Page 204 - Engineering Electromagnetics, 8th Edition
P. 204

186                ENGINEERING ELECTROMAGNETICS

                                     been adjusted so that the addition of this second set of lines will produce an array of
                                     curvilinear squares.
                                        A comparison of Figure 7.4 with the map of the electric field about an infinite
                                     line charge shows that the streamlines of the magnetic field correspond exactly to
                                     the equipotentials of the electric field, and the unnamed (and undrawn) perpendicular
                                     family of lines in the magnetic field corresponds to the streamlines of the electric
                                     field. This correspondence is not an accident, but there are several other concepts
                                     which must be mastered before the analogy between electric and magnetic fields can
                                     be explored more thoroughly.
                                        Using the Biot-Savart law to find H is in many respects similar to the use of
                                     Coulomb’s law to find E. Each requires the determination of a moderately complicated
                                     integrand containing vector quantities, followed by an integration. When we were
                                     concerned with Coulomb’s law we solved a number of examples, including the fields
                                     of the point charge, line charge, and sheet of charge. The law of Biot-Savart can be
                                     used to solve analogous problems in magnetic fields, and some of these problems
                                     appear as exercises at the end of the chapter rather than as examples here.
                                        One useful result is the field of the finite-length current element, shown in
                                     Figure 7.5. It turns out (see Problem 7.8 at the end of the chapter) that H is most
                                     easily expressed in terms of the angles α 1 and α 2 ,as identified in the figure. The
                                     result is
                                                                I
                                                          H =     (sin α 2 − sin α 1 )a φ             (9)
                                                               4πρ
                                     If one or both ends are below point 2, then α 1 is or both α 1 and α 2 are negative.



























                                                     Figure 7.5 The magnetic field intensity
                                                     caused by a finite-length current filament
                                                     on the z axis is (I/4πρ)(sin α 2 − sin α 1 )a φ .
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