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PHY 122 Prelim II (2001) - M l.W:ngnn“ V nrxlh a 2001...

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Unformatted text preview: M l .W. :ngnn“ V nrxlh a, 2001 - RECITATION DIVISI X: g > SHOW YOUR WORK. NUMERICAL ANSWERS SHOULD BE COMPLETELY PROCESSED. K ' :HSTRUCTIONS : l. The two batteries in the multiloo p circuit have zero internal resistance. -<—- 2.00 A "_\° 3.00 A m—a. «<——— 5.00 A a. Calculate the current in the 3.00 O resistor. J::8Ma f:— b. Calculate the value of resistor R. (SHOW YOUR WORK.) “Qafm —Hn_-?>n+(o.n_ 68:0 f “211 “(2V + 30‘, :O , W“ C. Calculate the value of emf, 3 (SHOW YOUR WORK.) . 8+ small)” essay) = C3 d. Calculate the electrical resistor. P=I£R P=(sg)2.(g.l= ISOW \S‘ Qower dissipated by the 6.00 O u guiuqflU...¢L¢i \c: Densitive current. illuludulng QEVlCE, w1tn resistance 15.0 ohms and full scale current 1.00 milliamp) and a shunt resistor. ‘ a) Calculate the voltage across the ammeter when it reads 50.0 milliam s. P 7 {seem ’ / / “ma —> El: 1mg @ \ ,7 .— \“-.\ mp1 4L“30b “3k .OOIRG '5.n_.= do‘fiv (L. b. Calculate the value of the shunt‘resistorff”_fi_—K 3. A proton, m : 1.67 X 10’27 kg, q z 1.60 x 10‘19 C has speed 5.00 x 106 m/s in the +X direction at point A, as shown. The I }fiyéfi proton moves in a circular path, in a uniform magnetic field \{m {A of 0.15 T. {"fi‘d a. The direction of the magnetic V ' XRJ field is A. the +X direction. B. the —X direction. C. the +y direction. the —y direction. (ii) out of the page. WA . into the page. J G. none of the abov choices. 0\3 ANSWER: E b. Calculate the radius of the circular path of the proton. "fivz. __ -B :: ,27 ' l gnv 2L F: E _[.Q><\O N \JJXto #3 (5K0 3) 2e; F b . 15-”: fiwxm :fiVB :\.bmo'qc.-5MD M's ‘1 I Lme'BN ‘EeHéB /’ . UL 4.uu a, counLer ciocKWise. The loop l;es in the xy plane as shown. +y1 B o c (9% a. Calculate the magnitude of the magnetic moment of the wire loop/£12164 2 Q‘OOH‘ 6‘wcm.8 00¢vi b. The loop is in a uniform magnetic field that points in the +y direction. What is the direction of the tor e on the The ~x direction. . The +y direction. D. The «y direction, E. Out of the page. F. Into the page. G None of the above choices. AhlSWER: E A Circular conducting loop of resistance 4.00 ohm and area 0.300 m2 is in a uniform magnetic field directed into the plane of the page, as shown. loop? _ r. The +x direction. 7:\“/fi'*E , Q: Lin. A: 0.20mi X X X X a. The magnitude of the magnetic field is given by B(t) = 1.50 + 0.10 t2 where B(t) is in Tesla and t is in seconds. Calculate the magnetic flux in the loop at t z 2.50 S. b. Calculate the induced emf at t = 2.50 s. :O‘EB _~Q\(\.ee+.\tllfi d(l.50A*-‘Rffl k‘ ‘flp’li’fi,i_i 333. at ‘ xv mM/fm . 4,45 / : “Zmlt ii J.vv cm anu reSiscance lD.U onm, that maKes cc::acc at points A and B with a U—shaped conductor of negligible resistance. The device sits in a constant uniform magnetic field of 0.60 T, directed into the page, as shown. The conducting rod is pulled to the right at constant speed, 0.30 m/s. 5.00 cm 0.30 m/s E k I conducting bar a. The magnetic flux in the conducting loop is A. constant. = increasing . EBA '. decreasing. D. 08 '1lating. E. ro // S ANSWER: b. Calculate the induced voltage (motional emf) across the sliding rod. e : m .i €~VBL .3 Ignogm» (m - .063le a ‘ / C. The induced current in the conducting loop is A. clockwise and constant. B. clockwise and increasing. C. clockwise and decreasing. D. counter clockwise and constant. /// [E. counter Clockwise and increasing. 5 F. counter clockwise and decreasing. ANSWER: 1:) d. Calculate the magnitude of the force required to pull the J”:];B” sliding rod at the constant speed of 0.30 m/s. ' 2. 1* 130°} g i'S-tl'w— .( / " , x “we 15:1th :(omb‘la'fiémgw Lame“ A closed curve encircles three current carrying conductors. The line integral around this curve —) a 9913 ‘dl is equal to 3.63 x 10'4 Tm. Calculate the net current in the conductors. $8 '011 3/1 1cm fl . -11 _ I ‘ g'togx‘o Tm ‘flojewc "/ ~.\ \ 1 ~ \\ z.\\\l **** USEFUL STUFF **** —+ —v -o #0]: F = qv x B B = Zn —0 —~ (pa = B.A B = ponI —+ a —v [LON]: F z 11 x B B = 2a u = IA —v a _. _. _. §Bod1 = MOIenc r = u x B d¢8 )3: = 0 a = — ——-— dt EV = O 6 = VBL RE=R1+R2+R3+ -¢ -o d¢a fiE‘dl = - —- dt 1 1 1 1 __=._.__.+.—-—-+—-.—+.., P=eI R R, R2 R ° 3 p = 12R ...
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