PHY 122 Prelim II (2005)

PHY 122 Prelim II (2005) - n ' ' «J .- PHY 122 MARCH 24,...

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Unformatted text preview: n ' ' «J .- PHY 122 MARCH 24, PRELIM II a ;- . g 2005 J/"' ’ NAME: ‘KU;' §_5“w‘ fig RECITATION SECTION: ’7 “I. READ EACH PROBLEM CAREFULLY. is being asked, If you do not understand what raise your hand for assistance. II. SHOW YOUR WORK. Present your answers as a decimal number, with appropriate units. 1. The batteries in the multiloop circuit below have zero internal resistance. lJm A 2.00A 20.0 V 7mm Q a) Calculate the current in the 5.00 0 resistor. ‘b) Calculate the power supplied to the Circuit by the 20.0 v battery. //' Q : £3 i 20 r: c} Calculate the value of the emf e2. ‘ f. 1 ELC> “r r?r-f l: 2. The resistance of a galvanometer coil is 25.0 Q and the current required for fullnscale deflection is 3.00 mA. a) Show, in a labeled diagram, how to convert this galvanometer to an ammeter. (fh b) Calculate the value of the resistor needed to convert this galvanometer to an ammeter which can read_50.0 mA at full—scale. "E/ 'Y (“l "mgr; ‘. - " ' {(4.251}; 'L—J— \ r I 3/?“ ' i c ‘- r‘ "N, r - | -. ‘- a ? Li 3. A positively charged particle with velocity 9 A V = 6.00 X 103 m/s i enters a velocity selector, a region of uniform electric field and uniform magnetic fieldn The uniform a A _ :”"\ r i. magnetic field is given by B = 1.50 T j. +igy ' s a) For the particle to pass through the velocity selector undeflected, the magnitude of the electric field must be (g) 9.00 x 103 N/c. i if? a: 1'!“ if; B. 4.00 x 103 N/C. {rqu‘ri r“/ c. 1.11 x 10’4 N/C. l - D. not enough information given, need to know the mass of the particle. {R} B c D (CIRCLE ONE) b) For the particle to pass through the velocity selector undetected, the direction of the uniform electric field must be A A. + i A B. - l i, A f C. + j 5 j A L,- D. — j A 0 E. + k A F. — k G. none of the choices given. A B c D “#3” E ‘ G (CIRCLE ONE) 2.00 cm W 3AM A 4. A rectangular loop of wire, 2.00 cm by 6.00 cm, lies in the x—y plane and carries a constant clockwise current of 3.00 A, as Shown. The loop is in a uniform magnetic field in the positive a A y direction, given by B : 0.25 T j. i a) Calculate the force on side AB of the loop. I) J EH | f} 2 a r . “‘3 j J a. . L magnitude: '0 jg “J direction: +X ,—x +y -y {12: —z other {CIRCLE ONE) XJ/ b) Calculate the force on side AD of the loop. 2" [M 0 J ' M magnitude: 1 I) direction: +X ~X +y —y’ +2 72 other (CIRCLE ONE) c) Calculate the magnetic moment of the loop. out)" ‘ ' *-. r f“ ’3. 4’: - ‘t. magnitude: v09 In /%;-5AW‘ rrfx direction: +X f?;\ +y —y +2 (:E9j other (CIRCLE ONE) d) Calculate the torque on the loop. [V H“ /[A i: L) “E if- ,b‘“ - I lira 13*.rfllgs <:>\M magnitude: q ¥ J~ ' jfiefi l L’ direction: —x +y fly +z other (CIRCLE ONE) \_7_,_/' +X 5. Two long straight wires are perpendicular to the page and located on the x—axis as Shown. Wire 1 carries current I1 : 3.00 A out of the page and wire 2 carries 12 = 3.00 A into the page. a) Calculate the magnitude of the magnetic field at point A, 10.00 cm from the origin, 0. ;__um‘ i L g r A .. U W - [/ \‘ *2»! b) What is the direction of the magnetic field at point A? +X -X +y FE? out of page into page (CIRCLE ONE) 5 \‘ k c) Calculate the magnitude of the force per unit length exerted by wire 1 and wire 2. t? ‘ fl, d) What is the direction of the force by wire 1 on wire 2? 1 2%) —x +y —y out of page into page (CIRCLE ONE) \ 6. A slidewire generator consists of a conducting rod, length 12.00 cm and resistance 5.00 ohm, that makes contact at points A and B with a U—shaped conductor of negligible resistance. This device sits in a uniform constant magnetic field of 0.80 T, directed into the page, as shown. The conducting rod is pulled to the right at constant speed, 0.25 m/s. x 5 x x F x 12.00 cm “‘ K B a) The magnetic flux in the conducting loop is //////» A. zero. B. constant. C. increasing. D. decreasing. U, A B D (CIRCLE ONE) ._ ff 1”" b) Calculate the value of the current induced in the f Conducting loop. _r n’J—) “ ._ "' ;:(y% 13' ” f— . t _ ” '1’ p . r r, v r k r ‘ —4” JV“ ; s- i c} The current induced in the conducting loop is clockwise and constant. clockwise and increasing. K r. 1 clockwise and decreasing.,*' // j counter clockwise and censtant. ml counter clockwise and/increasing. counter clockwise/and decreasing. \ ~ /- *ljf-UUOUJP F (CIRCLE ONE) V 03 ('1 d) Calculate the magnitude of the force required to pull the conducting rod at the constant speed of 0.25 m/s. 1 \ f a /i! j l ‘. 7. BONUS QUESTION (5 Points) The figure shows three long straight wires, each with current I, two parallel (up) and one antiparallel (down). Four Amperian paths are labeled A, B, C and D. Rank these paths —>-> according to the value of @B-dl for each. Rank greatest to least. *** IMPORTANT STUFF *** —> a a #1 F = qv x B B = O 4hr —>—) $8 : B. B : nonl a —9 -—3 F:les 32% 2a a a —9 ——) ,1 = IA SEE-d1 : #01 m -3 —) w) or," T=#XB {1:7 YB dt 21 = O ‘ e = VBL Z‘VZO , B Re—R1+R2+R3+ gfiE-dl:—~—e * dt 1 l l 1 : —__ + 5—" ‘ “” + P = El Re R1 R2 R3 P=I2R h Tm uh = 4n x 10" l__ ...
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This test prep was uploaded on 04/19/2008 for the course PHY 122 taught by Professor Smith during the Spring '06 term at University of Maine Orono .

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PHY 122 Prelim II (2005) - n ' ' «J .- PHY 122 MARCH 24,...

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