Page 183 - Electromagnetics
P. 183

and by integration

                                             I 2         ∞     y 2         2  µ 0
                                     F 1 = µ 0  2  ˆ x  dz  2    2  2  dy = I   ˆ x  dz.
                                             2π        −∞ [y + d /4]        2πd
                        The resulting force per unit length agrees with (3.175)when I 1 = I 2 = I.

                        Torque in a magnetostatic field.  The torque exerted on a current-carrying conduc-
                        tor immersed in a magnetic field plays an important role in many engineering applica-
                        tions. If a rigid body is exposed to a force field of volume density dF(r), the torque on
                        that body about a certain origin is given by

                                                      T =    r × dF dV                        (3.177)
                                                           V
                        where integration is performed over the body and r extends from the origin of torque.
                        If the force arises from the interaction of a current with a magnetostatic field, then
                        dF = J × B and

                                                    T =    r × (J × B) dV.                    (3.178)
                                                         V
                        For a line current we can replace J dV with Idl to obtain


                                                     T = I   r × (dl × B).

                          If B is uniform then by (B.7)we have

                                                 T =   [J(r · B) − B(r · J)] dV.
                                                      V
                        The second term can be written as


                                                               3

                                                 B(r · J) dV = B    x i J i dV = 0
                                               V               i=1  V
                        where (x 1 , x 2 , x 3 ) = (x, y, z), and where we have employed (3.27). Thus
                                                   3         3           3     3


                               T =    J(r · B) dV =  ˆ x j  J j  x i B i dV =  B i  ˆ x j  J j x i dV.
                                    V             j=1   V   i=1          i=1  j=1   V
                        We can replace the integral using (3.144)to get
                                            3
                                                  3
                                       1     
   
                      1
                                   T =         ˆ x j  B i [x i J j − x j J i ] dV =−  B × (r × J) dV.
                                       2  V  j=1  i=1                   2  V
                        Since B is uniform we have, by (3.145),

                                                         T = m × B                            (3.179)
                        where m is the dipole moment. For a planar loop we can use (3.147)to obtain

                                                        T = IAˆ n × B.




                        © 2001 by CRC Press LLC
   178   179   180   181   182   183   184   185   186   187   188