Conservation of charge. Kirchhoff s current law. Current density. Conduction current Convection current Displacement current

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1 5. TEADY ELECTRIC CURRENT Chrgs in mtin Currnt [ A ] Cnsrtin f chrg. Kirchhff s currnt lw. Currnt dnsity A/ m Thr typs f currnts Cnductin currnt Cnctin currnt Displcmnt currnt 5- Cnctin Currnt A currnt Tim rt f chng f chrg, I = dq / dt Chrgs crss rfrnc pln pr unit tim A currnt dnsity Chrgs pr unit r pr unit tim C / m i sc r A/ m. Ttl chrg crssing Δ in Δ t Δ Q = ρ cs ( ) Δt Δ α = ρ Δt Δ i = ρ i ΔΔt ΔQ Th currnt pssing thrugh Δ Δ I = = ρ iδ Ji Δ Δt Th currnt dnsity J ρ (5-) Th ttl currnt crssing I = Ji d s Th cnductin currnt rigints frm th chrg ming in fr spc r in gs. Chptr 5 5- Prpritry f Prf. L, Yn H

2 5- Cnductin Currnt Cllisins f fr lctrns Cnstnt drift lcity in cnductr. Vlcity f fr lctrn = + E t m, lcity ftr cllisin t, tim until nxt cllisin nd m, lctrn chrg nd mss Arg f lctrns, drift lcity N E E = j + tj = t N j = m m t, mn tim btwn cllisins In mtllic cnductrs = μe : μ, lctrn mbility[ m / V i sc] [ m / V i sc] fr cppr Cnductin currnt J = ρμe : ρ = n, lum chrg dnsity n, numbr dnsity f lctrns 9 =.60X0 [ C], lctrn chrg Pint frm f Ohm s lw J = E : = ρμ, cnductiity [ / m ] [ / m] fr cppr Exmpl 5-7 A cppr wir f dimtr mm hs th cnductiity = [ / m] nd th mbility μ = 0.003[ m / V i sc]. Find () th dnsity f cnductin lctrns (b) th drift lcity fr th mximum currnt f 5[ A ] lutin () Th chrg dnsity is ρ = n (5-0) Th cnductiity is Chptr 5 5- Prpritry f Prf. L, Yn H

3 = ρμ (5-) Insrting Eq. (5-) int Eq. (5-0) ρ n = = = μ μ r n = =. 0 9 ( 0.003)(.60 0 ) 9 [ lctrns] m 3 (5-) (b) Th currnt dnsity in th wir 5 6 J = = A/ m 3 π 0 ( ) Frm Eq. (5-) J J = = ρ n 6 Aim 9 9 = i.6 0 C r = i.6 0 m s = m s 4 [ / ] [ / ] Ohm s lw in circuit thry Vb = R I : V b, ltg btwn tw trminls Ttl currnt f wir Vltg btwn tw trminls Frm Ohm s lw V L = I J I = i d s = J =E Vb = Ei d l = EL b Th rsistnc L R = : Unit f hms [ Ω ]. ρl R = :ρ, rsistiity Chptr Prpritry f Prf. L, Yn H

4 Cnductnc G = = : Unit f imns [ ]. R L Prcdur fr rsistnc. Assum quiptntils V nd V t tw lctrds. Chs crdint systm 3. l th bundry lu prblm by Pissn s, Lplc s, r thr mthds 4. Find th lctric fild by th ngti grdint f V 5. Find th currnt dnsity nd th ttl currnt 6. Find R using th ltg-currnt rltinship Exmpl 5-3 A mdium f cnductiity fills th gp btwn cncntric cnducting sphrs f rdii nd b. Find th rsistnc f this hllw sphr. lutin W ssum V = V nd V = 0 nd chs sphricl crdint systm Lplc s qutin in th prsnc f sphricl symmtry V V = R 0 = R R R Th gnrl slutin is C V = + C R Applying bundry cnditins Chptr Prpritry f Prf. L, Yn H

5 V C = + C 0 = C + C b ling th qutins C b = V, b C = V b Th ptntil in th rgin R b b V = V b R Th lctric fild b E = V = V b R Th currnt dnsity b J = E = V b R Th ttl currnt t R = 3 4π b I = 4π J = V R= b Th rsistnc V R = = I 4π b R R Th rsistnc incrss s th innr rdius dcrss r th utr rdius b incrss. 5-3 Cntinuity Equtin Chrgs cnnt b crtd r dstryd Chrgs r gnrtd nly in pir f psiti nd ngti chrgs Ttl currnt flwing ut f I = Ji d s dq d Dcrs f lum chrg I = = ρd dt dt V Frm tw qutins d Jids = d = ρ ij V V d dt Cntinuity qutin Chptr Prpritry f Prf. L, Yn H

6 ρ i J = t In systm f stdy currnts, ρ / t = 0, i J =0 Frm cntinuity qutin i J = J id s = J id s + J id s + J id s +... = 0 V 3 urfc intgrl f J = Currnts j I = 0 j Exmpl 5-4 A lum chrg dnsity ρ is injctd int cnductr f cnductiity. Find th chrg distributin s functin f tim insid th cnductr. lutin Fr nt fr chrg ρ insid th cnductr ρ i E = ε Th cnductin currnt dnsity du t E J ρ i = ε r i J = ρ (5-6) ε Insrting Eq. (5-6) int th cntinuity qutin Chptr Prpritry f Prf. L, Yn H

