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I CONFIDENTIAL I FINAL EXAMINATION SEMESTER I SESSION 2009/2010 COURSE CODE COURSE LECTURERS PROGRAMME SECTION TIME DATE SEE 2523 / SZE 2523 ELECTROMAGNETIC FIELD THEORY ASSOC. PROF. DR. NORAZAN BIN MOHD KASSIM DR. YOU KOK YEOW MDM. FATIMAH BT MOHAMAD SEC / SEE / SEI / SEL / SEM !SEP / SET / SEW 01 - 03 2 HOURS 30 MINUTES 9 NOVEMBER 2009 INSTRUCTION TO CANDIDATE ANSWER FOUR (4) QUESTIONS ONLY. THIS EXAMINATION BOOKLET CONSISTS OF 9 PAGES INCLUDING THE FRONT COVER

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Page 1: See 2523 sze 2523 2

I CONFIDENTIAL I

FINAL EXAMINATION SEMESTER ISESSION 2009/2010

COURSE CODE

COURSE

LECTURERS

PROGRAMME

SECTION

TIME

DATE

SEE 2523 / SZE 2523

ELECTROMAGNETIC FIELD THEORY

ASSOC. PROF. DR. NORAZAN BIN MOHDKASSIMDR. YOU KOK YEOWMDM. FATIMAH BT MOHAMAD

SEC / SEE / SEI / SEL / SEM !SEP / SET /SEW

01 - 03

2 HOURS 30 MINUTES

9 NOVEMBER 2009

INSTRUCTION TO CANDIDATE

ANSWER FOUR (4) QUESTIONS ONLY.

THIS EXAMINATION BOOKLET CONSISTS OF 9 PAGES INCLUDING THE FRONTCOVER

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SEE 25232

Q I (a) Determine the electric field intensity on the axis of a circular ring of uniform

charge Pi (C1m) with radius a in the .xy plane. Let the axis of the ring be

along the z axis. (7 marks)

(b) At what distance along the positive z axis is the electric field from Ql(a)

maximum, and what is the magnitude of this field? (8 marks)

(c) Find the absolute potential at any point along the z axis referring to Q 1(a).

(5 marks)

(d) Compare the result of electric field using the gradient concept.

(5 marks)

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SEE 25233

Q2 (a) With the aid of suitable diagrams and Maxwell 's equations, develop theboundary condition equations for two dielectric materials with differentpermittivities.

(5 marks)

(b) Two concentric spheres are shown in Fig. Q2(b). The spheres are built fromtwo different dielectric materials . Each sphere has been covered with a thinlayer of conductor, in wh ich the th ickness can be ignored .

Fig. Q2(b)

By assuming negative charge on the inner sphere surface:

(i) Determine the electric field intensities in material #1 and material #2.(4 marks)

(ii) Obtain the energy stored in the structure.(5 marks)

(iii) Obtain the resistance between the spheres if the conductivity factor for

the material #2 is assumed to be (Yl !1S/m.(4 marks)

(iv) Given the radius r a"" 10 urn. Determine the range of structural radiu s(range of rb) such that the minimum and maximum charge perunit voltthat can be stored in material #2 are 3 F and 10 F, respectively.

(7 marks)

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SEE 25234

Q3 (a) Compare the usefulness of Ampere's Circuital Law and Biot Savart Law in

determining the magnetic field intensity Bof a current carrying circuit.

(5 marks)

(b) A coaxial cable as shown in Fig. Q3(a) consists of a solid cylindrical inner

conductor having a radius a and an outer cylinder in the form of a cylindrical

shell having a radius b. If the inner conductor carries a current of I A in the

form of a uniform current density J == IIrra2 i (Alm2) and the outer

conductor carries a return current of 1 A in the form of a uniform current

density Js = -Il2rrb i (Aim). Determine:

(i) the magnitude of the force per unit length acting 10 split the outer

cylinder apart longitudinally. (7 marks)

(ii) the inductance per unit length of the coaxial cable .

(8 marks)

(e) An infinite cylindrical wire with radius a carries a uniform current density

J == l ltta? i (A/m2) , except inside an infinite cylindrical hole parallel to the

wire's axis. The hole has radius c and is tangent to the exterior of the wire.

A short chunk of the wire is shown in the accompanying Fig. Q3(b) .

Calculate the magnetic field everywhere inside the hole, and sketch the lines

ofBon the figure . (S marks)

· · · ·~· · · · · : Ia

-r----rFig. Q3(a)

y

3D view

Fig. Q3(b)

2D view

x

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SEE 25235

Q4 (a) Within a certain region, e = 1O- 11F[m and f.-l = lO-sHlm. If

B; = 2 x 10- 4 cos lOSt sin 10-3 y (Testa):

(i) Use the appropriate Maxwell's equation to find E. (5 marks)

(ii) Find the total magnetic flux passing through the surface

x = 0,0< Y < 40m, 0 < z < 2m, at t = lfiS. (4 marks)

(iii) Find the value of the closed line integral of E around the perimeter of

the given surface at t = lfiS. (5 marks)

(b) A voltage source is connected by means of wire to a parallel-plate capacitor

made up of circular plates of radius a in the z =0 and z = d planes, and

having their centers on the z-axis. The electric field between the plates is

given by

_ n:TE = £0 sin 20 cos wt z

Find the total current flowing through the capacitor, assuming the region

between the plates to be free space, and that no field exists outside the region.

