SPECIAL Electromagnetic Field Analysis and Its Applications to Product Development Tomohiro KEISHI Electromagnetic field analysis, one of the numerical analysis, is now an indispensable method for designing and de - veloping electromagnetic application products. Such advanced analysis techniques including finite element methods and faster, higher-capacity analytical hardware, such as personal computers, enable even the most complex electro - magnetic phenomena to be investigated. Depending on the frequency of an object, an analysis is carried out differently; products with high frequency must be analyzed in the electromagnetic field, while products with relatively low frequency can be studied in the either field: the electric or magnetic field. This paper describes the outline and purposes of the electromagnetic field analysis, introducing some examples of the experiment. Keywords: electromagnetic field analysis, electric field analysis, magnetic field analysis, finite element method, product design q 1. Introduction div E = — …………………………………………(3) e0 Numerical electromagnetic field analysis has become div B = 0 ………………………………………………(4 ) an essential tool for the design and development of elec - tromagnetic products (1)-(16) . Advances in numerical analy - Here E is the electric field, B magnetic flux density, sis techniques, including the finite element method and i current density, t time, q charge density, e0 permittivity in high-speed, high-capacity analytical hardware such as the a vacuum, and µ0 permeability in a vacuum. personal computer have made it possible to investigate Within a substance of permittivity e and permeability numerically even the most complex electromagnetic phe - µ, the following equations hold: nomena. The frequencies involved in the operation of a product determine how the analysis must be carried out: D = eE…………………………………………………(5 ) in some cases, the electric field and magnetic field may B = µH ………………………………………………(6 ) be studied separately, as in electric field analysis and mag - netic field analysis; while in other cases it is necessary to Hence, Equations (2) and (3) can be represented as study both the electric and magnetic fields simultane - Equations (2)’ and (3)’ (17) . ously, as in electromagnetic field analysis. This paper out - ∂D lines and describes the purpose of electromagnetic field rot H =i + — ……………………………………(2)’ ∂t analysis, and presents examples of product design and de - velopment. The future outlook for electromagnetic field div D =q ………………………………………………(3)’ analysis is also discussed. Here, D is the electric flux density, and H is the mag - netic field. If the divergence of both sides of Equation (2)’ is 2. Electromagnetic Field Analysis taken, div(rot H)=0, and taking Equation (3)’ into ac - count, the law of conservation of charge (Equation (7)) 2-1 Fundamental Equations of Electromagnetic Field may be deduced. Analysis ∂q The electromagnetic fields E and B that are pro - div i =− — …………………………………………(7) ∂t duced by the electric current i and the electric charge q can be represented using Maxwell’s equations (1)-(4), When dealing with a phenomenon in the high-fre - where Equation (1) is Faraday’s law of electromagnetic quency range, it is necessary to solve an electromagnetic induction, Equation (2) is Ampere’s law with Maxwell’s field problem involving electric and magnetic fields at the correction, Equation (3) is Gauss’s law, and Equation (4) same time. In this case, Maxwell’s equations (1)-(4) or is Gauss’s law for magnetism (17) . (1), (2)’, (3)’ and (4) must be solved. When dealing with a phenomenon in the low-fre - ∂B rot E =− — …………………………………………(1) quency range, on the other hand, it is often sufficient to ∂t solve either an electric field problem or a magnetic field ∂E rot B =µ0i+e0µ0 — …………………………………(2) problem. In this case, it is necessary to solve a quasi-static ∂t electromagnetic field equation relating to either the elec - 4 · Electromagnetic Field Analysis and Its Applications to Product Development tric field or magnetic field (see Table 1 ). satisfied. In the case of an electrostatic field problem, B’s When dealing with the electric field only, Equations time derivative is zero and rot E = 0. Since rot H = i and (3)’, (7), and (1)’ in Table 1 (17) may be solved. 