Solutions to Midterm Physics 201


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1 Solutions to Midtem Physics. We cn conside this sitution s supeposition of unifomly chged sphee of chge density ρ nd dius R, nd second unifomly chged sphee of chge density ρ nd dius R t the position of the cvity. At the cente of the big sphee, Guss lw tells us tht the field due to the lge positively chged sphee will be zeo, nd thus the only contibution to the field comes fom the negtively chged sphee. Clling up the +y diection, we thus get Q E = R e y () 6 3 = π R 3 ρ e R y () = ρr 6 e y (3). At the time the pticles ente the mgnetic field, they hve enegy E = qv (4) nd thus momentum mv = me = mqv (5) Once in the mgnetic field, the pticles tvel in cicul obits of dius = d. Equting the mgnetic foce with the totl centipetl foce thus gives mv = qvb (6) mv = qb (7) = mv (8) qb mqv = (9) qb d = mv () B q 3. We know tht the potentil V b t the sufce of the inne sphee is given (up to constnt) by V b = Q b ()
2 nd the potentil V t the sufce of the oute sphee by V = Q () Fom the definition of cpcitnce, we find C = Q V (3) Q = V b V (4) = b (5) b = b (6) The totl enegy stoed in the cpcito is given by U = E d 3 (7) But we lso hve = = Q 8π b b = Q b 8π b Q ( ) 4 d (8) d (9) () U = Q C = Q b 8π b () () Thus we see explicitly tht the two fomuls fo enegy give the sme esult. 4. We cn conside this wie s supeposition of n infinitely long stight wie with cicul wie. At the cente of the cicle, Ampee s lw tells us tht the mgnetic field B due to the infinite wie is diected out of the pge nd hs mgnitude B = µ I π (3) Similly, the BiotSvt lw tells us tht the field B fom the cicul wie is diected
3 3 out of the pge, nd hs mgnitude B = µ I = µ I = µ I dl (4) π dθ (5) (6) The totl field is given by the sum B = B + B, nd theefoe hs mgnitude B = µ I + π (7) nd is diected out of the pge. 5. The mgnetic moment of the cuent loop hs mgnitude µ = ILw (8) nd is pointed downwd. Thus, the toque on the loop due to the mgnetic field is diected towds point A nd hs mgnitude τ B = ILwB (9) In equilibium, this must equl τ m = mgw, the mgnitude of the toque due to gvity. Equting nd solving fo B, we find B = mg IL (3) 6. Ampee s lw tells us tht t hoizontl distnce x fom the wie, the mgnetic field hs mgnitude B(x) = µ I (3) πx nd is diected into the pge. The mgnetic flux though the loop thus hs mgnitude Φ= +w L = µ IL π +w = µ IL π ln + w B(x)dydx (3) dx (33) x (34)
4 4 nd is diected into the pge. If I(t) =I cos ωt, then the mgnitude of the EMF E induced in the loop is given by E = dφ dt (35) = µ LI ω ln + w sin ωt (36) π which hs mximum vlue of E mx = µ LI ω π Finlly, the mutul inductnce M is given by ln + w M = Φ I = µ L + π ln w (37) (38) 7. Using the BiotSvt lw, we get tht the field due to the two stight segments is zeo. Fo the segment of dius, we find B = µ I = µ I = µ I 4 dl (39) π dθ (4) (4) diected into the pge. An identicl integtion gives B b = µ I 4b (4) diected out of the pge. Summing togethe these contibutions, we find tht the totl field is thus diected into the pge nd hs mgnitude B = µ I(b ) 4b (43) With the numbes given, B = µ A 8 =3.4 7 T (44)
5 5 8. The enegy dissipted by the esisto is, by consevtion of enegy, equl to the diffeence between the enegy E i stoed in the cpcitos in the initil configution nd the enegy E f stoed in the finl configution. Initilly, ll chge is on the cpcito C, nd thus E i = C V (45) To find E f, we use the fct tht cuent ceses to flow when the voltge coss both cpcitos e equl. By consevtion of chge, this occus vi tnsfe of chge q fom cpcito C to cpcito C Since C is initilly unchged, it will hve finl chge of q, while C will hve finl chge of C V q (46) Equting the voltges, we find Thus, the finl voltge coss C is given by C V q = q (47) C C V = q + (48) C C q = V C C C + C (49) V f = q C (5) = V C C + C (5) which, s we know, lso equls the finl voltge coss C. Thus, E f = (C + C )Vf (5) = V (C + C ) C (53) (C + C ) C = V (54) C + C
6 6 Thus, the esisto dissiptes s desied. W = E i E f = C V V (55) C + C = V C C (56) C + C = V C C (57) C + C = V C C (C + C ) Note tht we cn find E f n ltente wy, by elizing tht since the voltge on the two cpcitos is the sme fte they dischge, thei equivlent cpcitnce is C (58) C eq = C + C (59) nd thus by chge consevtion E f = Q C eq (6) V C = (C + C ) (6) 9. When the switch hs been closed fo long time, the cuent is no longe chnging nd thus the inducto cts s shot cicuit. Thus, no cuent flows though R nd the cuent leving the bttey is given by which is lso the cuent though the inducto. I bt = V (6) If the switch is suddenly open, the only complete cicuit is the loop contining only the inducto nd R. Since the sum of the voltge ound this loop must be zeo, we find Solving this eqution, we find L di dt R I = (63) I(t) =I e R L t (64)
7 7 Since the initil cuent is I =I bt,weget I(t) = V e R L t (65) The initil voltge coss the inducto is then V () = L di() dt = R V (66) Fo this to be less tht V, we must hve R < (67)
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