Electromagnetic Field Theory: A Problem Solving Approach Part 36. Electromagnetic field theory is often the least popular course in the electrical engineering curriculum. Heavy reliance on vector and integral calculus can obscure physical phenomena so that the student becomes bogged down in the mathematics and loses sight of the applications. This book instills problem solving confidence by teaching through the use of a large number of worked problems. To keep the subject exciting, many of these problems are based on physical processes, devices, and models. This text is an introductory treatment on the junior level for a two-semester electrical engineering. | Magnetic Field Due to Currents 325 with distance rQP 22 r2 1 2 6 The magnetic field due to this current element is given by 4 as _ Mo Zi dz it xiQp _ MoArdz . aa a 2 2 . 2 fr 4tt rQp 4ir z r The total magnetic field from the line current is obtained by integrating the contributions from all elements Hol it f dz . 2 2 4 r J- z r z 4tt r2 z2 r2 l 2lI -0O 8 Ttr If a second line current 1 of finite length L is placed at a distance a and parallel to I as in Figure 5-86 the force on 12 due to the magnetic field of It is i L 2 Z2 dz it x B -C 2 i L 2 r L 2 Sira - 9 rra If both currents flow in the same direction ZiZ2 0 the force is attractive while if they flow in opposite directions ZiZ2 0 the force is repulsive. This is opposite in sense to the Coulombic force where opposite charges attract and like charges repel. 5-2-3 Current Sheets a Single Sheet of Surface Current A constant current X0i2 flows in the y 0 plane as in Figure 5-9a. We break the sheet into incremental line currents Ko dx each of which gives rise to a magnetic field as given by 8 . From Table 1-2 the unit vector is equivalent to the Cartesian components 4 sin f ix cos 6ij 10 326 The Magnetic Field Figure 5-9 a A uniform surface current of infinite extent generates a uniform magnetic field oppositely directed on each side of the sheet. The magnetic field is perpendicular to the surface current but parallel to the plane of the sheet. 6 The magnetic field due to a slab of volume current is found by superimposing the fields due to incremental surface currents c Two parallel but oppositely directed surface current sheets have fields that add in the region between the sheets but cancel outside the sheet d The force on a current sheet is due to the average field on each side of the sheet as found by modeling the sheet as a uniform volume current distributed over an infinitesimal thickness A. Magnetic Field Due to Currents 327 lim Joû Ko Jo- A-0 B B I .
