PHYS 351 Final Exam 2008

PHYS 351 Final Exam 2008 - 1. (25 points) A very long wire...

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Unformatted text preview: 1. (25 points) A very long wire of radius R is carrying a current density a? = a (R2 — 32) it}, Where a is a constant and s is the cylindrical coordinate that measures the distance fiom the z-aacis. (a) What is the total current carried by this Wire? @ (b) Determine fi-field at the point (x = g, y = 523, z = 0). 2. (25 points) ' ‘ “529 g :32; (w) 4%.) The charge density on the surface of a sphere of radius R is given by cr = an 0082 6. (a) Find the electric potential inside and out side the sphere. g g 1-13- I. 3 anan‘acm V:..=V..,+lr-gw VM: 5» + E Maw) 0-013sz aim-5:5 T r3 , 60 am. _§gfi]=%% 916,550) + J3. a; ROM) 31- Y R 64:20,: PZQM) + £2.03 + 3% 1915000)] :- 0;[3FLCW) 3 o 20'; in " Gigi.J5-, 20334-3023 =%%i =v a}: m 0 2' 1.. “' TR 1—91. Ease-4 Vain fi+‘seok (1 AL) -U'Rz 2079!?“ 3 1.9") vow“ aer+ Ibeofiiew’ " ‘ "" d zrof 6570"; D = -' VVI“\(F/I-l "/11 ’7‘ = V 3] A _ _“ 4431‘ 3 m‘9_ IQ [filer £3036 snag _o n... Y' _ .- ilse,k(" 21)] r lam;~ 08m i”) 1? ,. A {‘7' R” U‘r "'" R Tr): o = O E(1'r£‘); +12 .55.? y we. 7 I, A A " '-" TL -l- 3o + -'- J] x3 r1 5 3. (20 points) Using boundary conditions: A planar sheet (2 = O) carries uniform surface charge density or and a steady surface current K = K053. (a) If Fields E" and B; on two sides (very near the surface) are given by Etc}, 2 Emir. + Elygj + E122; EDP 2 Bus + 313,3": + B132“: and Ebottom = Egmfi + Egyz? + E292; 3.130th = ngfi + Bgyé + Bgyé, respectively. Determine relations between components of fields Etop, Btop and Ebottom, Bbomm. (b) Potential inside a spherical cavity (in a conductor) of radius R is given by V[r, I9) = jag—£0 f5 (1 — {3-35) cos 9. y 3, aging _cr 55mm E” 9 ominous-M anal-345,: /‘0 a art'lji; @ Determine the surface charge density and the total charge on the inner surface of the cavity. W k B. 4. (15 points) Show that Maxwell’s equations in free space, i.e., in absence of charge and current deneities, lead to the following two wave equations: 6. (25 points) From Coulomb’s Law, the potential due to a uniformly charged ring of radius 3% and charge g, at a point on the axis (a distance z from the center of the ring) is given by _ q 1 V(0101z) _ 41r60(3%2+z2)1/2. '(a) For large values of z, expand the above potential in powers of %; (1)) Compare the above expansion with 2:0 F355 Pdcos 9) evaluated on the z-axis, and determine first three non-zero coefficients by. Hint: 133(1) = 1 for every 3. (0) Write first three non-vanishing terms for the potential at (33, R, 1, I Q 7- "/2 Won’t) =' ( "’3'. ) 417603 a =_L_(.L_J%z1-§—3+- ) 417750 3 25 q 00 30 z. _‘_3_ _ J— .. .. .3; _ C536" 43 g V(I§R’9-tau'\|i:¢) 4mk I 7 Sim: S'ow o—j‘flu‘hm Vawisk, in n’nifie 01: Stu-Md aka" WWA—vak M1" is T5 3d” Wm, Howgvg~,1 w)" «gab-f4“; on {Anya}. 34. Mfl-VWW‘ mi». 5. (25 points) 555$ = amt-H 2... A very long conducting cylinder of radius 3:21 is surrounded by a charged thin cylindrical shell of radius éRg. The charge density on the shell is given by :70 + 0'1 cos2 qt. The system is z—independent. Pol-mHfi-l 4"-N-J- 5‘- a. (a) Write down general solutions valid for following regions: a. 3% < r< 3‘22 (region II) (3 points) HM 7" "' r b. 3'32 < r (region III) (4 points) a“. (b) Determine undetermined constants of part a) using continuity of potential and the diSCOntinuity of the normal component of E field. (10 points) {c} Determine the charge density on the conducting cylingr.(8 points) Q i) VI: = (d.°+boinr)+ 2(anv”+h.f")cg1n15 "W n —“ n 3'.) Va :- (cudobr) ... '1 «relay-m) a: ‘5. a.) TM charge on Mt l-cwfih = CL: 5(a-,+cr,oa‘¢a) a“ a: Mag-r“? R2) = "ghee-r) .. d 2‘ H Madge-OT) ‘_ __L Ez(2d';,+0"|) -. o .— Efia 2' {-0 W2 Claws: Cur—o 510%,»th :: ~37éo$nkz -'2. a, a: + b, 2?: 4,, 2t ...
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PHYS 351 Final Exam 2008 - 1. (25 points) A very long wire...

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