written in the form of
magnetic energy as
L

2 wH N

.
2
I
I
The current inside conductor creates the magnetic flux inside the
material texture. This flux causes an internal inductance which
combines with the external inductance to get the total
inductance. Normally, the internal inductance can be neglected
due to its small value compared to the external one.
31/01/08
18
Mutual inductance

Mutual inductance M is the inductance that is caused by the flux
linking to the different circuit. The mutual inductance between
circuit 1 and circuit 2 can be expressed as
M 12
N 212

I1
N121
M 21 
I2
where M12 = M21
12 = flux produced by current I1 that is linked to current I2
21 = flux produced by current I2 that is linked to current I1
19
N1, N2 = number of loops in circuit 1 and circuit 2 respectively.
31/01/08
Ex4 Calculate the inductance for the following
configuration:
a) A coaxial cable with the length l = 10 m, the inner radius a = 1
mm, and the outer radius b = 4 mm. The inserted magnetic
material has r = 18 and r = 80.
31/01/08
20
b) A toroid with a number of turns N = 5000 turns with in = 3 cm,
out = 5 cm, and the length l = 1.5 cm. The inserted material has
r = 6.
c) A solenoid has the radius r = 2 cm, the length l = 8 cm, and N =
900 turns. The inserted material has r = 100.
31/01/08
21
Ex5 Two solenoids with the square cores are placed as shown
below. The inner dimension is 1.2x1.2 cm. The outer
dimension is 3x3 cm. The solenoid has 1200 turns and the
length is 25 cm. Given r1 = 6.25, r2 = 1, determine the
following
a) Given the current I = 1 A, determine H inside the solenoid when
the inner solenoid is removed.
31/01/08
22
b) Determine the resulting self inductance.
c) Determine the mutual inductance between two solenoids.
31/01/08
23