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

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

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

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

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

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

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.

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

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 1, 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

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

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

Ex1 The interface between two magnetic materials is defined by the equation shown. Given H1 = 30 A/m, determine the following a)

b) c) d)

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

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)

c) d)

Potential energy of magnetic materials J/m3.

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

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.

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.

Inductance for a coaxial cable total flux

Inductance for a solenoid with N turns

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.

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.

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.

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.

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.

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.

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