1. ENE 325 Electromagnetic Fields and Waves
Lecture 6 Permeability, Magnetic Boundary Conditions, Inductance and Mutual Inductance

2. 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

3. Review (2) Magnetic force on a moving charge N.
Magnetic force on a current element N.
For a straight conductor in a uniform magnetic field or F = ILBsin N.
Torque on a closed circuit in which current is uniform can be expressed as Nm.
For current loop that has uniform current and magnetic field, torque can be expressed as

4. Magnetization and permeability
how do the magnetic dipoles act as a distributed source for the magnetic field? The result will look like Ampere’s circuital law, the current will be the movement of bound charges, and the field, which has the dimension of will be called the magnetization
From
the total magnetic dipole
where n = number of dipole moment per unit volume.
Define the magnetization
A/m

5. Bounded current
Consider the alignment of the dipole when there is an external magnetic field,
in a volume of , there will be n dipoles aligning. Each dipole has the current I, therefore there will be the increase in the total current in that small volume which can be shown as or

6. The relationship between and in mediums other than free space (1)
From Ampère’s circuital law,
where If is free current, then

7. The relationship between and in mediums other than free space (2)
Define or
therefore, we can write
Note: this is the magnetic field in any medium.

8. The relationship between and in mediums other than free space (3)
For a linear, isotropic media
where m is the magnetic susceptibility,
therefore
which shows that

9. The relationship between and in mediums other than free space (4)
Therefore we can write
where
 = r0.
If r  1, diamagnetic material
If r  1, paramagnetic material
If r , ferromagnetic material
The magnetic susceptibility of some materials
hydrogen = -210-5
 copper = -0.910-5
 germanium = -0.810-5
 silicon = -0.310-5

10. Magnetic boundary conditions (1)
Gauss’s law for magnetostatics

11. Magnetic boundary conditions (2)
Use Ampere’s circuital law or

12. Ex1 The interface between two magnetic materials is defined by the equation shown. Given H1 = A/m, determine the following a)

13. b) c) d)

14. Ex2 From the interface shown, given mT, determine and the angle that it makes with the interface.

15. 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 A/m at x = 0, and if A/m, find a) b)

16. c) d)

17. Potential energy of magnetic materials
J/m3.

18. Duality of magnetostatics and electrostatics
Electrostatics Magnetostatics
V = IR Vm = 
 =

19. Inductance and mutual inductance
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,
henrys or Wb/A.

20. A procedure for finding the inductance
1. Assume a current I in the conductor
2. Determine 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.

21. Inductance for a coaxial cable
total flux

22. Inductance for a solenoid with N turns

23. Inductance for a toroid that has N turns and current I.
If the wired is tightly wounded, 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.

24. More about inductance
The definition of the inductance can be written in the form of magnetic energy as
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.

25. 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
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
N1, N2 = number of loops in circuit 1 and circuit 2 respectively.

26. 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.

27. 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.

28. Ex5 Two solenoids with the square cores are placed as shown below
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.

29. b) Determine the resulting self inductance.
c) Determine the mutual inductance between two solenoids.