Tutorial 5 Solution Emagnet

TUTORIAL 5

(SOLUTION)

EEE 3133 ELECTROMAGNETIC FIELDS AND WAVES DEPARTMENT OF ELECTRICAL AND ELECTRONIC ENGINEERING 1. Explain the difference between convection and conduction current density. Identify the unit of measurement for both. Ans : Convection current Conduction current Occurs when current flow through an Current flow through a conductor. insulating medium such as liquid, vacuum. Does not involve conductor. Conduction current density :    Convection current density :
    

                  

2. A perfect conductor (σ= ∞) cannot contain an electrostatic field within it. Explain why. Identify the electric field intensity, charge density and potential inside a conductor under static conditions. Ans : A perfect conductor is an isolated conductor. When an external electric field Ee is applied to an isolated conductor, the positive free charges are pushed along the same direction as the applied field, while the negative charges move in the opposite direction. The free charges accumulate on the surface of the conductor and form an induced charge. The induced charges set up an internal induced field Ei, which cancels the externally applied field Ee. 3. Consider a 1-mile length of #16 copper wire which has a diameter of 0.0508 inch. The wire can safely carry current, I = 10A dc. The conductivity of the wire is assumed to be 5.80 x 107 S/m. Calculate a. The diameter and length of the wire in meter.Note : 1 in = 0.0254 m; 1 mile = 1.6km b. The resistance of the wire. c. The current density of the wire. d. The potential difference between the two end of the wire. e. The electric field intensity inside the wire. Ans: a. !  " "#"$ %0.0254 m) = 1.291 x 10-3m and &'   ( )%( *"+,-  (*"+ /012  :( : 415/16 -%/ 7145/189 ; /1?   A BCADAE8FG 7.65 A/mm2 <=>< -H @%

b. .   %3 c.  

B

d. I  J.  (" K:( :  :(:I e.  

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Ans : (a) D = 1.291 x 10-3m; length = 1609m; (b) 21.2 ohm ; (c) 7.65 A/mm2 ; (d) 212 V; (e) 0.312 V/m

4. Define polarization. Identify unit of measurement for polarization. Explain the phenomena which lead to polarization. Identify the charge density produced from polarization. Ans : Polarization P is the total dipole moment per unit volume of the dielectric. Unit measurement is Coulumbs per meter squared (C/m2). When an electric field E is applied on a dielectric, the positive charge is displaced from its equilibrium position in the direction of E by force F+=QE, while negative charge is displaced in the opposite direction by force F-=QE. A dipole results from the displacement and the dielectric is said to be polarized. Where polarization occurs, QR SQ  is formed over the surface and throughout the dielectric respectively. 5. Explain the difference in terms of flux density when E is applied to dielectric material; and to free space. Ans : The application of E to the dielectric material causes the flux density D to be greater in the dielectric material as compared to the free space. In free space, P = 0. Ans : Dielectric material T   UV W X Y ; Free space T   UV W 6. Explain the relation between these two equation: T   Z1 W X Y and T  ZW. Find the polarization P in a dielectric material with Z=  : $ if D = 3.0 x 10-7 a C/m2. Y   [> Z\ W T ]

W  

7 1^/1_` AE8C 4 F9a

=

Y  T b Z1 W = 1.93 x 10-7 a C/m2 Ans : 1.93 x 10-7 a C/m2 7. A parallel-plate capacitor with plate separation of 2mm has a 1 kV voltage applied to its plate. If the space between its plates is filled with polystyrene (Z=  : ##- find E,P, and cR . Assume that the plates are located at x = 0 and x = 2mm. Ans: W  bdI efghijkilmnjifomphnjiqnjx = 0 and x = 2mm, the displacement is 2mm is the x-direction.. Therefore, L

  r s5 = 1000 V/ 2mm = 500 ax kV/m

Y   [> Z\ W = (2.55 – 1)Z\ 500 ax kV/m = 6.853ax µC/m2 cR = Ps5 = 6.853µC/m2 Ans : 500 ax kV/m, 6.853ax µC/m2, 6.853µC/m2

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8. Two point charges in a dielectric medium where Z=  # : interact with a force of 8.6 x 10-3 N. What force would be expected if the charges were in free space. Ans: u u

Coulumbs law, t   v@]A ] BrB y shows that the force is inversely proportional to Z= In the free w x

space, the force will have its maximum value. z/ z t  Z  $ *}("_7%# :-  { {~}("_  {|Z\ Z=  = Ans: 4.47 x 10-2N 9. A homogeneous dielectric (Z=  : #- fills region 1 (x < 0) while region 2 (x > 0) is free space. (a) If D1 = 12ax – 10ay + 4az nC/m2, find D2 and € . (b) If E2 = 12 V/m and € = 600, find E1 and € Take €/ and € as defined in Example 5.9 page 170 in the text book. Ans : (a) 12ax – 4ay + 1.6az nC/m2 , 19.75o, (b) 10.67 V/m, 77o Ans : NEXT PAGE

Prepared by: Pn. Elya M Nor

Prepared by: Pn. Elya M Nor