University:
Washington State University
Course:
Physics 103 | Problem Solving for Physics 101
Academic year:

2022
Views:

140

Pages:

16
Author:

Janor

Static Magnetic fielddH12Biot Savart Low :->R12P2 (Field point)P,diI,dl : differential current element at P,->,->top,OP2 = r2according to Biot Savart low theIrdifferential Magnetic field intensitycurrentat P2 i.e d'H12 due tocaryingorigin,conductorIdlatP,#R12= P2 - P,isgivenas(Field)(Source Pt)^dH12 = Idl X R12d H12 =Idl X R124/ 12121234 II R12TotalHatP2duetoICurrentgoinginaclosedline is :^aH12ofI all x R12R12 = R12 like }unit=(R12 vector4TT 1R121HaIi.eHisLinearw.rotCurrentSource.Types of Current Source :I1. Line current />20 Surface currentK = CurrentVCross-Section width# / conductor surfaceIlaz(or)ITR I az ( A/mSheet thickness isw&-wzero]Ksurface current density.A/m3. Volume current:J= Currentcross sectional AreaI az (A/m2)whVIJ=Volumecurrentdensity(A/m2) 0 Idl = kds = J dv2Hat P2 due to surface Current having K1 (A/m) SurfaceCurrent density at P, is^nR12 = R12H12 =K,ds X R12R1241 IR123T.T at P2 due to volume current having J, (A/m )volume Current density at P, isAH12= J, dV x R124T /R12/2 Magnetic field at a point due to current carryingZ" Conductorwa "IFieldIPointL2=dzaZRyPfield point)Px^H=SIdlxRIIIdlxR122aR12L49IR124TT I R12)ZePde = dz azA^R12=- Z az + e apR12/ =Z2 + p2^dl x R12 = e dz ap^H=/Iedzap; P = Constant4TTP= tanxZZ = Pcot xHe S I e(-ecosec dx)asdz = -Pcosecadd -4k(p2+ 3/2X2^H =/I Cosec2 daag ^ = - ISinxda ase cosec 34TTPX, x2H=-Isinada of = - I4TP4TP[-cosk]apaa,42 aH =I[cosaao4TPL,^H=I [cos X2- - (os x)]as4THP011 Find the H in all regions due to infinite long linecurrent carrying I along Z- axis25to/x2-oSofI^X2H= I [COS K2- - (os x ) aHazP4THwhere ; x2 O &x, -> 180) x,T^AA^z->- DaH = az X ap = as)x,14a,180H=I COSO - COS 180 ] as4TheaAA= I 1- (-1) ]asan = alxa,the^= az X ap = as^IT = I asae2TTP^ap^az Note . # Tdotout of the page.xCrossin the page.2Right hang thumb ruleIf I is in R.H.S thumb direction then H is inR.H closed finger direction.If H is in R.H. thumb direction then I is inRH closed finger direction.3H in all regions due to infinite long linecarrying Corrent I along axis or parellel to anyaxis isH =I^aHwhere P is L distance from2THCurrent line to field point^^aH = a X a4 H at (0,0, h) due to I current going in aclosed loop of radius a loop is onxyplane with Center coincides with Z- axis asShown.2AI a=(az)(o,o,h)at(o,0,4)2( a2 f n2) 3/2IIIf I is Anticlock wiseyaIH =I a 2(-az)xpla, 0,0)(a2+th2) 3/2de = edo ao#It I is clock wiseas = add as Note/H in all regions due to infinite area Surface havingK (A/m) Surface Current density is^^an is normal unit vectorHKXaN2from Current Surface tofield point.2Hinside N turn Solenoid Shown below.