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Unformatted text preview: Sd/Q'H'on ECE604 Homework 3
Out: Tuesday, January 25, 2005
Due: Tuesday, February 1, 2005 Note: There is now a class website at . Problem sets,
solutions, and other handouts will be posted there. These problems are from Chapter 1 of the text by Ramo, Whinnery and Van Duzer (3rd edition, 1994) 1)
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6) 7) 7.9a 7.11a parts (i) and (ii) only. Assume both functions have period L. 7.1 lb 7.12a 7.12b 7 .12e Note: Make sure you can identify all the important boundary conditions before you jump into the math. A conductor on a circuit board has width W and carries a current I. The conductivity of the conductor is 0'. There is a small hole of radius a at the center of the conductor. The
thickness t of the conductor is small compared to a. (Therefore this problem may be
treated as a twodimensional problem in cylindrical coordinates.) Assume the current is V
DC. You may also assume that the width of the circuit board W is large compared to the
hole radius (W>>a). Let the coordinate origin be at the center of the hole. (i) Sketch qualitatively the sheet current density J S (units are A/m) in the vicinity of the
hole. (The sheet current density is the uniform current density J integrated over the thickness of the conductor, that is J S = J Xt ) (ii) Derive an expression for J 8. Express your answer in terms of I, W, r and 4). (iii) What is the maximum magnitude of JS? Express your answer in terms of I and W.
(The answer does not include the hole radius a.) Note that the maximum should be
located along the x = 0 line (where the current is maximally constricted). (iv) Give an expression for the sheet surface charge density (units Coulomb/m) along the
edge of the hole. . I . 3. = 'L‘la In calindrical coordinates WtUn :9, O,
‘ L . ‘  Letting 31‘: 20') Rat); ’ves R3? ‘ ' u: J¢z I a" c. We Have two AWecential eQuaﬁons . ., 31, F; + n‘F‘4, = o (harmonic «2%. Wrist» solqtion )._
=¢=.c3¢os'nd> + C4sﬁnnd) ' 4H] _ _
 ' ' dz 'JPA [c.rnr"'+c,,r(n)e _ .
Also ésww  a a? ., . tEhzc.r"+ ht. 50 given ‘For'ms u"; Wow1:5: ‘ I Illa (.i) we“) #M)=Va(I—3ﬁ‘_—), 0<x<‘% %
=V93é—I), géxcL .
__.;_ o % L 3%; ¥wiscven30 b,.=o {’Oralln 1
4(1)=o.o+,.§ancos3='—EEZ‘ L ‘
an = t Mum = afﬁx)“ = 2.;3fvm ELL) dx = N5 g
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~11“): 0.. + imam”? + mam”? )dx
(3.. = t fan)“ = t3 vaIL—Ax — 3% M = Zargmwwsxax =%<:%)‘J:2n—Lumz%lae—%x—) = +£auaw§xﬁ= 2X1m=<ww —w ..I All 0m=0 except ﬁlo48) isodd except {70.— a level surPt o1: 5%)
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= 41—?)me = _ __ i‘”: [it EJ521153)
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\ I / Iao=kjiﬂka1dx= “f?
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5%,1QIICI4) wzun c» V0 and o.>b~, 9(5):" Eguabing tine M7“ term; 4,95: an CO‘hD%a’  o MELFE" ‘ Kg
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SQ—g‘fﬂ‘éjfﬁddnxw .. I I
' @rur ﬂ ‘.' ' : ~ ~. I
= g (’7? ' 5/0,?) was; ) w my): gm? ) . a
' Ir l' I?" 7Lg~u(a,'7) CFC: . 'HZb it 3 .av, '5‘” ,
§ § Obtain “total sduﬁon as §= §.+§; when: £13
I‘=() ‘=O
so‘ubion 09 Rob. 'Ilza— and E is soltglﬁon 09 POHern shown
“" a: °' in 4.6%. Far 7.5;, need repeated. zeros in x direction . I §m = (Aeoskx +Bsuhlgx)(c.coshk5 +Dsu'nh kg)
_B_.__C__. Potarb'al 53mmetﬁc in 7‘ so 830. §=o at x =10.
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§=o is“ it" =' (A (:05ka + Bsa'm kx)CCwosh L5 +‘ 195th us)
Since §(0.5)=o'and §Cx,o)=o,‘~A=.:c.=.o ’
O '5 i=0 5 cas‘mg'xsﬂﬂ I m:' N
 ‘%L»?k,¢"_°°‘k“s‘f"hk5=9. Thus lax3:; (VI 066)
:.§ =:..§°”$3”§'§*5‘”“€% . .
To expand b,c~. ox ab; (52 _ .1ncm with 0.929. and.
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 Fall '08
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 Electromagnet, Electric charge, Van Duzer, j S, Ramo, c» shah

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