1. President UniversityErwin SitompulEEM 12/1 Lecture 12 Engineering Electromagnetics Dr.-Ing. Erwin Sitompul President University http://zitompul.wordpress.com 2014

2. President UniversityErwin SitompulEEM 12/2 Stokes’ Theorem Previously, from Ampere’s circuital law, we derive one of Maxwell’s equations, ∇ ×H = J. This equation is to be considered as the point form of Ampere’s circuital law and applies on a “per-unit-area” basis. Chapter 8The Steady Magnetic Field Now, we shall devote the material to a mathematical theorem known as Stokes’ theorem. In the process, we shall show that we may obtain Ampere’s circuital law from ∇ ×H = J.

3. President UniversityErwin SitompulEEM 12/3 Stokes’ Theorem Consider the surface S of the next figure, which is broken up into incremental surfaces of area ΔS. If we apply the definition of the curl to one of these incremental surfaces, then: Chapter 8The Steady Magnetic Field or

4. President UniversityErwin SitompulEEM 12/4 Stokes’ Theorem Let us now perform the circulation for every ΔS comprising S and sum the results. Chapter 8The Steady Magnetic Field Therefore, where dL is taken only on the perimeter of S. As we evaluate the closed line integral for each ΔS, some cancellation will occur because every interior wall is covered once in each direction. The only boundaries on which cancellation cannot occur form the outside boundary, the path enclosing S.

5. President UniversityErwin SitompulEEM 12/5 Stokes’ Theorem Chapter 8The Steady Magnetic Field Example Consider the portion of a sphere as shown. The surface is specified by r = 4, 0 ≤ θ ≤ 0.1π, 0 ≤  ≤ 0.3π. The closed path forming its perimeter is composed of three circular arcs. Given the magnetic field H = 6r sin  a r + 18r sinθcos  a φ A/m, evaluate each side of Stokes’ theorem.

6. President UniversityErwin SitompulEEM 12/6 Stokes’ Theorem Chapter 8The Steady Magnetic Field

7. President UniversityErwin SitompulEEM 12/7 Magnetic Flux and Magnetic Flux Density Chapter 8The Steady Magnetic Field In free space, let us define magnetic flux density B as where B is measured in webers per square meter (Wb/m 2 ) or tesla (T). The constant μ 0 is not dimensionless and has a defined value for free space, in henrys per meter (H/m), of The magnetic-flux-density vector B, as the name weber per square meter implies, is a member of the flux-density family of vector fields. Comparing the laws of Biot-Savart and Coulomb, one can find analogy between H and E that leads to an analogy between B and D; D = ε 0 E and B = μ 0 H.

8. President UniversityErwin SitompulEEM 12/8 Magnetic Flux and Magnetic Flux Density Chapter 8The Steady Magnetic Field If B is measured in teslas or webers per square meter, then magnetic flux H should be measured in webers. Let us represent magnetic flux by Φ and define Φ as the flux passing through any designated area, We remember that Gauss’s law states that the total electric flux passing through any closed surface is equal to the charge enclosed. This charge is the source of the electric flux D. For magnetic flux, no current source can be enclosed, since the current is considered to be in closed circuit.

9. President UniversityErwin SitompulEEM 12/9 Magnetic Flux and Magnetic Flux Density Chapter 8The Steady Magnetic Field For this reason, the Gauss’s law for the magnetic field can be written as Through the application of the divergence theorem, we can also find that Fourth Maxwell’s Equation, static electric fields and steady magnetic fields.

10. President UniversityErwin SitompulEEM 12/10 Magnetic Flux and Magnetic Flux Density Chapter 8The Steady Magnetic Field Collecting all equations we have until now of static electric fields and steady magnetic fields, The corresponding set of four integral equations that apply to static electric fields and steady magnetic fields is

11. President UniversityErwin SitompulEEM 12/11 Solved Example Example Find the flux between the conductors of the coaxial line we have discussed previously. Chapter 8The Steady Magnetic Field

12. President UniversityErwin SitompulEEM 12/12 Exercise Problems 1.A dielectric interface is described by 4y + 3z = 12 m. The side including the origin is free space where D 1 = a x + 3a y + 2 a z μC/m 2. On the other side, ε r2 = 3.6. Find D 2 and θ 2. (Scha.7.29 ep.117) Answer: 5.14 μC/m 2, 45.6° 2.A coaxial capacitor consists of two conducting coaxial surfaces of radii a and b ( a < b), and length l. The space between is filled with two different dielectric materials with relative dielectric constants ε r1 and ε r2. (a) Find the capacitance of this configuration; (b) Assuming l = 5 cm, b = 3a = 1.5 cm, and oil and mica are used, calculate the capacitance. (Ina2.4.33 S.386) Answer: (a) C = π(ε r1 +εr2)ε 0 l/ln(b/a); (b) 9.75 pF Chapter 8The Steady Magnetic Field

13. President UniversityErwin SitompulEEM 12/13 Exercise Problems 3.The regions, 0 < z < 0.1 m and 0.3 < z < 0.4 m, are conducting slabs carrying uniform current densities of 10 A/m 2 in opposite directions, as shown in the next figure. Find H x at z = –0.04, 0.06, 0.26, 0.36, and 0.46 m. (Hay.E5.S261.13) Answer: 0, –0.6 A/m, –1 A/m, –0.4 A/m, 0. Answer: (a) 0.213 μWb; (b) 2.85 μWb. 4.A solid nonmagnetic conductor of circular cross section, ρ = 2 cm, carries a total current, I = 60 A, in the a z direction. The conductor is inhomogeneous, having a conductivity that varies with ρ as σ = 10 5 (1+2.5 × 10 5 ρ 2 ) S/m. Find the total flux crossing the radial plane defined by φ = 0, 0 < z < 1 m, and: (a) 0 < ρ < 1 cm; (b) 1 < ρ < 2 cm. (Hay.E5.S261.31) Chapter 8The Steady Magnetic Field

14. President UniversityErwin SitompulEEM 12/14 End of the Lecture Chapter 8The Steady Magnetic Field