UNIVERSITI MALAYSIA PERLIS EKT 241/4: ELECTROMAGNETIC THEORY UNIVERSITI MALAYSIA PERLIS CHAPTER 4 – MAGNETOSTATICS PREPARED BY: NORDIANA MOHAMAD SAAID dianams@unimap.edu.my

Chapter Outline Maxwell’s Equations Magnetic Forces and Torques The total electromagnetic force, known as Lorentz force Biot- Savart’s law Gauss’s law for magnetism Ampere’s law for magnetism Magnetic Field and Flux Vector magnetic potential Properties of 3 different types of material Boundary conditions between two different media Self inductance and mutual inductance Magnetic energy

Maxwell’s equations Maxwell’s equations: Where; E = electric field intensity D = electric flux density ρv = electric charge density per unit volume H = magnetic field intensity B = magnetic flux density

Maxwell’s equations For static case, ∂/∂t = 0. Maxwell’s equations is reduced to: Electrostatics Magnetostatics

Magnetic Force B = Magnetic Flux Density B B q q I q B B

Magnetic Torque on a Current- Carrying Loop Applied force vector F and distance vector d are used to generate a torque T T = d× F (N·m) Rotation direction is governed by right-hand rule.

Magnetic Forces and Torques The electric force Fe per unit charge acting on a test charge placed at a point in space with electric field E. When a charged particle moving with a velocity u passing through that point in space, the magnetic force Fm is exerted on that charged particle. where B = magnetic flux density (Cm/s or Tesla T)

Magnetic Forces and Torques If a charged particle is in the presence of both an electric field E and magnetic field B, the total electromagnetic force acting on it is:

Magnetic Force on a Current- Carrying Conductor For closed circuit of contour C carrying I , total magnetic force Fm is: In a uniform magnetic field, Fm is zero for a closed circuit.

Magnetic Force on a Current- Carrying Conductor On a line segment, Fm is proportional to the vector between the end points.

Example 1 The semicircular conductor shown carries a current I. The closed circuit is exposed to a uniform magnetic field . Determine (a) the magnetic force F1 on the straight section of the wire and (b) the force F2 on the curved section.

Solution to Example 1 a)

The Biot–Savart’s Law where: The Biot–Savart law is used to compute the magnetic field generated by a steady current, i.e. a continual flow of charges, for example through a wire Biot–Savart’s law states that: where: dH = differential magnetic field dl = differential length

The Biot–Savart’s Law To determine the total H:

The Biot–Savart’s Law Biot–Savart’s law may be expressed in terms of distributed current sources.

Example 2 Determine the magnetic field at the apex O of the pie-shaped loop as shown. Ignore the contributions to the field due to the current in the small arcs near O.

? = -dl For segment AC, dl is in φ direction, O  A C  O A C ? For segment AC, dl is in φ direction, Using Biot- Savart’s law:

QUIZ!! z ε1=2ε0 ε2=8ε0 Ө2 E2 xy E1 Ө1 Find E1 if E2 = 2x -3y +3z with s = 3.54 x 10-11(C/m2) And find Ө1 and Ө2

Gauss’s Law for Magnetism Gauss’s law for magnetism states that: Magnetic field lines always form continuous closed loops.

Ampere’s law for magnetism Ampere’s law states that: true for an infinite length of conductor H C, +aø dl true for an infinite length of conductor I, +az r

Magnetic Field of an infinite length of conductor From then re-arrange the equation in terms of Hφ: Hence, the magnetic field vector, H:

Example 3 A toroidal coil with N turns carrying a current I , determine the magnetic field H in each of the following three regions: r < a, a < r < b,and r > b, all in the azimuthal plane of the toroid.

Solution to Example 3 H = 0 for r < a as no current is flowing through the surface of the contour H = 0 for r > b, as equal number of current coils cross the surface in both directions. For a < r < b, we apply Ampere’s law: Hence, H = NI/(2πr) .

Magnetic Flux The amount of magnetic flux, φ in Webers from magnetic field passing through a surface is found in a manner analogous to finding electric flux:

Example 4 An infinite length coaxial cable with inner conductor radius of 0.01m and outer conductor radius of 0.05m carrying a current of 2.5A exists along the z axis in the + az direction. Find the flux passing through the region between two conductors with height of 2 m in free space.

Solution to Example 4 inner conductor radius = r1 0.01m outer conductor radius = r2 0.05m current of 2.5A (in the +az direction) Flux radius = 2m Iaz=2.5A z aø Flux,z xy r1 r2

Solution to Example 4 where dS is in the aø direction. So, Therefore,

Vector Magnetic Potential For any vector of vector magnetic potential A: We are able to derive: . Vector Poisson’s equation is given as: where

Magnetic Properties of Materials Diamagnetic Paramagnetic Ferromagnetic The behavior of magnetic dipole moments & magnetic susceptibility, of its atoms with an external magnetic field is used as a basis for classifying magnetic materials.

