An iron ring of circular cross sectional area Of 3.0 cm? and mean diameter of 20cm is wound with 500 turns of wire and carries a current of 2.09 A to produce the magnetic flux of 0.5 m Wh in the ring. Determine the permeability of the material. Solution : The given values are : 2 = 3cm’=3x 10m, d= 20cm, N=500, I= 2A, $ = 05mWb Now, 1 = nx d= mx 20 = 628318 cm = 0.628318 m ue 0.628313 _ 1.6667 x 10? a HoRra = 4nx 107 xp, x3 x 104 By f= mmf _NI ss NI __500%2 9 x 105 aT / Wb w Q) % 05x 103 Equating equation (1) and equation (2), 9 1.6667 x10? sy ass.ga4 Br 2x 10° = Sanned with CamscannerAn iron ring 8 cm. mean diameter is made up of round iron of diame Lem and permeability of 900, has an air gap of 2 mm wide. It consists of winding w 400 turns carrying a current of 3.5 A. Determine, i) mmf. éi) Total reluctance iii) The flux iv) Flux density in ring. Solution : The ring and the winding is shown in the Fig. 1.185. Diameter of ring d = 8 cm, ” Length of iron = md- Length of air gap l= mx 8x107?)-2*10% amt ae gop Fig. 1.18.5 = 0.2493 m. While calculating iron length do’ not forget to subtract length of air ¢ total mean length. = Ue es I, = Length of air gap = 2mm = 2x10 m Diameter of iron = 1 cm Tt -2\2 52 = Area of cross section a=5@ = F(xto 2)° = 7:853x10~° m gap and ring is to be assumed same. Area of cross section of air i) Total mmf. produced = Ni = 400x3.5 = 1400 AT (ampere turns) ii) Total reluctance S; = S; +S L : Ss = —i— ... Given p, = 900 t pobra us 5 a = 2806947.615 AT/ Wb 4nx10~7 x900x7.853x10™ Sanned with Camscanner'y R Hoa n a an pty 1 for air 2x10" An? x7.853 «10-5 a a = 20.2667 x10 AT / Wb . Sp = 2806047.615 +20.2667 x10 = 23.0797 «10% AT Wb ¢ = mmf NI 1400 5 = — = a = ————_ = 6.067 x10 Wb Reluctance “ Sp ~ 23.0737 x10 % _ 6067 x10°S a 7853x105 ii) iv)Flux density = = 0.7725 Wb / m= [ example 1.18.3 BN magnetic circuit is excited by three coils as shown in the Fig. 1.19.6, Calculate the flux produced in the air gap. The material used for core is iron having relative permeability of 800. The length of the magnetic circuit is 100 cm with an air gap of 1, 2 mm in it. The core has uniform Ne=100 4 EE Fig. 1.18.6 Solution : Given, , = 600, L=6A, N,=100, 1=5A N, = 800, =1A, 1p=100cm=1m 1m-2x 10° = 0.998 m : B = 2x1073 m, Ht, = 800,a=6 cm>=6x 104m Now total reluctance S = S, + S, 0.998 4nx10~7 x800x6x107* = 1654548.263 AT/Wb s, = —& =__“""__ = 659582.385 AT/Wb 5 Hoa 4nx10-7 x6x10-4 “ S = 430719N 448 AT/Wh Sanned with CamscannerLet us find the direction of flux due to various coils using right hand thumb rule. As shown in the Fig. 1.18.6 (a) m.m.f. of coil (1) and (2) are in same direction while mmf. of — coil ) is in opposite direction, Net mmf. = (N, 1) + (N, 1) - (N, I) (600 x 6) + (100 x 5) = (1 x800) NI = 3300 AT mmf. _ NI_ 3300 Reluctance = S — 4307130.648 +. Flux in air gap 9 = 0.7661 mWb For the magnetic circuit pee ern SOE shown in Fig. 1.18.7 determine the current required to establish a flux I density of 0.5 T in the air gap. ; N= 1000} Iron core : thickness = 2 cm Ht core = 5000 up Fig. 1.18.7 Solution : N = 1000, B=0.5T,a=2x2=4cm?,J, =1cm |, = Length of iron path = 8 +8 +6 +5 = 27 cm Hoore = 5000 Wg ie. m, = 5000 9 = Bxa = 0.5x4x 10-4 = 0.2 mWb eee yee 27x1072 ss Gd teens Holra 5000x4n x10"? x4 x10~4 = 107.4295 x103 AT/Wp S, = ig 1x 19-2 ast Baden Hoa -4mx 10-7 x 4x 49-4 Sanned with Camscanner" 19.8943 x 106AT/Wb Si +S_ = 20.00173x10° AT/Wb ie. 0.2x1073 0.2x 10-3 x 20.00173x 106 _ 1000 ~ NI 4a 20.00173x 10° ‘Scanned with CamScannerion 2.5 cm x 2.5 cast steel structure is made of a rod of square me mx 25 on shown im the Fig. 1.18.14. What is the current that should be passe in hs ea coil the icf limb so that a flux of 2.5 mWb is made to pass in the right limb. Ass permeability as 750 and neglect leakage. Fig. 1.18.14 This is parallel magnetic circuit, Its. electrical equivalent is shown in Fig. 1.18.14 (a), Reluctance of right side ——____ Sanned with Camscanner1, ¢ 1 Bx 10°? Woy yam 107 x 750% 25% 2.5% 10-4 i 4 = 424413 109 AT/Wb - (oy) (oy 2 2 oS ego 40x10? cea gue 43 ae - © Hobe 142x107 x750%2.5x 25x10 { = 679.061 103 AT/Wb . 7 . For parallel branches, m.m,f. remains same. t & For branch AB and CD, m.m. is same. Fig. 1.18.14 (a) mmf. = 1 S)= 62S; And 62 = 25 mWb .- Given i 6) = 22:82 _ 25x 107 x 679.061%10° _ 4g oy Sy 424.413x 10° @ = 01+. =254+4=65 mWb Total mmf. required is sum of the mmf. required for AEFB and that for either central or side limb. Sperp = Sp = 679.061x103 AT/Wb mmf. for AEFB = Sagpg X = 679.061x 10° x 65x 10= 413.8965 AT + Total mmf. But NI 4413.8965+1 Sy 4413.8965+4x 107 x 424.413x 103 = 611.548 AT Total mmf. a 6111.548 " = 12.223 A 500 Sanned with Camscanner