Physics 712 Electricity and Magnetism Solutions to Final Exam, Spring 2016

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1 Physics 7 Electricity nd Mgnetism Solutions to Finl Em, Spring 6 Plese note tht some possibly helpful formuls pper on the second pge The number of points on ech problem nd prt is mrked in squre brckets t the beginning of the problem [3] A cpcitor consists of two coil cylindricl conductors, the inner one with outer rdius, nd the outer with inner rdius b, nd hving length L much lrger thn or b () [] Find the cpcitnce of the cpcitor b Cpcitnce is defined by the eqution Q C V If we hve chrge Q on the inner conductor, then the electric field (wy from the ends) will be rdil outwrd E E ˆ ρ, nd we then we cn use Guss s lw with cylinder of rdius nd length L, Q LE, Q E L We cn then integrte this to get the potentil b Q bd Q V Ed ln b, L L Q L C V ln b (b) [7] A bttery with voltge difference V is now ttched to the cpcitor Find the mount of chrge the bttery sends to the cpcitor, nd the mount of energy required to chrge the cpcitor Of course, QC V, nd the energy cn be found from U Q V, so we hve LV QCV, ln b U QV ln b L V

2 (c) [3] While still connected to the bttery, dielectric mteril of permittivity is slid between the two conductors How much work does the bttery do mintining the potentil, how much does the cpcitor energy increse, nd how much work must be done to push the dielectric in? Obviously, doubling the permittivity doubles the cpcitnce, which since the voltge stys the sme, doubles the chrge Therefore, the chrge chnges by n mount Q Q, nd since it is moving cross constnt voltge difference, the work done is LV WB QVQV ln b The cpcitor styed t constnt voltge, so the new cpcitor energy is so the chnge in energy is L V U U U U U U ln b U Q V Q V, This is only hlf s lrge s the energy put in Where did the blnce in energy go? It went into drwing the dielectric into the cpcitor The work done by us on the dielectric is L V L V L V WD U WB ln b ln b ln b It is negtive, indicting tht, in fct, the dielectric is being drwn into the bttery

3 [5] In the presence of chrged plsm, it is possible to crete electromgnetic wves tht re longitudinl, hving electric polrition prllel to the direction of propgtion The fields tke the form E, t E ˆ cos kt, but with no mgnetic field B () [6] Show tht this electric field (nd lck of mgnetic field) cn be written purely in A,t, without the use of the sclr potentil, so tht terms of vector potentil Since, we hve EA, so we cn get A by performing time integrl of E: A, t E, tdt E ˆ cosk tdt E ˆ sin k t The constnt of integrtion is irrelevnt, since we re simply trying to find vector potentil A tht works (b) [9] Convince yourself tht this is not Coulomb guge Find guge trnsformtion tht cuses A to vnish Find the sclr potentil in this guge cos, but Coulomb guge is defined by A Clerly, not Coulomb guge To mke A vnish, we need to find guge function such tht A A, so we need For this vector potentil, A E k kt A E ˆ sin k t Since we wnt the result to be in the -direction, we clerly wnt function of tht when differentited yields sine It s pretty obvious tht cosine will work In fct, if we pick E, t cos kt, k then when we perform guge trnsformtion, we will hve E E A A ˆ sin k t cosk t, k E E cosk t sin k t t k k

3 [] An electromgnetic wve with wve number mgnitude k is incident from vcuum on perfect conductor t n ngle with respect to the norml, s sketched t right Let the verticl direction be nd the horiontl

4 3 [] An electromgnetic wve with wve number mgnitude k is incident from vcuum on perfect conductor t n ngle with respect to the norml, s sketched t right Let the verticl direction be nd the horiontl direction Let the incident wve hve electric field mplitude E () [6] Write, in comple nottion, the generl form for the incident electric nd mgnetic fields s functions of spce nd time You must write ny constnts you use, like, in terms of given constnts In generl, wve tkes the form i i t, i e i k k E E B ke t e The vector E is restricted to be perpendiculr to k The direction of k is specified by the problem, nd in vcuum, ck, so k ksinˆ kcos ˆ nd ck (b) [8] Write the generl form of the reflected wve E nd B s functions of spce nd time in terms of the reflected electric field mplitude E Eplin logiclly why you were forced to pick the vlues you pick for the reflected wve number vector k (ie, why is the lw of reflection vlid?) The reflected wve will tke the sme generl form, so i i t e, i i k k E E B ke t e The reflected wve will hve to stisfy pproprite boundry conditions t =, so to mke it mtch with the incoming weve, we re going to hve to pick the sme frequency Also, to mke it mtch t ll, we re going to hve to mke k k Finlly, we re going to still hve ck, which implies k k, so tht k k k k, or since k k, we will hve k k, so tht k k But we wnt the wve to be outgoing, so we see tht k ksinˆ kcos ˆ This implies the lw of reflection, tht the reflected wve hs the sme ngle compred to the norml s the incident wve (c) [6] Assuming tht the incident wve hs E E ˆ y, find E The prllel electric field must vnish on the boundry =, which implies ikit ikit Ee E e t lest prllel to the surfce of the conductor We therefore hve E E E ˆ y, since this is prllel to the conductor