Magnetic Properties of Materials Magnetization in a material is associated with atomic current loops generated by two principal mechanisms: Orbital motions of the electrons around the nucleus, i.e orbital magnetic moment, mo Electron spin about its own axis, i.e spin magnetic moment, ms

Magnetic Permeability Magnetization vector M is defined as where = magnetic susceptibility (dimensionless) Magnetic permeability is defined as: and to define the magnetic properties in term of relative permeability is defined as:

Magnetic Materials - Diamagnetic metals have a very weak and negative susceptibility ( ) to magnetic field slightly repelled by a magnetic field and the material does not retain the magnetic properties when the external field is removed Most elements in the periodic table, including copper, silver, and gold, are diamagnetic.

Magnetic Materials - Paramagnetic Paramagnetic materials have a small and positive susceptibilities to magnetic fields. slightly attracted by a magnetic field and the material does not retain the magnetic properties when the external field is removed. Paramagnetic materials include magnesium, molybdenum, lithium, and tantalum.

Magnetic Materials – Diamagnetic, Paramagnetic However, the absolute susceptibilities value of both materials is in the order 10-5. Thus, can be ignored. Hence, we have Magnetic permeability: Diamagnetic and paramagnetic materials include dielectric materials and most metals.

Magnetic Materials – Ferromagnetic Materials Ferromagnetic materials is characterized by magnetized domain - a microscopic region within which the magnetic moments of all its atoms are aligned parallel to each other. Hysteresis – “to lag behind”. It determines how easy/hard for a magnetic material to be magnetized and demagnetized.

Process of Magnetic Hysteresis material is magnetized and can serve as permanent magnet! B material is demagnetize

Magnetic Hysteresis of Ferromagnetic Materials Comparison of hysteresis curves for (a) a hard and (b) a soft ferromagnetic material is shown. Hard magnetic material- cannot be easily magnetized & demagnetized by an external magnetic field. Soft magnetic material – easily magnetized & demagnetized.

Magnetic Hysteresis of Ferromagnetic Materials Properties of magnetic materials as follows:

Magnetic boundary conditions Boundary between medium 1 with μ1 and medium 2 with μ2

Magnetic boundary conditions Boundary condition related to normal components of the electric field; By analogy, application of Gauss’s law for magnetism, we get first boundary condition: Since , For linear, isotropic media, the first boundary condition which is related to H;

z xy By applying Ampere’s law

Magnetic boundary conditions The result is generalized to a vector form: Where However, surface currents can exist only on the surfaces of perfect conductors and perfect superconductors (infinite conductivities). Hence, at the interface between media with finite conductivities, Js=0. Thus:

Example xy (plane)

Solution: H1t = H2t thus, H2t = 6ax + 2ay Hn1 = 3az, but, Hn2 = ?? 6000μ0(3az) = 3000 μ0(Hn2) Hn2 = 6az thus, H2 =6ax + 2ay + 6az μr1 = 6000 ; μr2 = 3000 ;

Inductance An inductor is the magnetic analogue of an electrical capacitor. Capacitor can store electric energy in the electric field present in the medium between its conducting surfaces. Inductor can store magnetic energy in the volume comprising the inductors.

calculate magnetic field in solenoid Calculate Magnetic Energy Mutual inductance: produced by magnetic coupling between two different conducting structures. Calculate Magnetic Energy Self inductance: is the ratio of the magnetic flux linkage, Λ to the current I flowing through the structure. INDUCTANCE store magnetic energy

Inductance Example of an inductor is a solenoid - a coil consisting of multiple turns of wire wound in a helical geometry around a cylindrical core.

Magnetic Field in a Solenoid For one cross section of solenoid, When l >a, θ1≈−90° and θ2≈90°, Where, N=nl =total number of turns over the length l

(The ratio of the magnetic flux to the current) Self Inductance The self-inductance of a circuit is used to describe the reaction of the circuit to a changing current in the circuit, (The ratio of the magnetic flux to the current)

Self Inductance Self-inductance of any conducting structure is the ratio of the magnetic flux linkage, Λ to the current I flowing through the structure. Magnetic flux linkage, Λ is the total magnetic flux linking a given conducting structure.

Self Inductance Magnetic flux, linking a surface S is given by: In a solenoid with uniform magnetic field, the flux linking a single loop is:

Self Inductance – magnetic flux in solenoid

Self Inductance Magnetic flux, linking a surface S is given by: In a solenoid with uniform magnetic field, the flux linking a single loop is:

Self Inductance For a solenoid: For two conductor configuration:

Self Inductance for a solenoid Thus,

Mutual Inductance Mutual inductance – produced by magnetic coupling between two different conducting structures.

Mutual Inductance Magnetic field B1 generated by current I1 results in a flux Φ12 through loop 2: If loop 2 consists of N2 turns all coupled by B1 in exactly the same way, the total magnetic flux linkage through loop 2 due to B1 is:

Mutual Inductance Hence, the mutual inductance:

Magnetic Energy Consider an inductor with an inductance L connected to a current source. The current I flowing through the inductor is increased from zero to a final value I. The energy expended in building up the current in the inductor: i.e the magnetic energy stored in the inductor

Magnetic Energy Magnetic energy density (for solenoid): i.e magnetic energy per unit volume Magnetic energy in magnetic field:

Example: Magnetic Energy in a Coaxial Cable Derive an expression for the magnetic energy stored in a coaxial cable of length l and inner and outer a and b. The insulation material has magnetic permeability μ.