Electromagnetic Field Theory: A Problem Solving Approach Part 1 ppsx

... the field plots.Markus ZahnCartesianCoordinates(x, y, z)af. af afVf = -, + Oi, + i,ax ay OzaA, aA, aA,V . A= a+ +ax ay az(LAX, A AaA\ .a, OaA.ay z az Ox ... If -14 ar r •O r sin 0 a O A 1 (r 1 a( sin OAo) 1 oA*V" -A= (rPA,)+ +r ar r sin 0 ae r sin 0 a4 x 1 a( sin OAs) aA]r sin a8 0 04, a S 1 MA, a( rA,)) 1 [ra(rAo) ... aA. A, DA A. I(rAs) A r 1 xzazrr) 'L ar a4 JV~ f a0 l af\ 1 82f a 2f r-+ -2rr Ora r) 14 azSphericalCoordinates (r, 0, 4) a. af. 1 af.Vf= ,+ ae+I If -14 ar...

Electromagnetic Field Theory: A Problem Solving Approach Part 5 ppsx

... Section 1- 3 -1 and thevector A: aAx aA, aA,div A = V - A =-+ + (5)ax ay az 1- 4-3 Curvilinear CoordinatesIn cylindrical and spherical coordinates, the divergenceoperation ... becomes infinitesimally small, each of the bracketedterms defines a partial derivativeaA 3A ýAMzlim =( + + A V (4)ax-o \ax ay azwhere AV = Ax Ay Az is the volume ... originate at a source and terminate at a sink. 1- 4 .1 FluxWe are illustrating with a fluid analogy what is called theflux 4 of a vector A through a closed surface:D= f AdS...

Electromagnetic Field Theory: A Problem Solving Approach Part 10 ppsx

... incremental line charge alone has a radial field component as given by (5) that in Cartesian coordinatesresults in x and y components. Consider the line charge dA 1 , a distance ... total charge on the hoop is q = 27taXo so that (14 ) canalso be written asqzE.= 4reo (a +z )3/2 (15 )When we get far away from the hoop (IzI > a) , the field approaches ... givesrise to a field in the positive y direction:EPo yya (11 )E fa PO ,= poaa y > a (11 ) -a 2Eo EoFor any position y in region II, where -a y 5 a, the chargeto...

Electromagnetic Field Theory: A Problem Solving Approach Part 16 ppsx

... Probkms 12 943. A point charge q is placed between two parallel groundedconducting planes a distance d apart. (a) The point charge q a distance a above the lower planeand ... formPn =AAn(i) Show that the capacitance is of the formC _2 1 A A 2What are Co, 6, and A? (j) What is the capacitance when the two spheres areconcentric so that D ... Try power-law solutionsq, = AA", b, = Baand find the characteristic values of A and a that satisfy theequations in (d).(f) Taking a linear combination of the...

Electromagnetic Field Theory: A Problem Solving Approach Part 28 ppsx

... A capacitor has a moveable part that can rotate throughthe angle 0 so that the capacitance C(O) depends on 0. (a) What is the torque on the moveable part? (b) An electrostatic ... charge -Q at radius R2.48. A capacitor C is charged to a voltage V0.At t = 0 anotherinitially uncharged capacitor of equal capacitance C isSwitch closes at t = ... Strength of IrradiatedPolythene, Nature 17 6 (19 55) 12 25-26.]52. An electret with permanent polarization Poi, and thick-ness d partially fills a free space capacitor. There...

Electromagnetic Field Theory: A Problem Solving Approach Part 40 ppsx

... 0ik (20)5-8 MAGNETIC FIELDS AND FORCES5-8 -1 Magnetizable Media A magnetizable medium carrying a free current J 1 is placedwithin a magnetic field B, which is a function of ... Ax)-B(x).AxAY 1 Ax AxB.(y +Ay)-Bz(y) .B,(y +Ay)-B(y) ] (2)+ AY y A (2)Ay AyBYtkx, yl m = iA x Ayi,xFigure 5-26 A magnetic dipole in a magnetic field B.I•374 ... themagnetic dipole moment m = i Ax Ay iý:lim f = m[LB +-aB, + B,, (3) A O xLax x y iyAmpere's and Gauss's law for the magnetic field relate thefield...

Electromagnetic Field Theory: A Problem Solving Approach Part 57 ppsx

... situation where e1 = E2and P 1 #I 2:sin2 B= 1 tan O8 = (15 ) 1 +A/ II 2 A1 At this Brewster's angle, the reflected and transmitted powerflows are at right angles ... critical angle of (17 ), as can be particularlyseen when A =-L2 for the magnetic field polarized parallel tothe interface or when 81 = e2 for the electric field polarizedparallel ... directionparallel to the interfacial boundary and it varies as a functionof z.7-8-2 H Field Parallel to the InterfaceIf the H field is parallel to the conducting boundary, as...

Electromagnetic Field Theory: A Problem Solving Approach Part 62 ppsx

... constants 7= ZoCL and 7 = LL/Zo as thesolutions approach the dc steady state. The waveforms begin after the initial V. wavearrives at z = Iafter a time T= 1/ c. There ... formV+.= AA" (18 )which when substituted into (17 ) requiresAA"-' (A -F, FL) = 0= )A = rjFL (19 )The total solution is then a sum of the particular andhomogeneous ... = 0 as a new V+ wave given by (11 )again returns to z = I.(iv) Thus, the voltage at z = 1 only changes at times (2n- 1) T, n = 1, 2, , while the voltage at z...

Electromagnetic Field Theory: A Problem Solving Approach Part 72 ppsx

... Si(x) = 1. 5708x lim Si(x) - x.2 I x-O 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Figure 9 -13 The sine integral Si(x) increases linearly for small arguments andapproaches ir/2 ... for large arguments oscillating about this value for intermediatearguments.21R /17 L,2 ( L 1 2 3kLFigure 9 -14 The radiation resistance for a dipole antenna carrying a uniformlydistributed ... Figure 9-7e.(c) ArbitraryCurrent PhaseFor arbitrary current phase angles and dipole spacings, a great variety of radiation patterns can be obtained, as illus-trated by the...

Electromagnetic Field Theory: A Problem Solving Approach Part 75 pdf

... diode, 19 9Poisson-Boltzmann equation, 15 7Polariscope, 518 -520Polarizability, 14 3 -14 4and dielectric constant, 14 7Polarization:boundary conditions, 16 5 -16 6charge, 14 0 -14 2, 14 9cylinder, ... 14 9cylinder, 16 6 -16 8and displacement field, 14 6 -14 7electronic, 13 6force density, 215 - 219 ionic, 13 6orientational, 13 6in parallel plate capacitor, 17 6 -17 7by reflection, 546-547spontaneous, ... 546-547spontaneous, 14 9 -15 1of waves, 514 - 516 Polarizers, 517 -520Polarizing angle, 547Polar molecule, 13 6 -13 7Polar solutions to Laplace's equation,2 71- 272Potential:energy, 19 9retarded,...

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