PHY132_S10_M2_Solutions

# PHY132_S10_M2_Solutions - ID Number LAST Name First Name A...

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ID Number: LAST Name: First Name: A 1 I 2 I 1 3.0 mm 4.0 mm 4.0 mm A O C PHY 132 Midterm 2 Spring 2010 A Put legible ID# and Full Name in PRINT on every sheet! There are 3 problems . For multiple-choice ques- tions: mark your answers by circling the correct answers . For word problems: work the problems in the space with the problem and show all work for full credit! The blue booklet is for scrap calculations and drafts; NOT for hand-in! When modifying your answer make it unambiguous . This exam will last 2.5 hours. Success! 1. Concepts (65 points). a. Consider a long solid cylindrical copper wire of radius R with a uniform current density I / A , where A is the cross sectional area and I is the total current carried by the wire; see the figure. The magnetic field for the points P and Q in magnitude and direction is: direction in point P : (2 points) magnitude in point P : (3 points) none 0 1 2 I r A 0 1 2 I r 0 2 1 4 I r zero 1 2 3 4 5 6 7 12 3 4 direction in point Q : (2 points) magnitude in point Q : (3 points) none 0 2 2 I r A 0 2 2 I r 0 2 2 4 I r zero 1 2 3 4 5 6 7 1 234 b. Consider the force between two wires . The first wire carries a current I 1 =6.00 A and is located directly above a second wire carrying a current I 2 =5.00 A. i) Initially , the two wires are parallel and separated by 3.00 mm. Calculate the force-per-meter ex- erted on the upper wire by the lower wire in case the two currents are in opposite directions. The relative orientation of the wires is now changed , see the figure to the right: ii) The direction of the magnetic force on the lower wire in the points A , O , and C is: c. The plane of a flat wire coil makes an angle θ = 36.9º with a uniform magnetic field B= 1.5 T, see the figure. The coil has a cross section A= 80 cm 2 and consists of 5 turns of wire carrying a current I= 0.50 A. The direction is: (2 points) The magnitude is: [N/m] (5 points) none 2.0×10 –3 2.4×10 –3 3.2×10 –3 4.0×10 –3 6.3×10 –3 1 2 3 4 5 6 7 123 ±45 The force in O is directed: (2 pnts) The force in A is directed: (2 pnts) The force in C is directed: (2 pnts) none none none 1 2 3 4 5 6 7 1 2 3 4 5 6 712 ±3 ±4 56 7 Calculate the magnitude of the magnetic torque on the loop. (6 pts) Calculate the magnetic potential energy of the coil in this field. (6 pt) Calculate the total magnetic flux seen by the coil. (6 pts) 4 2 sin(90 36.9 ) 4 58 01 0 0 . 51 . 5 5 2.4 10 Nm NAIB          τμ B wrong angle: -3;, wrong sign=-1 4 2 cos(90 36.9 ) 3 . . 5 5 1.8 10 Nm UN A I B             μ B no angle: -4, N= 1: -1. 4 22 cos(90 36.9 ) 3 0 1 . 5 5 3.6 10 Tm B A Nd N A B     BA R r 2 r 1 Q P I B I θ

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ID Number: LAST Name: First Name: A 2 d. A copper wire loop lies flat on a table. A strong bar magnet is dropped north-side down towards the center of the loop from above .
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PHY132_S10_M2_Solutions - ID Number LAST Name First Name A...

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