PHYS 232 Exam #2 04-01-05 50 minutes. This quiz has 8 questions. Please show all your work. For multiple choice questions, circle the correct answer. If your chosen answer is not correct, you will get partial credit based on your work shown. You are allowed to have one hand-written, one-sided page of equations. There are six short questions, each worth 7 points, in part I, and there are two long questions each worth 29 points in part II. PART I 1. A straight wire is bent into the shape shown. Determine the net magnetic force on the wire when the current I travels in the direction shown in the magnetic field B. 2L .oC (DF 900 ------f--_n 900 L F::-O __n_t_n-- a. (f) c. d. e. 2. _;~t(n~ , I f$ ) L (aJ.h 'D) fa jL ! -----------i 'F 2IBLin the -z direction 2IBLin the +z direction ;:. - 4IBLin the +zdirection 4IBLin the -z direction zero r . ~LLt )( B C> r.-\~:J I.2.L-S (.ti) A deuteron is accelerated from rest through a 10-kVpotential difference and then to a uniform magnetic field with B = 1.6 T. What is the radius of the resulting circular path? (deuteron: m = 3.3 x 10-27kg, moves perpendicularly q = 1.6 X10-19 C) a. 19 mm b. 16mm €> d. e. / CVIS 13 mm 10mm 9.0mm . 1) t. R= -t .r; /' ~ 1 J2-~6 ~ ';;: I ..3 ). ~ I )(. ( 0- 2- (p ~(' -- Jt)4 )l 3'3 Y.{O - XfD '3 fIt\~ { '1 3. The segment of wire (total length = 6R) is formed into the shape shown and carries a current 1.What is the magnitude of the resulting magnetic field at the point P? ..."~p @ 4. 3/loI 8R b. 3/loI 2R c. 3/loI 4R d. 3/loI 2R e. 3/lo" I 8R B r: ~IJ fu II 0\,+ ,. 0-" ,ck LJ kP- + Equal currents of magnitu e I travel out of the page in wire M and into the page in wire N. Eight directio indicated by letters A through H. B . c ~ J~ .Jl Y:> f{ blV~t A @N H- E . G The direction of the magnetic field at point P is a. A. b. B. -7> @ d. e. C. D. E. 2 'F 5. A bar magnet is dropped from above and falls through the loop of wire shown below. The north pole of the bar magnet points downward towards the page as it falls. Which statement is correct? l N , ,\VJCt. ) ~~ I a. b. c. @ e. 6. a.r.,'vU( ,. t'~ S ~'IV1dM.~d) l'(I\~c.LJ The current in the loop always flows in a clockwise direction. The current in the loop always flows in a counterclockwise direction. The current in the loop flows first in a clockwise, then in a counterclockwise direction. The current in the loop flows first in a counterclockwise, then in a clockwise direction. No current flows in the loop because both ends of the magnet move through the loop. A 10-f!Fcapacitor in an LCcircuit made entirely of superconducting materials ( R = O.Q ) is charged to 100 J.1c.Then a superconducting switch is closed. At t = 0 s, plate 1 is positively charged and plate 2 is negatively charged. At a later time, Vab= +lOV. At that time, Vdcis b J- c tf bVI.- , a b. c. d. o 3.54V. 5.0V. 7.07V. -I Vdt 6V' z lbV \ I ---- IOY "-- 10V. 3 0 - 6 \iL- V L..61{" a. ov. &VL,'=- +- ) -) WUL (6 v 1 PART II 7. In the figure, the current in the long, straight wire is 11= 1.00A, and the wire lies in the plane of the rectangular loop, which carries the current Iz= 2.0 A. The dimensions are c = 0.100m, a = 0.150 m, and t = 0.50m. CD (J) @ a. Indicate the direction of the magnetic field due to current l1, at the locations of the four sides of the conducting loop (with arrows, arrow heads, or arrow backs) b. Calculate the magnitude of the magnetic field due to II at the sides A and C using Ampere's law. c. Indicate the direction of the force due to 11exerted on each side of the loop (with arrows, arrow heads, or arrow backs) and calculate magnitude and direction of the net force exerted on the loop. d. Now consider the situation where the loop is slowly turned around its long axis by 90°, so that sides A and C are equidistant from the long wire. Describe the forces, the net force and the torque on the loop (no calculations are necessary). Fb (1) DI@ B B:; )'''\> I :..--- ;l.7C~ .. ,.,. " J t' @ ~'GLL 0 L~ LiCe II! :::. J.. '1./ D- t, B! I" 8c r-c FA A r'u[-- :; .~ B FA , S\,,~ (~.J 4 B $< ~ ch fh.t I~ ' S Clb~ L @)< I 0-1- )( 'B A; 'eM\. (Q.f () -e.C(c t,c.rff'-lY s I.i S M ~ t' I~~ , c-J,f. c~ tR AA (-to B L cRJ ~ h.f1) o o '0 /Vuf F 0"'-~ ~vu.. '"-=- ]2A :J...>< ~) ~t. ~ {-l (B D .5 ><1.;;)..)( ( 0- , - -~ -1"2..YID II -15e ') -1-. t-J Va JoJ~} tt\-t "'~Qr" o . , , e .~ , vi- f '[k ==) A -toy 'tf'IA. ,\-No N IL-~fo t"'- x to Lv\. '( cJL rt:...t.. F0 L-"" vUhvt-s rtu. C-C tAl - d cP . '1- cL' rt uti .,-eL ~ . U\-j (et-,,<'I-\ , , 8. A long solenoid has 5000 turns per meter and has a radius of 10.0 cm. It carries a current given by 1=(1.0+ 100t)A , where t is time measured in seconds. Inside the solenoid and coaxial with it is a coil that has a radius of 6.00 em and consists of a total of 250 turns of fine wire. (J) a. Calculate the magnetic field inside the solenoid at t=O. b. Calculate the emf induced in the coil by the changing current. c. As indicated in the figure, the current in the solenoid flows in the @ @ clockwise direction, what is the direction of the current induced in the coil (CW or CCW)? Explain how you arrived at your answer. d. The maximum current is reached in 2 seconds, after that the curre~t in the solenoid decreases as given by 1=(lmax -100t)A; the direction of this current is still clockwise. What is the direction of the current induced in the coil now (CW or CCW)? Explain how you arrived at your answer. 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