(7 marks)

(c) By analyzing the expression of current in Q4(b) above, give two suggestions

to increase the magnitude ofthis current. (4 marks)

Hint: Judv = uv - Jvdu

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SEE 25236

Q5 (a) Derive the vector wave equation for electric field using Maxwell equation.

(4 marks)

(b) The propagation constant y can be written as y = a + jf3, where a is the

attenuation constant of the medium and f3 is its phase constant. Proof that a

and f3 can be expressed as

j1E' ( (E")2 )a=w T 1+"7 -1

j1E' ([ (E")2 )f3=WT~1+7 +1

where OJ is the angular frequency. E;' and E;" are the dielectric constant and

loss factor, respectively. J.1 is the permeability.

(8 marks)

(c) A 20 V 1m electric plane wave with frequency 500 MHz propagates in the z

direction and polarized in the x direction in a medium. The properties of medium

has relative permittivity, e r = 4.5 - jO .02 and relative permeability, J.1 r = 1 .

(i) Write a complete time-domain expression for the electric field, E(6 marks)

(ii) Determine the corresponding expression for the magnetic field.

(7 marks)

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SEE 25237

ELECTROSTATIC FIELD I MAGNETOSTATIC FIELD- J dQ " I - f~ XClRCoulomb's Law E::= ---2aR Biot-Savart Law H = 4;d(247!&oR

Gauss's Law db.ds=Qm Ampere Circuital law dH· a1 = len

Force on a point charge F=EQ. Force on a moving charge F = Q(u x B)

Force on a current element F = IdI x 7fElectric field for finite line charge Magnetic field for finite current

E=.--t1..-F(Sina2 + sinal) +~(cosa2 -cosa)}- I .H =-(sina2 +sina1)¢

47l"&o r r 47lr

Electric field for infinite line charge Magnetic field for infinite current

E=J:L,. - I·H=-¢

27l"&or 27lr

Electric flux density D = cE Magnetic flux density B=pH

Electric flux If/E=Q=QD. ds Magnetic flux If/m = JB .ds

Divergence theorem QD .ds= j (v»» Stoke's theorem c}H.dI = j (vx H) dss , I s

Potential difference VAB ::= -1E .dIB

Absolute potential V J dO= 47l"~R

Gradient of potential ~ E = -V' V Magnetic potential, (A)-1 B= V' x AEnergy stored in an electric field Energy stored in a magnetic field

WE =! J(15 .E)1v w; ::= ~ !(B. H)1v2 v

Total current in a conductor

l=fJ·ds where] = 0-£

Polarization vector P =D-8aE Magnetization vector M == XmHwhere Xm =u; -1

Bound surface charge density Magnetized surface current densitypso=P.n J,m=Mxn

Volume surface charge density Magnetized volume current density

Pvo::=-Y'P ]m =YxMElectrical boundary conditions Magnetic boundary conditions

.L1" - Dg, =p, and £'1 =E21 B... = Bz,. and ~I - H2t =J, --R . I Inductance L ::=!:.- where A= 11/,,/'/esistance R=-

as I

Capacitance C =gVao

Poisson's equation V'2V ::= _frE:

Laplace equation v'v = 0

Maxwell equation y. D = Pv. V' x E =0 Maxwell equation V'. B = 0, 'YxH=J

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SEE 25238

TIME VARYING FIELD

Maxwell equation V'. D== p, Gauss's Law for electric field

- a8\I x E = -- Faraday's Law

if\1·8=0

- - aD\lxH=J+­

if

Gauss's Law for magnetic field

Ampere Circuital Law

Characteristics of wave propagation in lossy medium (a*O,/-l=AA,E=fioE,.)

Electric field, E(z,t) == Eoe-a: cos(a>t - /lz)xMagnetic field, H(z,t) =~e-«" cos(aJt- fJz+ 8n)y

Attenuation constan! a"Q)~~ ~I+(:J -I

where tan 28 =.!!...n OJf:

Phase constant

Intrinsic impedance

Skin depth

Poynting theorem

Poynting vector

Average power

j;i&'7=1 FLBn1+(CT I (j)£)"

I ~ = Eon,o==l/a

.r~ -) - a III " 1 2} I 2I..!\.E xH ·ds=-- -6£ +-pH v- aE dvsa,. 2 2 .'

tJ=ExHP E; -2«" e

avg =-e cos n217

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SEE 25239

Kecerunan "Gradient"

Vj =xOf +Yoj+ Zojox By oz

Vj =Poj + i Of + Zojor r or/! 8z

Vj=p aj +~ Of +-L Ofar r ae rsine or/!

Kecapahan "Divergence"

- aA oA, aAV -A =-'+-"+-'

ax By oz

«: =.!.[acrAJ] +.!. OAf +aA,r or r or/! OZ

V -:4=~[O(r'A,)]+_I_ [O(Aa Sine)]+_I_OAfr' or r sine' 8e r sine' or/!

Ikal "Curl"

- _(OA, 8Ay \j _(8A, OA, ) _lOA, oA, JVx A =x 8;- az + Y a;-a; +Z a;--8;

- _[ 18A, oA; j\ "( 8A, OA,) Z[oCrA;) 8A,]VxA =r ---- +r/! --- +- ----r or/! oz oz or " r Or or/!

V x A = _P_[OCAfsine) _8Aa]+~[_1_8A, _8(rA;)] +i(8crAa) - oA, Jrsine ae ar/! r sine or/! or r or oe

Laplacian