Physics 217 Practice Final Exam: Solutions

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1 Physis 17 Ptie Finl Em: Solutions Fll This ws the Physis 17 finl em in Fll 199 Twenty-thee students took the em The vege soe ws 11 out of 15 (731%), nd the stndd devition 9 The high nd low soes wee 145 nd 75 If ny of these solutions seems obsue, plese ontt us so we n eplin it bette Poblem 1 (3 points) An infinite ylinde with dius R is hged unifomly, with hge density ρ, eept fo n infinite ylindil hole pllel to the ylinde's is The hole hs dius R nd is tngent to the eteio of the ylinde A shot hunk of the ylinde is shown in the ompnying figue E y Clulte the eleti field eveywhee inside the hole, nd sketh the lines of E on the figue ρ R Supepose now ylinde with hge density ρ on wide ylinde with hge density ρ to ete the hole; then supepose the fields fom these two hge distibutions The field inside unifomly hged infinite ylinde n be omputed by using Guss Lw, with ylindil Gussin sufe (dius s less thn the wie s dius, length ) oil with the wie: R y s s + E d = Qenlosed E πs = π ρπs E = πρs ( ) 4

2 Then, with E + s the field fom the wide ylinde nd now one, we hve (see figue): E s the field inside the E+ = πρs+, E = πρs, E= E + E = πρ s s = πρryˆ ( ) + + The field is onstnt, nd points stight up Multiply on the ight by 1ε to get the nswe in MKS units b An infinite ylindil wie with dius R ies unifom uent density J, eept inside n infinite ylindil hole pllel to the wie's is The hole hs dius R nd is tngent to the eteio of the wie A shot hunk of the wie is shown in the ompnying figue y B Clulte the mgneti field eveywhee inside the hole, nd sketh J the lines of B on the figue R Supepose now wie with uent density -J on wide ylinde with hge density J to ete the hole; then supepose the fields fom these two uent distibutions The field inside unifomdensity uent n be omputed by using Ampèe s Lw, with iul Ampèen loop (dius s less thn the wie s dius) oil with the wie: B d = Ienlosed πjs B πs= ( Jπs ) B = φˆ Then, with B + s the field fom the wide ylinde nd now one, we hve: B s the field inside the πjs+ πj πj B+ = φˆ+ = s+ ( sinφ+ ˆ + os φ+ yˆ) = ( y+ ˆ + + yˆ), π J B = ( y ˆ + yˆ ) But = = nd y = y + R= y (see oss setion figue in pt ), so + +

3 3 ( y ( y R) ) B πj ˆ ˆ ˆ ˆ πjr = B + + B = + + = ˆ y y The field is onstnt, nd points hoiontlly, to the left in the figue Reple the 1 with µ to get the nswe in MKS units An infinite ylindil flu tube with dius R ies unifom mgneti field B, pllel to the ylinde's is, eept inside n infinite ylindil hole pllel to the flu tube s is The hole hs dius R nd is tngent to the eteio of the flu tube A shot hunk of the flu tube is shown in the ompnying figue The mgneti field is eo inside the hole nd outside the flu tube, but whee it eists, it is inesing linely with time: B y E R t B( t) = B ˆ t Clulte the eleti field eveywhee inside the hole, nd sketh the lines of E on the figue Supepose now flu tube with mgneti field -B on wide flu tube with mgneti field B to ete the hole; then supepose the fields fom these two flu tubes The field inside unifom mgneti flu tube n be omputed by using Fdy s Lw, with iul Ampèen loop (dius s less thn the flu tube s dius) oil with the flu tube: 1 dφb E d = dt 1 db πsb Bs E πs= πs = E = φˆ dt t t Then, with E + s the field fom the wide flu tube nd now one, we hve: E s the field inside the Bs B E = φˆ = ˆ + yˆ, t t ( y ) B E = ( ( y R) ˆ + yˆ ), t

4 4 so ( y ( y R) ) E B B ˆ ˆ ˆ ˆ R = E + + E = + + = ˆ t y y t The field is onstnt, nd points hoiontlly, to the ight in the figue Eliminte the fto of 1 to get the nswe in MKS units Poblem (3 points) Clulte the eleti field t point on the is of unifomly-hged iul disk, distne fom the ile's ente The disk hs sufe hge density σ nd dius de Fo two infinitesiml elements the sme distne wy fom the is (s) on opposite sides ( φ nd φ+ π ), the hoiontl omponents of de nel nd the vetil () omponents dd; thus the ontibution to the eleti field fom ing with dius s nd width ds is dq osθ π sds de= ˆ =σ + s + s θ s ds σ Thus + sds du E = ˆπσ = ˆπσ = ˆ πσ u ( + s ) 3 3 u 1 + = πσ 1 ˆ + b The hged iul disk fom pt is set into ottion bout its is, with ngul fequeny ω Clulte the mgneti field B t point on the is distne fom the ente of the disk Agin, the hoiontl omponents nel nd vetil omponents dd, fo uent elements symmetilly pled bout the is, so db ω θ σ

5 5 ( σω)( π sds) 1 Kd 1 s db = os θ = s + + s Thus πσω sds B= ˆ 3 3 ( + s ) Not mny points would be lost by stopping hee, but we ll ssume we e tying fo pefet soe: πσω udu B= ˆ 3 Integte by pts, with ( ) 3 ( + u) dv = du + u, v = + u : πσω πσω u du B= ˆ uv vdu = ˆ + + u + u + dw πσω + w w + + πσω + ˆ πσω = ˆ + = ˆ + πσω = ˆ + + = + + Multiply on the ight by 1ε in pt, nd eple the 1 with µ in pt b, to get the nswes in MKS units Poblem 3 ( points) Conside the efeene point fo eleti potentil to be t infinity fo both pts of this poblem A onduting sphee with dius is in n infinite vuum Wht is its pitne? Conside it to y hge q; then, fo >,