5 4 [] How do the quntities E c B nd EB trnsform under () proper rottions, (b) improper rottions, nd (c) time reversl? Under proper rottions, the dot product of two vectors is sclr, so both of these re sclrs Under improper rottions, sy reflections, E does ectly wht you d think vector would do, which leves E unchnged, but B picks up n etr minus sign, which mens tht B is lso unchnged So E c B is still sclr under improper rottions However, EB will hve E reflect normlly, but B picks up minus sign, so EB will chnge sign Hence EB is pseudosclr Under time reversl, E stys the sme nd B chnges sign Hence E c B will be unchnged under time reversl, while EB stys the sme 5 [] How does the quntity EB trnsform under Lorent boost in the -direction? When we perform Lorent boost, the fields trnsform ccording to E E, B B E ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ E vb Ey E v By B Ey vb E vb y y y y v v v B c B ˆ ˆ ˆ ˆ ˆ ˆ ˆ B ve yy B Eyy E By E B E y c y c c We therefore hve v v EB E B EyBy E B EB Ey vbby E E vbyb E y c c v v v v EB EB y y vbb y EE y BE EB vbb y EE y BE y y c c c c v v v EB EB y y BE y y EB BE EB EB y y EB c c c EB EB EB EB y y So, in fct, this epression is unchnged

6 6 [3] A prticle with chrge q nd mss m, initilly t rest t = is plced in uniform electric field E E ˆ () [8] Find set of coupled differentil equtions for the derivtives of the fourvelocity U s functions of time We strt by writing the electric field, first with indices both up, nd then lower the second inde by chnging the three columns corresponding to the spce coordintes: We then use the eqution We then write these equtions out s F Ec Ec Ec Ec, F m du d qf U We therefore hve U E U d U q E U m d U c U 3 3 U U 3 du qe du qe du du U, U, d mc d mc d d (b) [] Find the solution of the equtions from prt () subject to pproprite boundry conditions t = At =, the prticle hs U c,,, The lst two equtions re trivil to solve, U U 3 To solve the other equtions, tke nother derivtive of the first eqution, which then yields du qe d qe U U d mc d m c This hs generl solution qe qe U Aep Bep mc mc We cn substitute these into, sy, the first of the differentil equtions, we find mc d qe qe U U Aep Bep qe d mc mc

7 Since U () =, we conclude tht A = B Since U () = c, we hve A B c, which then tells you A B c We therefore hve c qe ep c qe ep qe qe cosh, sinh mc mc mc mc U c U c 7 (c) [] A proton hs mss m673 kg 9383 MeV/ c nd chrge 9 qe6 C It is plced in constnt electric field E V/m How much proper time (in seconds) will it tke to rech 99% of the speed of light? The four-velocity is given by U c, v, so tht hve v U sinh qe mc qe tnh c U cosh qe mc mc We therefore hve qe v mc c tnh tnh , 647mc c c qe e V/m c v c U U, so in this cse, we ev/ m m 8 88 s 998 m/s

7 [35] A wveguide is in the shpe of 45-45-9 tringle, empty, with legs of sie, with boundries t y =, =, nd y = Consider potentil mode function of the form y B C y, cos cos () [5] By considering the

8 7 [35] A wveguide is in the shpe of tringle, empty, with legs of sie, with boundries t y =, =, nd y = Consider potentil mode function of the form y B C y, cos cos () [5] By considering the boundry t y =, determine if this would work for TE or TM mode By considering the other two boundries, find suitble conditions on, B, nd C E For TM modes, E, nd since on the surfce of conductor, nd E is prllel, but clerly the mode function does not vnish t y = We therefore epect we must hve TE mode, for which we demnd n In fct, t y =, we hve Csin y y n y So we re deling with TE mode We lso must hve the boundries t = nd = y work Looking t =, we hve B sin Bsin This suggests tht must be n integer multiple of pi, so n This only works when n is positive integer Finlly, we hve to get pproprite boundry conditions on the boundry = y The perpendiculr to this line is in the direction ˆ y ˆ, so we hve B sincsin CBsin n y y To mke this true ll long the boundry, we must pick B C (b) [7] Check tht it stisfies the mode eqution, nd determine The mode eqution is t So we hve cos cos y y B cos cosy cos cos y t B y n It is obvious from this formul tht

9 (c) [8] Write out eplicitly the wve function E or B (which ever is pproprite) s function of spce nd time, if it hs wve number k in the -direction, eplicitly giving n epression for the frequency The spce nd time dependence for the mgnetic field (since this is TE mode) will be ikit n ny ikit B, ye Bcos cos e The frequency is given by n k k n n k c k (d) [5] We now modify this wveguide by mking it into cvity by dding conductors t = nd = Wht restrictions does this impose? Find formul for the frequencies, nd find the lowest frequencies i t B y ke, where we chose sine becuse we need B on this boundry We lso need it to vnish t =, so gin, must be multiple of pi gin We therefore hve p k, where gin we hve p s positive integer The frequency for these sttes is therefore The mgnetic field B now tkes the form, sin The lowest stte hs frequency c n p c c k c n p

Jackson 2.26 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell

Jackson 2.26 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell Jckson 2.26 Homework Problem Solution Dr. Christopher S. Bird University of Msschusetts Lowell PROBLEM: The two-dimensionl region, ρ, φ β, is bounded by conducting surfces t φ =, ρ =, nd φ = β held t zero