Page 205 - Engineering Electromagnetics, 8th Edition
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CHAPTER 7   The Steady Magnetic Field         187

                         Equation (9) may be used to find the magnetic field intensity caused by current
                     filaments arranged as a sequence of straight-line segments.


                                                                                            EXAMPLE 7.1
                     As a numerical example illustrating the use of (9), we determine H at P 2 (0.4, 0.3, 0)
                     in the field of an 8. A filamentary current is directed inward from infinity to the origin
                     on the positive x axis, and then outward to infinity along the y axis. This arrangement
                     is shown in Figure 7.6.

                     Solution. We first consider the semi-infinite current on the x axis, identifying the
                                                    −1
                     two angles, α 1x =−90 and α 2x = tan (0.4/0.3) = 53.1 . The radial distance ρ is
                                        ◦
                                                                   ◦
                     measured from the x axis, and we have ρ x = 0.3. Thus, this contribution to H 2 is
                                         8                    2           12
                                                   ◦
                               H 2(x) =     (sin 53.1 + 1)a φ =  (1.8)a φ =  a φ
                                      4π(0.3)                0.3π         π
                     The unit vector a φ must also be referred to the x axis. We see that it becomes −a z .
                     Therefore,
                                                       12
                                              H 2(x) =−  a z A/m
                                                       π
                                                             −1
                     For the current on the y axis, we have α 1y =− tan (0.3/0.4) =−36.9 , α 2y = 90 ,
                                                                                       ◦
                                                                              ◦
                     and ρ y = 0.4. It follows that
                                           8                        8
                                                         ◦
                                  H 2(y) =     (1 + sin 36.9 )(−a z ) =− a z A/m
                                         4π(0.4)                    π






















                                 Figure 7.6 The individual fields of two semi-infinite
                                 current segments are found by (9) and added to obtain
                                 H 2 at P 2 .
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