Page 412 - Engineering Electromagnetics, 8th Edition
P. 412

394                ENGINEERING ELECTROMAGNETICS

                                     The time-average power loss is easily obtained, since the average value of the cosine-
                                     squared factor is one-half,

                                                                     1
                                                                         2
                                                                P L =  J bLδ                         (89)
                                                                        x0
                                                                     4σ
                                        Comparing (88) and (89), we see that they are identical. Thus the average power
                                     loss in a conductor with skin effect present may be calculated by assuming that the
                                     total current is distributed uniformly in one skin depth. In terms of resistance, we may
                                     say that the resistance of a width b and length L of an infinitely thick slab with skin
                                     effect is the same as the resistance of a rectangular slab of width b, length L, and
                                     thickness δ without skin effect, or with uniform current distribution.
                                        We may apply this to a conductor of circular cross section with little error,
                                     provided that the radius a is much greater than the skin depth. The resistance at
                                     a high frequency where there is a well-developed skin effect is therefore found by
                                     considering a slab of width equal to the circumference 2πa and thickness δ. Hence

                                                                   L       L
                                                               R =    =                              (90)
                                                                   σ S  2πaσδ
                                     A round copper wire of 1 mm radius and 1 km length has a resistance at direct
                                     current of
                                                                    10 3
                                                        R dc =               = 5.48
                                                                           7
                                                              π10 (5.8 × 10 )
                                                                 −6
                                     At 1 MHz, the skin depth is 0.066 mm. Thus δ 	 a, and the resistance at 1 MHz is
                                     found by (90),
                                                                   10 3
                                                  R =                              = 41.5
                                                                                −3
                                                                    7
                                                       2π10 (5.8 × 10 )(0.066 × 10 )
                                                           −3
                                        D11.7. A steel pipe is constructed of a material for which µ r = 180 and
                                                  6
                                        σ = 4 × 10 S/m. The two radii are 5 and 7 mm, and the length is 75 m. If
                                        the total current I(t) carried by the pipe is 8 cos ωt A, where ω = 1200π rad/s,
                                        find: (a) the skin depth; (b) the effective resistance; (c) the dc resistance; (d)
                                        the time-average power loss.

                                        Ans. 0.766 mm; 0.557 
; 0.249 
; 17.82 W

                                     11.5 WAVE POLARIZATION

                                     In the previous sections, we have treated uniform plane waves in which the electric
                                     and magnetic field vectors were assumed to lie in fixed directions. Specifically, with
                                     the wave propagating along the z axis, E was taken to lie along x, which then required
                                     H to lie along y. This orthogonal relationship between E, H, and S is always true for
                                     a uniform plane wave. The directions of E and H within the plane perpendicular to a z
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