Stokes’ Theorem

President University Erwin Sitompul EEM 12/1

Lecture 12

Engineering Electromagnetics

Dr.-Ing. Erwin SitompulPresident University

http://zitompul.wordpress.com

2 0 1 3


Stokes’ TheoremPreviously, 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 8 The 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.


Stokes’ TheoremConsider 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:


( )SN

d

S

H LH

( )SN

d

S

H LH a

( ) ( )S Nd S H L H a H S

or

or


Stokes’ TheoremLet us now perform the circulation for every ΔS comprising S

and sum the results.


( )S

d d H L H S

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.


Stokes’ TheoremChapter 8 The Steady Magnetic Field

ExampleConsider the portion of a sphere as shown. The surface is specified by r = 4, 0 ≤ θ ≤ 0.1π, 0 ≤ φ ≤ 0.3π. The closed path forming its perimeter is composed of three circular arcs. Given the magnetic field H = 6r sinφ ar + 18r sinθcosφ aφ A/m, evaluate each side of Stokes’ theorem.

sinrd dr rd r d L a a a

1 2 3

sind H rd H r d H rd H L

0.3

0

18(4)sin 0.1 cos 4sin 0.1d d

H L2288sin 0.1 sin 0.3 22.2 A


Stokes’ TheoremChapter 8 The Steady Magnetic Field

1 36 sin cos cossin

1 1 6 cos 36 sin cossin

rrr

r rr

H a

a

2 sin rd r d d S a

0.3 0.1

0 0

( ) (36cos cos )16sinS

d d d

H S

0.30.121

2 00

576( sin ) cos d

2288sin 0.1 sin 0.3 22.2 A


Magnetic Flux and Magnetic Flux DensityChapter 8 The Steady Magnetic Field

In free space, let us define magnetic flux density B as

0B H

where B is measured in webers per square meter (Wb/m2) or tesla (T).

The constant μ0 is not dimensionless and has a defined value for free space, in henrys per meter (H/m), of

70 4 10 H m

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 = ε0E and B = μ0H.



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,

WbS

d B SS

d Q D S

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.



For this reason, the Gauss’s law for the magnetic field can be written as

0S

d B S

Through the application of the divergence theorem, we can also find that

0 B• Fourth Maxwell’s Equation,

static electric fields and steady magnetic fields.



Collecting all equations we have until now of static electric fields and steady magnetic fields,

0v

DE

The corresponding set of four integral equations that apply to static electric fields and steady magnetic fields is

vol

0

vS

d Q dv

d

D S

E L

0 H JB

0S

S

d I d

d

H L J S

B S


Solved ExampleExample

Find the flux between the conductors of the coaxial line we have discussed previously.


( )2IH a b

00 2

I

B H a

S

d B S 0

0 2

d b

a

I d dz

a a

0 ln2Id b

a


Exercise Problems1. A dielectric interface is described by 4y + 3z = 12 m. The side including the

origin is free space where D1 = ax + 3ay + 2 az μC/m2. On the other side, εr2 = 3.6. Find D2 and θ2.

(Scha.7.29 ep.117)

Answer: 5.14 μC/m2, 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 εr1 and εr2. (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(εr1+εr2)ε0l/ln(b/a); (b) 9.75 pF



Exercise Problems3. The regions, 0 < z < 0.1 m and 0.3 < z < 0.4 m, are conducting slabs

carrying uniform current densities of 10 A/m2 in opposite directions, as shown in the next figure. Find Hx 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 az direction. The conductor is inhomogeneous, having a conductivity that varies with ρ as σ = 105(1+2.5 × 105ρ2) 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)



End of the LectureChapter 8 The Steady Magnetic Field

Stokes’ Theorem

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