^Z3=dmaterial [magnetic, u]raHd^N turnsH = NIazId3=0SH at e radius inside N turn toroid is.as^asH = NIa2KPXZ# # 10 out side toroidN turns"4Ampere's circuital law " It states that the closedline integral of magnetic field intensity is equal to theto tal Current enclosed by that closed line.while finding H using circuital law take a closedline at the location of H. 01/11Find IT in all regions due to infinite long coaxialcable along Z-axis carrying +I in inner condition-I - current in outer conductor as shown.Sol Inner Conductor current densityJ. = Ia2Current in IT e2 area = J. Cre )I= I (Tre)InaIenclosed for oseka is - Ie2-Ia2And outer conductor current density - Jo = - ICurrent in area = JoIenclosed = Iinner + Iouter ;for beece= I + (Jo (TRe-T62)) = It -IT(2)Ienclosed = I C 2 - e2-2[bF=QE + Q"Lorentz Force Equation"F = O (ve x B )In diff form :- dF = dQ (ve e xB)/I = dQ = do = I dt-2dtFrom , and 2:dF =I dt ( Ve xBeg")dF = I ( Vedt X B )& due = dldtdF = I ( dl x B )Find the force on current ( (I2) carrying conductor l2 due tocurrent carrying (I,) conductor l,.LB,I,78,XR12P,al,I'I2Line 1I,Line 2l,l2 #I,dl, X R12->H12=4T 1 R1212B12 = le I 4T dl x R12| R12B12It is total magnetic flux density at P2 due to I, CurrentDifferent ForceI,dl,duetoB12isdF12 = I2 dl2 X B12Total force on Line2F12=SI2dl2X B12^line2F12 = S I2 dl2 X / UI, di, XR124TT IR1212line1F12=leI,I2Idl2IIt,XR12X4TJline2R12F12 = j I2 dl2 X B12 = - - I2 I B12 X dl2 2linezIfBis uniformthen+12=-I2B12xS212=-- I2B12Xl2=Il2XB12=F12In general^F =Idl xB = I anFIIIIIIIBI I Sin OeB I aimi-=IlBSinO z#H12Magnetic field.xIntensity at line-2Due to line - \X(yI,H12 = I, aH ^-AI22kedline- 2line IB12 = UH12 = UI, air ^2581312 =u I ( - ax)2x(d)AI2 dl2 = Izdz azdF12 = I2 diz X B12^dF12 = I2dz az X (III(a)2T((d)dF12 = uI,I2 dz (-ay)2ndTake force on line- - 2 for h height is2thF12 = le I, I2 dz (-ay)2ndZ=OieDirection ofF12 = (-aj)ll I, I2hForce on line 22nddue to line-is towardsline - 1 Note1.dF=UI, I2h(N/mI,2ndI2F a I, I2&FxdForcebetweentwo parallel conductors carryingCurrents in the same direction is 7 tothe conductors and attractive .Force between two parallel conductors carryingcurrents in opposite directions is 1 to theconductors and repulsive.Q If Force on loop 1 is F then find force on=loop. 2 in terms of F.ZZI2=I2a2aLoop II2 =ILoop - 2aIITTaaayXXI,= IyI, = ISolB12 = UIAaH =le I ax ^2TTP2FT(2) F12 = - I2 of 312 X dl2 2 = -If 2T(2) lI ax X^ dlzF12 = -ue 2TT pazz x dlzZoaF12 = -IIa" X dy ay + ax 2 x dz az2 TTay=oz=a+ j a: x dy3ay=a 2aF12 = I2 & dlz X B12 = I2 - dizxist j. dl2 xB12y=02=aa+ iiz X B12 + I dirbi+y=a2=2aa30aF12 -- I2 S B12 xdl + of Blzxdi + j B12xdi + I B12 xdlz=3ay=o z=a y=az=a const 2=3acousta3a0N2==S2TZ 4I ax xdyay + Sa+2FTZjMAIL2TZaxxdyayyouz=a2=a ,y=a y=a,z=3aa+"II 2THZ ax X dz asZ=3a,y=0F12========= 2TI [af dy a3 + A + 39 1/6 j dy az Ay=oy=a= - "I22/1az + (4:)4F12 = - 2TT 3For Loop 1 i.e FSimilarly for Loop2= 2 TT "I2 3 A = F Note: 1. Magnetic flux density/B/= 101&B = 4 H (wb/m2)Areau = lott2. GivenBthen magnetic flux crossing open surface0= ll B.ds3. Inductance isL = NO(Henry).INO is total magnetic fluxNis number of tornsFor Single Loop N = 1 :-L=Iobservation -1. Find inductance of coaxial Cable betweene = a and b for "h" heightH I as for=25 PB=llHB -> = u as2 P0 = IIB.dsfl2ripuI ais). - (dedz as)bh= u/de2n/dzeP=a 2=0hh0=leI[emp]PZ=le Ih ln (b/a)2TT2TTP=a 0L = ofLI2TT le ln1b/a) In Henry. Q = Inductance of N-turn Solenoid Shown:-Sol = H = NI a3 Ad^B = le H = UNI azd= IIB.ds =UNI) d a3 (P de do az)a2T0= leNId fede SdpP=80=0a270= leN Id FEP 2UNIa2=(2/T)d2e=o 0=0$ = UNT (Ta2)dL=NO I = N2 henry.L= N2Rewhere RL = Reluctance of solenoidRL=& u lly lloee TTa2l= Length of conductor coill = N 1(2xa).