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106-20082-MT2

# 106-20082-MT2 - PROBLEM{1(25 points A copper wire with...

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Unformatted text preview: PROBLEM {1) (25 points) A copper wire with length L : 0.30 in has a varying thickness so that the area of the cross section of the wire at a distance x from the left end is given by AU) 2 10"] (20.: + 0.50)'1 m2 where x is in meters. A current of I : 0.40 A enters into the Wire through the left end. Assume that the current is distributed uniformly over the cross sectional area along the length of the wire. The resistivity of copper is ,0 = 2.0 X 10'8 gm. The charge carriers of copper are free electrons (charge q = —l.6 >< l0“lg C and density n = 9.0 X l028 Iii—3). (a) (10 pts) Find the electric field EU?) at a distancex from the left end. ............................................................... M Etx) : 0.80 x10‘2(2.0x+0.50) Wm Definition Of “:51“in is T The electric field as a function ofx is z; ______.—_.___ 9:7 ; lEU):8.0x10_3(2.0,r+0.50) Wm] (Ans) EU) : pﬂx) 2,0 I . Where x is in meters. A(X} Substituting the given values, we find 0.40 A 10'6(2.0x+ 0.50)’I £(,r):(2.0><10‘8 Q.m) (13) (i0 pts) Find the resistance of the wire. ................................................................................................ ——30i”[i0“1 V: 8.0x10’3(0.50) v Potential difference as an integral of 2; ' 54 _ w V=4.0x10‘3V V — V : 5-47: I 0 t l. laECOSW ; Ohm's law V: [R gives The potential difference between the left and right ends of V 4 0 X1073 V . 1?: — = —'——- the copper'wire is ; I 0.4-0 A V:ﬁfttjair:fg'SOSDX10'3(2.0.r+0.50)cir 1 ’ If‘=1.0><10'2 t2 (Ans) 0.50 V: so x 10—3 [(162 + 0.50x)]0 (c) (5 pts) Find the drift speed of the electrons at the left end (x = 0) of the wire. ................................................... Solution: 1 So. The current densny lS given by J: may. 80, the drift speed . J05: O) : 6 0.40 A—l 7 : 0.20X106 A/mz - 10' (0+05) m' is equal to .. . V a = Al“: ‘ The drift speed is .12 . ‘ ' 0.20x106 Arm? 1.0x10’4 , t/L?:4-~—————L 28 16 10719 C :-——-—7,) 11115 When x = 0, the current density is 0‘0 x 10 N. - X 'J “‘ f l ' Vd:l.4><l0W5 mt’s (Ans) . ﬁxzmz— . A(.r=0) l Phys i06 Second Midterm Examination Page 3 Saturday, May 02, 2009 PROBLEM (2) (25 points) Consider the circuit shown in the figure. [1 : 0.30 A a 13 (a) (10 pts) Find the current [3 and the emf 82. Solution: To determine [3, we apply the loop rule to the outer loop. Starting at the lower left corner and going clockwise, we find Eliﬁﬂl+j3ﬂ3"g3:o 1 _ _L 7 i ”j [3 :w : 80+? ! To determine 82, we may apply the loop rule to any of the J loops, at the right or left, of the given circuit with the given I“ I "0-50 A (A113) directions of currents. Starting at the lower left corner and J The negative value of 13 tells us that its direction is opposite going clockwise through the loop at the left, we find to the direction given rn the figure. : 81 _ [1R1 , 12192 _ 82 : 0 In the given circuit, applying the junction rule to point a, the 8 _ 8] _ [1R1 _ AR? 2 c _ o current 12 111 R215 52 : 30 _12 _{_020)(30) 122134— 132030 A—O.50A=—0.20 A. _._ . i E = 24 Vi (Ans) The negative value of [2 tells us that the current [2 in R2 flows |2_._J ' upward. t (b) (3 pts) Find the potential difference Vab. ................................................................................................ Solution: ‘ Taking the path at the right To find Vab, the potential at a with respect to b, we start at b Var: I 53 ‘13}?3 : 8-0 ’(‘0'50X'2m 50 i Vaib =18 V (Ans) and add potential changes as we go toward a. There are three :. Taking the path at the middle possible paths from b to a; taking the path at the left th = 82 + [2}? = 24 + (-0.20)(30) so 5 Vab =18 V (Ans) Va!) : 817/1131 = 30—(0.30){40) so th =18 v (Ans) ‘ (c) (6 pts) Find the total rate of energy supplied by the three emf devices. ............................................................ Solution: terminal of the 8.0~V battery. So, its power output is The currents [I and I2 run from the negative terminal to the P3 = 83/3 : (SUN—050) 30 102 Z _4-0 W positive terminal of the 30-V and 24~V batteries. So, their; The negative value of power P3 means that the 8.0—V battery power outputs. respectively, are is being charged by the other two batteries. P1 2 6111 = (30M)- 30) 50 Pt : 9-0 W : The total rate of energy supplied by the three emf devices is 3 z 5212 : (24)(0.20) so P2 = 4.8 w ‘ P: P1+P2 +103 : 9.0+48 —4.0 so P: 9.8 w (Ans) The current [3 runs from the positive terminal to the negative ' (d) (6 pts) Find the total power dissipated by the three resistors. Compare your answer with that of part (c). .................. Solution: P3 = @2133 : (050)3(20) so P} = 5.0 w A resistor always dissipates electrical energy. The power. The total power dissipated in the three resistors is dissipated in a resistor is P = FR. So, the powers dissipatedi P: Pr +102 t P3 : 3.6+l.2+5.0 30 w (ANS-l in the three resistors, respectively, are The total power supplied by the batteries is equal to the total _ 2 _ 2 , . _ a I _ _ Pa “ f1 fh ’ (0-30) (40) 50 Pi “ 3'6 W power dissipated by the three resistors. This 13 an expected w 2 _ 2 a ‘ g . ‘1 If)? _ [3}?er (0'20) (30) 50 Pl _1‘2 W . result becauseof the law of conservation of energy. Phys 706 Second Midterm Examination Page 4 Saturday. May 02, 2009 PROBLEM {3) (25 points) Consider“ the circuit shown. The capacitors are initially uncharged. The switch 5' is closed at time I: 0. s00.) (Take 8 : 2.72; 62 : 7.40; e3 (a) (4 pts) Find the final charges Q; and Q2 on the capacitors C1 and (:2. respectively, when they become completely charged at time t: 03. Solution: The voltages across C, and C3 are equal to E : 20 V when they become completely charged. So, the final charges are Q1 : C8 = (1.0 x10‘6)(20) EQ :210x10r6 c=2011c (Ans) (b) (8 pts) Find the charge ql on the capacitor on the left at time t = 6.0 ms. Solution: The time constant of the circuit is an R1 + [P2 1'21?qung ((31+C'2) 7.0 T=(§—5— kQ){3.0 ,uF) so 7:2.0 ms =2.0><10‘3 s (c) (8 pts) Find the instantaneous current through R1 at time t: 6.0 ms. Solution: The total charge on the capacitors at time t is a +4): =tQi no )0 a a”) The instantaneous current i in the main branch (in the battery) is just the time derivative of the above equation. So, 0’ d . z'=— + 2— + 1—67” dzta a) deg :21 )( 1] 60;,tC . . Qt +92 if =(30x10’J f 18*” :t 230%) 1':( 2.0 ms (d) (5 pts) Find the instantaneous rate of energy dissipated by R2 at time 1‘: 6.0 ms. Solution: The junction rule gives the current 1'; through R2 as rig : f—Ji :l.5—l.0 : 0.5 mA The instantaneous rate of energy dissipated in R3 is P: i; 2 :(0.5x10'3)2(2.0x103i Phys 106 Second Midterm Examination l l l Page 5 [3121.0 kg}? R3 : 2.0 k-.. E: 20 V C: = 1.0 #17 ,3 = 210 yF 5 Q2 = €35 = (2.0 ><10‘6 )(20) l Q2 = 40x106 c : 40 “Ci (Ans) The capacitor C1 is charging. The charge ql on C1 as a ’ function of time is g] = Q16 e 57’”). So, 9,1 2 (20 ”(2)0 _ 845.0 msx’2.0 ms) ~ 1 a =(20 #CXI—e” :(20 uC)(t—%) 41:19 ,uc:19><10~6 cE (Ans) ’ :'=I.5 mA The instantaneous current 1'] through R; is R2 2.0 = r=—(1.5 mA) at +82 i.0+2.0 ’1 1‘1 : 1.0 mA = 1.0 ><1{)'3 A ' (Ans) Notice: Using the result of part (b) you can find the instantaneous voltage at t z 6.0 ms across C1 and subtracting this voltage from the emf of the battery you get the voltage across R}. Then ,using Ohmls law you find i1 through R1. . l—1::0.50><10‘3 w:0.50 mW (Ans) Notice: The voitages across R1 and R2 are equal at any .instant. Using the result of part (c) you can find the instantaneous voltage at t 2 6.0 ms across R2 and use this voltage to find P dissipated in R3. Saturday, May 02, 2009 PROBLEM (4) (25 points) The rectangular coil abcd shown has N = 2000 turns of a wire which carries a current of I : 0.50 A. The coil is initially oriented in such a way that its magnetic dipole moment is given by ,1] :tt(e0.80f +0.60j) where tt is the (positive) magnitude with units Anni. A uniform magnetic field [9: (4.0? + 3.0)) r is present everywhere in the space. (a) (8 pts) Find the magnetic torque, in unit vector notation, on the coil in the initial position. .................................... Solution: i The torque exerted by the magnetic field 3’ is The magnetic dipole moment has magnitude ti = NIA. So, . — = a x 3;: (_12}+ 90;.) x (4.0; + 30;) y = (2000)(0.50A)(0.10 n]><0.15 m)215 A1112 I f2e36{;’><}')+36(}'XE)=—362—362 The magnetic dipole moment (vector) [A is —n- .~ E: —72kN.m (Ans) it =(1sirai.802+0.60}) A.m2 a : (712? +9.0}3 Amz (b) (8 pts) First show the direction of the current on the given figure and find the magnetic force, in unit vector notation, on the side be of the coil in the initial position. Solution: , The magnetic force on a current can’ying Wire is 1:7: [I x 15’. The right-hand—rule determines the direction of the magnetic The magnetic force on the side be (with N wires) is di ole moment 4 of the coil. and ﬂ is e endicular to the . 4 - u p M y p rp : F Nfle plane of the 0011. The current I in the corl 18 shown on the where 2: 4115211]. So, given figure. A A A a F (2000){0.50)(—0.15km) >< (4.0i—i— 30/) I! I? (—150/Em) X (4.01% 3.0)) = (—600}+4502) IV: (45021 600)) N (Ans) (c) (9 pts) When the coil rotates untii it is in stable equilibrium position, find the change in the coil's potential energy, AU. between its initial and final positions. SO_1UUOJ ‘ The coil‘s initial potential energy is The potential energy of a coil in a uniform 6’ is U-=-#,-’1§:*(i12;+9-0}3'(4-0;+3-0h U: 1113’: —tt15’cos o where (D 2 angle between )7 and E". [4:48 J72? I:21 J The coil has stable equilibrium position when its final; The change in the coil's potential energy is magnetic dipole moment is directed parallel to the magnetic field 3’ So, when 1;“) : angle between [if and 3’ = 0, we ‘ AU: Uf_[/i:_75 JTEI J have . i AU: 796 J (Ans) a. : —i1f-8:W¢ = we we 60:45 A.m2){5.0 T):i751 ,, _ Phys 106 Second Midterm Examination Page 6 Saturday, May 02, 2009 ...
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106-20082-MT2 - PROBLEM{1(25 points A copper wire with...

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