Page 254 - Engineering Electromagnetics, 8th Edition
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236                ENGINEERING ELECTROMAGNETICS

                                        D8.3. The semiconductor sample shown in Figure 8.1 is n-type silicon, hav-
                                        ing a rectangular cross section of 0.9 mm by 1.1 cm and a length of 1.3 cm.
                                                                                       2
                                        Assume the electron and hole mobilities are 0.13 and 0.03 m /V · s, respectively,
                                        at the operating temperature. Let B = 0.07 T and the electric field intensity in
                                        the direction of the current flow be 800 V/m. Find the magnitude of: (a) the
                                        voltage across the sample length; (b) the drift velocity; (c) the transverse force
                                        per coulomb of moving charge caused by B;(d) the transverse electric field
                                        intensity; (e) the Hall voltage.

                                        Ans. 10.40 V; 104.0 m/s; 7.28 N/C; 7.28 V/m; 80.1 mV


                                     8.3 FORCE BETWEEN DIFFERENTIAL
                                            CURRENT ELEMENTS
                                     The concept of the magnetic field was introduced to break into two parts the problem
                                     of finding the interaction of one current distribution on a second current distribution.
                                     It is possible to express the force on one current element directly in terms of a
                                     second current element without finding the magnetic field. Because we claimed that
                                     the magnetic-field concept simplifies our work, it then behooves us to show that
                                     avoidance of this intermediate step leads to more complicated expressions.
                                        The magnetic field at point 2 due to a current element at point 1 was found to be

                                                                    I 1 dL 1 × a R12
                                                              dH 2 =
                                                                           2
                                                                       4πR 12
                                     Now, the differential force on a differential current element is

                                                                dF = IdL × B

                                     and we apply this to our problem by letting B be dB 2 (the differential flux density at
                                     point2causedbycurrentelement1),byidentifying IdLas I 2 dL 2 ,andbysymbolizing
                                     the differential amount of our differential force on element 2 as d(dF 2 ):

                                                             d(dF 2 ) = I 2 dL 2 × dB 2

                                     Because dB 2 = µ 0 dH 2 ,we obtain the force between two differential current
                                     elements,

                                                                  I 1 I 2
                                                      d(dF 2 ) = µ 0  dL 2 × (dL 1 × a R12 )         (13)
                                                                     2
                                                                 4πR 12
                   EXAMPLE 8.2
                                     As an example that illustrates the use (and misuse) of these results, consider the
                                     two differential current elements shown in Figure 8.3. We seek the differential force
                                     on dL 2 .
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