6 6 q E= ˆ, qd q CV V = E d = = = C = Multiply on the ight by 1ε fo the MKS nswe b The sme sphee is pled in n infinite, wekly-onduting medium with esistivity ρ Wht is the esistne between the sphee nd infinity? If the ondutivity is smll enough, the eleti field is given simply by the eletostti vlue, so Fo ll But V >, then, π π σ q J = σ E= ˆ 1 I = J d = σq dφ dθsinθ = σq ( π)( ) = 4 πσq = q still, so, sine V lso = IR, R V 1 ρ = = = I σ Divide on the ight by 1ε fo the MKS nswe Poblem 4 (1 points) Clulte the mgneti field inside long solenoid with dius, N tuns pe unit length, nd uent I Use Ampèe s Lw with etngul loop, two sides pllel to the solenoid s is: B d = Ienlosed Bh = INh B= NIˆ,

7 7 whee I is the uent in eh tun of the solenoid, nd the dietion is long the oil s is in the dietion given by the ight hnd ule Reple the 1 with µ to get the nswe in MKS units b Inside the long solenoid nd pllel to it, thee is shot solenoid with dius b, N b tuns pe unit length, nd length Wht is the mutul indutne of the two solenoids? b It s esiest to wok out the flu in the smll oil fom the field in the lge one: Φ = = Bb BA b NIN b π b (flu theds Nb b NN b = I = MI, so M = NN b b loops) Reple the 1 with µ to get the nswe in MKS units Poblem 5 ( points) An infinite, hged, stight wie (hge pe unit length λ ) lies pllel to the is, nd pllel to two infinite, gounded onduting plnes whih inteset t 9 ngle The hged wie lies distne fom eh plne Clulte the eleti field E t point A, fom eh plne on the sme side s the line hge, nd t point B, fom eh plne but on the side opposite the line hge y A λ 1 λ λ 4 B 3 λ

8 8 Point B is inside the onduto, so the eleti field is eo thee Fo point A, use the method of imges Thee e thee imges of the line hge in the onduting plne, two with hge pe unit length λ nd one with λ, lbeled -4 in the figue bove The eleti field fo n infinite line hge n be lulted fom Guss Lw, using Gussin ylinde with dius s nd length h: E d = qenlosed λ E πsh= πλh E= s ˆ 4 s So we n supepose the fields fom the fou line hges t point A: E= E + E + E + E ˆ ˆ ˆ ˆ = λ + y + y ˆ ˆ ˆ ˆ + y + + y λ ˆ = ˆ y λ = ( ˆ + yˆ ) 3 Poblem 6 ( points) A sphee with dius R ies polition distne fom the ente P = K, whee K is onstnt nd is the veto Find the sufe nd volume bound hges σb = P( R) ˆ = KR, 1 d ρb = P = ( K ) = 3 K d Chek: the totl bound hge should ome out to eo, nd it does: q = R σb + πr ρb = KR KR = 3 b Clulte the fields E nd D eveywhee

9 9 Sine the hges e distibuted with spheil symmety, we n use Guss Lw with spheil gussin sufes With one of those dwn outside the sphee > R, we enlose no hge (see pt, bove), so ( ) E d = E =, D = E + P = Fo the inside of the sphee ( R) popotionl to its volume:, Gussin sphee enloses hge E d = qenlosed E = π ρb = π ( 3K) 3 3 E= K = 4 πp, D= E+ P= Note tht we ould hve done the whole poblem with D, beuse thee e no fee hges nd the polition is dil (ie P =, so D= ) Poblem 7 ( points) A iul wfe is mde of vey wekly onduting mteil with esistivity ρ nd dieleti onstnt ε Its dius is nd its thikness is ( ) Highly ondutive, metlli eletodes ove the iul fes Unde the ssumption tht the mteil s ondutivity doesn't ffet the pitne, lulte the esistne nd pitne of the wfe with its eletodes ρ, ε ρ ρ R = = A π εa ε C = = 4, Reple ε by ε to onvet to MKS b Suppose tht the wfe is hged up with bttey with voltge V, nd tht the bttey is disonneted t t = Wite one diffeentil eqution desibing the hge on the eletodes, nd the simil equtions desibing the potentil diffeene V ( t ), nd the eleti field E ( t) within the wfe

10 1 (Hint: The wfe's esistne nd pitne hve the sme potentil diffeene, nd n theefoe be epesented by two elements in pllel) Kihhoff s ule #, with the loop dwn ountelokwise, gives us Divide this esult though by C nd use q q + IR = ; C dq 1 dq + q = + q= dt RC dt ερ = CV to get dv + V = dt ερ + I dq = q dt + R C V The field is onstnt between the pltes; divide this lst esult by nd use V = E to get de + E = dt ερ Solve the diffeentil eqution to obtin the eleti field between the eletodes s funtion of time de de + E= dt ερ = E ερ E t ln = E ερ Et = Ee () t ερ d In tems of the esistivity nd dieleti onstnt of the mteil, wht is the time onstnt? Comment on the genelity of this esult, onsideing the dependene of the time onstnt on the shpe of the wfe The esult of pt n lso be witten s t τ ερ () = Ee, τ = Et Reple ε by ε to onvet to MKS Note tht the time onstnt doesn t hve ny ftos o vibles in it tht hve to do with the shpe of the wfe; this must dt

11 11 theefoe be genel esult fo the eltion of n eleti field in medium tht onduts