1 1 From Equation (1)’, the relation between E and elec - H = —µ B = —µ ro tA, all that remains is to find the A that tric potential V can be defined in Equation (8). satisfies Equation (13). E = – grad V …………………………………………(8) 1 …………………………………… rot (—µ rot A) = i (13) If electrical conductivity s or resistivity r are used, the relation between i and E is represented by Equation (9). In the case of a quasi-steady magnetic field problem, or an eddy current problem, both electric scalar potential i = E = —E ………………………………………(9) ø and magnetic vector potential A are introduced to rep - s r resent the electric field E and magnetic flux density B in When dealing with the magnetic field only, Equa - Equations (14) and (15), respectively. tions (2)”, (1), and (4) in Table 1 (17) may be solved. A From Equation (4), the relation between B and vec - E = – gradø – —∂ …………………………………(14) ∂t tor potential A can be defined by Equation (10). B = ro tA ……………………………………………(15) B = rot A ……………………………………………(10) If Equations (14) and (15) are used, Equation (2)”can be represented in Equation (16), with Equations (6) and (9) also being taken into account. Table 1. Maxwell’s equations (Quasi-static electromagnetic field equations) 1 A rot — rot A = s – gradø – —∂ ………………(16) The electric field system The magnetic field system ( µ ) ( ∂t ) divD ＝q (3)’ rotH ＝ i (2)” Since div(rot H)=0, Equation (17) holds. ∂q ∂B divi ＝- (7) rotE ＝- (1) A ∂t ∂t di v s – gradø – —∂ =0 …………………………(17) ( ∂t ) rotE ＝0 (1)’ divB ＝0 (4) { } E ＝-gradV (8) B ＝rotA (10) D ＝εE (5) B ＝µH (6) If Equations (16) and (17) are solved, a solution to E E the eddy current problem may be attained. i ＝σEE＝ρ (9) i ＝σ ＝ρ (9) 3. Electric Field Analysis 2-2 Electromagnetic Field Analysis Using Numerical Analysis 3-1 Purpose of Electric Field Analysis Several numerical techniques have been developed Cables and coils are often used in electric power trans - for electromagnetic field analysis. Among these, one of mission circuits. High voltage is applied between internal the most frequently chosen is the finite element method, and external conductors of the cable and between the coil as it is highly versatile and thus applicable to most prob - conductor and the surface of an insulator, thereby placing lems (18)-(20) . a high electric field stress on the insulator. When the in - As shown in Table 1 , in the case of equations relating sulator is subject to excessive electric field stress, dielectric to the electric field, if the scalar potential V which satisfies breakdown occurs, terminating the transmission of elec - E = – grad V is introduced, rot E = 0 is identically satisfied. tric power. It is therefore necessary to determine the elec - In the case of an electrostatic field problem, the time de - tric field accurately so as to design electrical insulation to rivative of charge density q is zero. Since div i = 0, it is not include sufficient tolerance, avoiding excessive electric necessary to solve this equation. Since div D = q and D =eE fields across the insulator. If, like cables, the insulator’s = –e grad V , all that remains is to find the V that satisfies cross section is circular, it is possible to determine the elec - the Poisson’s equation (11). tric field through analysis. Because most insulators have a complex shape and more than one type of electrical insu - div( e grad V) = – q……………………………………(11) lator is used, however, it is necessary to conduct electric field analysis to determine the electric field. If q = 0 , all that remains is to solve the Laplace’s equation (12). 3-2 Element Discretization to Determine the Maxi - mum Electric Field div( e grad V) = 0 ……………………………………(12) The finite element method was used for electric field analysis of a parallel plate electrode with a spheroid at its As shown in Table 1 , in the case of equations relating top (see Fig. 1 ) to determine the maximum electric field to the magnetic field, if the magnetic vector potential A at the tip of the projection. The distance between the par - which satisfies B = rot A is introduced, div B = 0 is identically allel plates is 3 mm, and a voltage of 50 Hz, 38 kV is ap - SEI TECHNICAL REVIEW · NUMBER 69 · OCTOBER 2009 · 5 plied. The major axis radius of the spheroid is 1,325 µm, ducted by alternating finite element discretization near and the radius of curvature at the tip is 50 µm. As shown the tip of the projection. Shown in Fig. 3 is how the max - in Fig. 2 (a), (b), and (c) , electric field analysis was con - imum electric field at the tip of the projection relates to the thickness of the element. If these three points are used to determine a quadratic regression line to approx - imate the maximum electric field, it is assumed to be High-voltage electrode (38kV applied) 238.6 kV/mm (element thickness = 0 mm), with a theo - retical value of 236.4 kV/mm (21) . With an error ratio of less than 1%, the value estimated from the finite element method and the theoretical value are in close agreement. Crosslinked polyethylene insulator In this way, it is necessary to cut thin slices of the ele - (3mm thick, relative permittivity 2.3) 1325µm 38kV (50Hz) voltage applied ment at the point where the maximum electric field is cal - 50µm culated. The thickness of the element must be determined Earth Electrode (0V) so that the required accuracy for the maximum electric field is achieved.