Static Magnetic Field, Questions and Answers

Q Find it at the centre of the Sq. loop shown in=fig.y. 2xlJ,-ySolFIatthecentreduetotheIst lineaH = axai = az => - -axx-ayIl/22245()x,=135 H,- Ip=1/24TT.+2H, = 4ap Lts + 1/2 ] azH=H, + H2 + H3 + Hy = 4H,H1= I (52) GAN4x l/2IT=252IazH1 = ITlaz52 TTmyQFin H at centre oftheSalloopOH= az as inpreviouscase.z=TleH, = I^lCosar -the9/2 * 2/2c/2 = lan go = P=1536060120zeX2=00x1=120;H1=I/2 TT (530)[21az=Ia 325 it a Find H at (3,4,0) in the fig.Sol air= a'x as =) tyans8A) sfy= -ay x ax = a38A lx=35an = axx ay => + x - amslyzy#(3,4,0)tx= a3Hy=I4th Px[cos az Cosa, osxi] day = 44 I (3) JazHy=4T(3) I az - 20TT 3Iaz1xx=455ly I a 4/((4) I /caso Coso - (os(90+B) 93IHx I4T(4)3 = 2I 2071 az->H = Hx Hy = J 20T [s+27" = H / S azH=4T 8 az =Qk = 30 x10 1 (A/m) a is on y=o plane . Find it at (1,20,-2)SolH= ->Z= 1/2 1x30 x103 (93xay)properan = ay7=15 x 103 (-ax)(1,20,-2)H = - 15 ax m (A/m)x QIn which loops total magnetic flux iszeroXIxx.Ans brop.xx2 & 3=34I.xxx.xxxQB = x2 ax - xy aj - (K x 2) azIf B represent magnitic field then find K. -SolV.B=0dx2 - (xx) K (2x2)=0ox dydz2x -x-kx - = 0(1-K) =0K=1Q Two infinite current carrying sheets are placed parallel to eachother in Such as that they carry current in the oppositedirecting with the Same current density The magnetic fluxdensity in the Space between The sheets will be ?Sol3 = B,+B2 the (H1+H2)(az)-k,ckoasTa'sazB1/2^K2 = Ko az= to 1/2 Kpaiz + 1/2 IIII= komeleo az x ag + x3= 2 boko [ ax + (a))xB= = - Ms ko a xAns Relation between Z andr.Z60tan 60° = Z(d+6-r)k(d+b-r) -3.From eq in12& 3^P21 =dleo Iz as . ( 53) ( b+d-r) dr as.2K rQ21 =Sno I2(53) (-(r- (bid))) dr.2TTXd+6- 53lio I2 S r - b+d ) dr2 15dotd-J3 lio I2r b+dlur-271After solving P21 = 53 ho I2(64d) ln (b+d) d - b]211Mutual Inductance = Q21 S3 lio=I22TI(b+d) lu(btd) d - b]a1In a material the magnetic field intensity in H - 1400 A/mWhenmagnetic field flux density B =2 wblm2 when His reduces to 700 A/m at B = 1.4 wb/m2 the changein permeability isSol H, 1400 A/m=B, = 2 wb/m2 = u,H2=700 AlmB2 = 104 wo/m2 = l22B- le H leyllo= 31 H => lr - 31 = 2 x 107 & Ur2 = 104x1071400x4700x40Day= ler, ear, = 454.73 Ans QA non- magnetic wire of circular cross-section has a radiusof 10 mm and a length of 100 m. It carries atotal Current of 1000 A in az direction and has potentialdifference 10V between ends. Let H = kp2 as wherek is an unknown constant The value of Kx 108 A/m3.azrelommSof H=Ke2 as10VThinkingV X H = J1/ = Vxit =aerapa3Iaa&ede00ozI=1000 AOkeO7=1 e [ 3ke2 az] a = 3ke az^J = 3ke azI SS J.ds3kl a3 . e de do 030.010.01I = 3k se"dedo3ke3=(2TT)3Oo1000 = 29K (10 x 10 3 3 331000 = 2K K ( 10 X 109)108K=1000== 1.592 X 10215x1062Th QIn a certain material for which us - 4 the magneticfield is given by .H=2ax + 4ay-8az - Alm.The corresponding magnetic energy density isu I/wiSolWm = 1 2 S H.B dv Joulewm = 1 u H2 J/W22Wm = 1 (4) x lio ( ut 16+64)2= x(45) X 10 -7 (84)=211.1 u J / m2#Were =1/ folfi dvImp#The laws related to Static magnetic field.1. Stokes circuital law V x H = J-->2. Gauss law of magnetic fieldJ.B =03. Ampere's law $ 11.00 = I Q = A Solenoid with N1= 1000 , 8, 21 (m and l, = 50 cmis concentric with in a Solenoid coil of N2 = 2000r2 = 2 cm and l2 = 50cm The mutual iductance assumingfree space conditions ism H.Sof Given N, = 1000 : / l, = 50cmIst coilN2 = 2000 / r2 - 2cm, l2 - Soccato = free SpaceE2 = - N2d 012 di,= - N2 d012 di,dtdi,di, dtIndcoifi Mdi,M=dt(i)Now if we consider on mag 1H1 = NI = N,II = 1000 XId40.5B12 les (2000 I1)12 = B/2(area)2 = (2000 leo ID (T(2)2x154)412 = 0.8 T le I,M - N2(012) =2000 ( 0.811 lio I) 1600 X leo H=I,I1M = 6.31 MHa=Determine the mutual inductance b/w a conducting triangularloop and a very long straight wire as shown in fig?circuit-Sol B2 due to Circuit-2 B2 = leo I2 as I2Kd2Fr(long line at a distance 'r')&dcircut 2I,Flux linkage 021 = S B2.ds, - 1ZsoS1>look6ds = Zdr as -2r| MislaneausProblemQAportionofmassinisprojectedintoauniformfield B = Bo azwith aninitialvelocity^aax + Bay aThedifferentialequationssuchthatthepositionvectorr=xax+yayt2 azmustSatisfyare(w represent constant)solF = ma = m dv:B = B az/V o = Laxt BaydtForcethepartialinmagneticfieldisgiven asonchargeFema=qVB.d'x^dy+d zan ay a3maxaig+a3=qdt2dtataOoBoI'x2an + d2y ay +d20a }=dt2dt2allIandt2(BBO)-inay(aBo)+93(0)d2xax+d'yanyd2^+qBdt2azBoanqx=-dt2inBodt2aym22dx2 3 Body = - 280 qdz-l;Odt2mdtindt2dx = q B Bot + C,dy =xBoqt+C2 dz =o+C3drindtindtIVy= - Boq t2 + C2t + d2x qBBo t2=+Cit + d,m2m2dx- qfBot =IAnsdyxBoqt = C22Ansdtdfmm