# Exam 2 - UV? 7:6&amp;quot;? PHYS 1214 SECOND HOUR EXAM...

This preview shows pages 1–9. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: UV? 7:6"? PHYS 1214 SECOND HOUR EXAM SPRING 2011 1111 a [wax 9 M f“: '- ‘ 3 ’ Laboratory Section Number SIT IN A SEAT SO THAT YOU HAVE ONE EMPTY SEAT ON EITHER SIDE! This is a closed book exam. You should have only your calculator, two pencils, and the exam paper at your seat. The last two pages of the exam are blank to provide scratch paper. The test consists of 7 multiple—choice questions/problems (10 points each) and two problems to be solved (40 points each). USEFUL FORMULAS AND CONSTANTS (Vectors are indicated with arrows) K; 2 Ad) P:;</\g..| {\leR F=k 1 g: NJ B=“_0£ "J 2 At 2'lT l' r l, 1 =E°£ 1 =2 d =B-€- X :wL X =— \/;b rev Tl' ra s E 7 [V L C NC ” IE I48 1ev=1.6x10'19J E = —. 1:3 w=21Tf At q R. k _1_:__1._+__1_+_i. Ezﬂ IREC‘I:R1+[22+IQ3 E:V.B 1 9 N'm2 12 02 19 k = E 9x10 so = 8.85x10’ e=1.6x10‘ c z' = Icos(wt) 4rr€o o2 N-m2 p Z - I - I ‘4 p N2 A _ =—°—1—2 F:Ij-<;z;-B-sin(cp) L: 0 §= N-B-A-w-srn(wt) 2n r ‘ 9* ‘ E ,ﬂ‘w . l VS NS |JO _ -31 :Eﬁzq-v-B-smw) ——=———=—-— B: me—9.11x10 kg p0 =4TrX1OT7T‘m/A (DB =Bi-A=B-A.coscp _i _ﬁ I rms “f “L “"75: PART ONE: MULTIPLE-CHOICE(1O POINTS EACH) /- s 1. Along, straight wire produces a magnetic field of 1.1 x 10'3 T at a distance of ~ 0.020 m from the center of the wire. The current in the wire is j/. .5 j .1 \E’i‘x”. Q, ‘ //:>l...xx (a) 25A f (b) 110A j (c) 405A (d) 220A V’Max (e) 79A 9,; M43 V 2. An electron moves at right angles to a magnetic field of 0.50 T. If the force on it Q is 8.9 X 10'15 N, the speed of the electron is t/ 4.7 x 104m/s (b) 7.2 x 104m/s (c) 2.8 X105 “1/3 .7 (007-3 X 106 "V5 (e) 5.3 x 105 m/s (a) (e) (a) 120 v (b) 2800 v (c) 3600 v {éo 9:» 9332 mg & % K A 30.0 cm X 60.0 cm rectangular circuit (see below) contains a 15 Q resistor and is perpendicular to a magnetic field that starts out at 2.65 T and steadily ' decreases at 0.50 T/s. The induced current read by the ammeterpis g 6.0 mA (b) 3.0 mA (c) 14.5 mA /‘/(d) “55.3 mA J g: 1.8 mA WWW” ‘ ‘9 5:“; ::>;.:;. 3 6—600 cm%ll 30.0 .,- g V: j: «g; _ g V r C Q“ . l_ gr, 0 9;; ‘><: ﬂ: 7 CwyvighlazlJWPemsonEﬂuuawan:pubﬁshingus y nWeslBy w, \/ I r 4:21 . A step down transformer has a voltage of 12.0 V across the secondary'coil when the voltage across the primary coil is 120 V. If the voltage across the secondary is increased to 120 V, the voltage across the primary coil is W 0 (e) 360 v w... «.1...» V; - r MV,K_‘;:».I.,,,.,V 1.; gal l ‘x I: it; i} f 42,...” My r f i. 5. (a) (d) i (a) The magnitude and direction of the force between two parallel wires 50 m long and 4.0 cm apart, each carrying 20 A in the same direction is 0.8 N, repulsive ‘(b) 0.1 N, attractive > (c) 1.6 N, attractive 3.7 N, repulsive (e) 3.3 N, attractive ___...> WWWW WWW who cf" “M? f I x: :7; {9:15 lam Pf“ 4‘; a H .02 (2 LC) (0»? The reactance of a 4.00 pF capacitor at a frequency of 80 Hz is fmzmuwz 1.540 [(5) 497 9/] (c) 18.0 o (d) 9350 (e) 6370 \ M. 7. ‘ Two different wire loops are concentric and lie in the same plane. The current in the inner loop is clockwise and decreasing with time. The induced current in the outer loop (a) is zero. ’ \x‘x . ‘@) is clockwrse. (c) is counterclockwise. Z (d) None of the above. PART TWO: TWO PROBLEMS (40 POINTS EACH) ’ Work out the following two problems. Put boxes around all answers and make sure they are clearly labeled. Be sure to show all work clearly. Answers with no clear reasoning will receive little or no credit. 1. in the circuit shown below, find; (a) the current going through the 4 Q resistor. (b) the unknown emf £1. (0) the unknown emf £2. G\> . éw’lnff 13‘" Rm“; 1‘? Q V 6 f3?” Biff? —WvJ l ﬂ”: 1.00 A ‘ ‘13 —,—-.. l \/ 4.00 9 1'00; Q. 51 + V‘_ M” r a: y .c/ 2. The moving rod ab in the figure below is 50.0 cm long and generates an emf in the circuit abcd. The circuit is perpendicular to a downward directed uniform magnetic field of 0.400 T. The rod is moving to the right at a speed of 5.0vm/s. E (a) Calculate the emf induced in the rod. (b) In what direction does the current flow in the circuit? Explain your answer. (c) If the resistance of the circuit is 1.8 Q, calculate the current in'the circuit. (d) Calculate the magnitude and direction of the force required to keep the rod moving to the right with a constant speed of 5.0 m/s. ...
View Full Document

## This note was uploaded on 01/10/2012 for the course PHYS 1214 taught by Professor Staff during the Fall '08 term at Oklahoma State.

### Page1 / 9

Exam 2 - UV? 7:6&amp;quot;? PHYS 1214 SECOND HOUR EXAM...

This preview shows document pages 1 - 9. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online