7 ρ ρ = ε t r ρ + ρ = 0 t ε (5-7) ling Eq. (5-7) using th initil cnditin ρ ( t 0) ( / ) t ρ ( t) = ρ ε = = ρ Th initil chrg dnsity ρ dcrss xpnntilly insid th cnductr s functin f tim bcus it sprds ut t th cnductr surfc. Th chrg dnsity rducs t / r 36.8% f its initil lu in th rlxtin tim f ε τ = (5-8) 7 In slid cppr with = [ / m] nd ε ε = [ F / m ], th rlxtin 9 tim is τ =.53 0 [ sc]. Lt us cmpr this with nthr tim cnstnt in physics t s hw shrt it is. Whn UV(ultrilt) light, gnrtd insid flurscnt lmp, striks phsphr tm dpsitd t th insid wll f th lmp, th tm is ltd t th xcitd stt. Aftr shrt prid f tim th phsphr tm rlxs t th grund stt mitting isibl light 9 spntnusly. Th liftim f th xcitd stt is f th rdr f 0 [ sc ]. 5-4 Pwr Dissiptin nd Jul s lw Ptntil nrgy f E Kintic nrgy f Thrml nrgy f th lttic E xpnds th nrgy Ei Δl in distnc Δl Pwr is nrgy pr tim Ei Δl p = lim = Ei Δ t :, drift lcity 0 Δt Ttl pwr dlird, by th lctric fild, t th lctrns in lum d dp = p j = ( nd ) Ei = Ei Jd : n, numbr dnsity f lctrns. j Ttl pwr dissiptd in lum V P = E i J d : Jul s lw [ W ] V Vlum pwr dnsity Chptr Prpritry f Prf. L, Yn H

8 dp d = E i J : Pint frm f Jul s lw. P = V I tting d = dl ds ( ) ( ) P = Ei Jd = Edl J ds = VI V L Ohmic pwr P = I R : Pwr dissiptd in R. Exmpl 5-5 Find th dimtr f silr fus fbrictd fr dc currnt rting f 0A. Mtril 3 cnstnts fr Ag: spcific ht 0.33 J / gi K, dnsity 0.49 g / cm, rsistiity [ Ωi m], mlting pint 35 K. Assum th prmtrs t b indpndnt f tmprtur. A slw blw fus is dsignd t discnnct th circuit in 0 scnds t twic th currnt rting. lutin Lt th fus h crss sctin f m nd lngth f Lm. [ ] W cn btin th nrgy rquird t ris th tmprtur f th fus frm rm tmprtur 93K t th mlting pint 35K J g 6 3 W = ( L 0 ) cm ( 35 93)[ K] g K cm i 9 = ( L ).30 0 [ J ] Th lctric nrgy xpndd t th fus L 4 L W = I R 0[ sc] = 40 0[ J] =.54 0 [ J] Equting tw qutins = ( π D / 4) = = r D = 0.64mm 5-5 tdy Currnt Dnsitis t n Intrfc Hlmhltz s thrm : Dirgnc nd curl f J fr uniqu J Frm th cntinuity qutin i J = 0 Frm Ε = 0 nd J = E ( J / ) = 0 Chptr Prpritry f Prf. L, Yn H

9 Th bundry cnditins fr J J = J n n J t J = t J n is cntinuus, but J t is discntinuus t th intrfc. Exmpl 5-6 Th z = 0 pln is n intrfc btwn tw lssy dilctrics f prmittiitis ε nd ε, nd cnductiitis nd, rspctily. In rgin ( z 0 ), stdy currnt dnsity xists s J = Jy y + Jz z. Find () J in rgin ( z 0 ) (b) E nd E (c) th surfc chrg dnsity t th intrfc lutin () Frm th bundry cnditins J = J z z J J = y y Thus J = Jy y + Jz z (5-37) (b) J J J E = = + z (5-38) y z y J J J E = = + z (5-38b) y z y (c) Th surfc chrg dnsity frm th bundry cnditin fr D r D D = ρ ε D = ε E = J z z z ε D = ε E = J z z z z z s Thus ρ = D D ε ε = J s z z z (5-39) Chptr Prpritry f Prf. L, Yn H

10 Th surfc chrg dnsity is prprtinl t th diffrnc f th rlxtin tims f tw mdium. 5-6 Th Anlgy Btwn J nd D Dilctrics Cnductrs ε Anlgy btwn D nd J () D = εe J = E () Di d s = Q Ji d s = I (3) E, D nd J Cnductr surfc RC-cnstnt(Fig-5-5) Ptntil diffrnc V b = E b i d l V d b b Rsistnc Ei l R = = I Jids Q Dids Cpcitnc C = = V b Eidl b Chptr Prpritry f Prf. L, Yn H

11 Fr cnstnt ε nd RC = ε Exmpl 5-7 Find th lkg rsistnc, pr unit lngth, f cxil cpcitr f th innr nd utr rdii nd b, rspctily, tht is filld with mdium f prmittiity ε nd cnductiity. lutin Frm Eq. (4-55), th cpcitnc pr unit lngth f cxil cpcitr πε C = ln ( b/ ) Frm Eq. (5-45), th lkg rsistnc ε ln ( b/ ) R = = C π W will btin th sm rsult if w clcult R fllwing th prcdur fr th rsistnc in sctin 5-, bcus th fild lins f D nd J r xctly th sm btwn tw cnductrs f th cxil cpcitr. Chptr 5 5- Prpritry f Prf. L, Yn H