PHY2049 Exam 2 solutions Fall 2016 Solution:

Transcription

1 PHY2049 Exam 2 solutons Fall 2016 General strategy: Fnd two resstors, one par at a tme, that are connected ether n SERIES or n PARALLEL; replace these two resstors wth one of an equvalent resstance. Now you have a crcut s one resstor less. Repeat the procedure untl you have one equvalent resstance left ths s the answer.

2 General strategy: 1) Defne currents n each segment of elements connected n seres (drectons can be arbtrary). See how many unknown currents do you get (three n ths crcut). So, we need three equatons to fnd three unknown currents. Count number of loops (two n ths crcut), each of whch wll gve you a loop equaton for sum of potental dfferences. The remanng equaton (thrd equaton n ths problem) wll come from the juncton rule for currents. 2) For each loop, wrte down equatons for potentals: walk along each loop (n any drecton) and sum all dfferences of potentals as you jump over elements of a crcut; when you come back to the start pont, the sum must be zero. - left loop (walkng clockwse from the left-bottom corner): ε R 2 r 2ε R = 0 - rght loop (walkng clockwse from pont b): 2ε + 2 r 3 R ε = 0 3) For each juncton of wres, wrte down an equaton for currents: sum of ncomng currents = sum of outgong currents. Dscard equatons that are not ndependent. - Pont a: = In ths problem, we can stop: we already have three equatons for the three unknown. For curous mnds: Pont b gve the exact same equaton for currents as pont a,.e =. So, we have only one ndependent equaton for currents. 4) Solve the three equatons: ε R 2 r 2ε R = 0 2ε + 2 r 3 R ε = 0 = 2 + 3

3 Power dsspated by the resstor: P = 2 R In a dschargng RC crcut: (t) = q 0 τ e t/τ, where τ = RC Therefore: P(t) = (t) 2 R = q 0 P(t) P(0) = e 2t/τ 1 2 = e 2t/τ t = ln2 2 τ 2 τ 2 Re 2t/τ Power dsspated by the resstor r nsde the battery: P = 2 ε r, where s a current n the crcut, = R + r

4 Use the rght hand rule (do not forget that the charge s negatve!) Consder drecton of forces on each of the four wre segments makng the square loop. Left and rght wres carry the current parallel to the magnetc feld: force s zero. The top wre (use the rght hand rule): the force s out of page. The bottom wre: the force s nto the page. Hence, the loop wants to turn around the x-axs. In the gap wth the electrc feld, ons are accelerated and gan energy: K f K =U U f mv 2 2qV = qv v = 2 m In the space wth the magentc feld, ons follow a crcular path of radus r: m v2 r = qvb r = mv qb = m qb 2qV m ~ V B

5 Potental energy of a magnetc dpole n a magnetc feld: U =! µ! B = µbcosθ Torque of a magnetc dpole n a magnetc feld:! τ =! µ B! = µbsnθ! τ Therefore: U = tanθ (Keep n mnd that an angle between two vectors s between 0! and 180! ) Magnetc feld made by the wre wth current =12A: B = µ 0 2π r Consder forces on each sde of the loop wth current 2 = 2A!! The left segment of length b = 8cm, dstance a = 4cm away from the wre: F L = 2 b B(a) = 2 b µ 0 2π a.!! The rght segment of length b = 8cm, dstance 2a = 2 4cm away from the wre: F R = 2 b B(2a) = 2 b µ 0 2π 2a. These two forces are n opposte drectons, so the magntude of the net force along the x-axs s F L F R = µ 0 2 2π 2a b The forces on the top and bottom part of the loop are same n magntude (by symmetry) and opposte; so ther sum s zero.

6 Quadrant 2: The magnetc feld made by the wre wth current s out of the page B = µ 0 2π r = µ 0 ; 2π y wth current 2 nto the page B = µ 0 2 2π r = µ π x So the two felds would cancel each other, when ther magntudes are the same: µ 0 2π y = µ 0 2 2π x,.e. along a straght lne, gven by y = 2 x Quadrant 2: The magnetc feld made by the wre wth current s out of the page; wth current 2 out the page. So the two felds cannot cancel each other. Quadrant 3: Smlar to Quadrant 1 Quadrant 4: Smlar to Quadrant 2 The fgure represents an nfntely long wre wth current and a crcular loop, also wth current. The magnetc feld made by the current n the straght wre at the center of the crcle (.e. at a dstance d from the wre) s B = µ 0 and nto the page. 2π d The magnetc feld made by the current n the loop at the loop center s B = µ 0 (2π ) and out of the page. 4π d The magntude of the net feld s the dfference between the two,.e. µ 0 4π d (2π ) µ 0 2π d = µ 0 2π d π 1 ( )

7 Magnetc feld nsde a solenod of length l wth densty of wndngs n = N l (N s number of wndngs) and carryng current s B = µ 0 n = µ 0 N l. Hence, N = Bl. The nformaton on the radus of the solenod s rrelevant. µ 0 The nductance of the orgnal solenod s L = µ 0 n 2 N Al = µ 0 l The nductance of the new solenod: L new = µ 0 π (3N)2 r 2 = 9 2l 2 L 2 ( πr 2 )l = µ 0 π N 2 r 2. l The nduced EMF n an nductor s ε = L d dt, where d d at t =1 ms can be read off the graph: dt dt Therefore: ε = 900 V = 2A 8A 2ms = 3000 A s Potental on the rght s hgher than on the left, therefore V rght V left = 900 V

8 The magnetc feld produced by the central loop n places where loops 1 and 3 s out of the page. The magnetc feld s ncreasng. Therefore, the magnetc feld flux n loops 1 and 3 s out of the page and ncreasng,.e. dφ s also out of the page. dt Induced EMF n loops 1 and 2 s then ε = dφ,.e. clockwse, whch wll result n clockwse currents (n both loops). dt (Algn your rght hand thumb wth the drecton of dφ and curl your fngers to get the drecton of nduced EMF.) dt A col of area A wth N turns rotatng wth angular frequency ω = 2π f n magnetc fled B wll generate an nduced EMF ε(t) =ε max sn(ωt), where ε max =ANBω. If the col has resstance R, the current s then (t) = ε(t) =ε max R R sn(ωt) = max sn(ωt), where max = ANBω R. RMS value of current s rms = max 2. The energy stored n an LC crcut s conserved. Hence: q2 2C L = q 2 max 2C. (At the tme when the charge s maxmum, the current must be zero)

9 The decay tme constant n an RLC crcut s τ = 2L R. If both L and R are doubled, the decay tme does not change. (The nformaton on capactance C s rrelevant.) The ampltude of current n an RLC crcut drven by an external EMF s ε max max = X, where X = 1 R2 + ω d C ω L d 2 If a capastor C s connected to an external EMF of freqency f, RMS value of current n the crcut s rms = ε rms X C, where X C = 1 ω d C = 1 2π fc US rms =ε US rms 2π f US C UK rms =ε UK rms 2π f UK C Hence: US rms rms US f US ε UK rms f ε = rms = 0.6 UK UK