HW18Sol
5 Pages

HW18Sol

Course Number: PHY 303L, Fall 2005

College/University: University of Texas

Word Count: 1307

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Platt, David Homework 18 Due: Oct 21 2005, 4:00 am Inst: Ken Shih This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 5) 10 points The current loop ABCDA carries current I in the direction indicated in the figure, where the segments AB and CD are concentric arcs with radii a and...

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David Platt, Homework 18 Due: Oct 21 2005, 4:00 am Inst: Ken Shih This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 5) 10 points The current loop ABCDA carries current I in the direction indicated in the figure, where the segments AB and CD are concentric arcs with radii a and b respectively. Conversion from degrees to radians: 60 = radians 3 1 Solution: For current along arc CD, ds is perpendicular to ^. Therefore, if we are not r concerned with the field's direction, the BiotSavart law gives dB = 0 I ds 4 b2 D A y 60 C B for current along arc CD. Integrating over the segment yields BCD = D 0 I ds 4 b2 C 0 I (b ) = 4 b2 0 I 0 I = = 4b 4b 3 0 I = , 12 b x O What is the magnitude of the resultant magnetic field B at the center of curvature O due to the current segment of arc CD? 1. BCD = 2. BCD 3. BCD 4. BCD 0 I 12 0 I = 6 0 I = 24 0 I = 6 1 correct b 1 b 1 b2 1 b2 a 5. BCD = 0 6. BCD = 7. BCD 8. BCD 9. BCD 0 I 12 0 I = 3 0 I = 3 0 I = 24 1 b2 1 b 1 b2 1 b Explanation: Basic Concepts: Biot-Savart law: dB = 0 I ds ^ r 2 4 r b where the pseudo unit of should be radians. Radian denotes a pure number, and is not a unit. 002 (part 2 of 5) 10 points What is the direction of the resultant magnetic field B at the center of curvature O due to the current segment of arc CD? 1. pointing into the paper 2. positive y direction 3. undetermined 4. negative x direction 5. positive x direction 6. pointing out of the paper correct 7. negative y direction Explanation: Using the right-hand rule, for ds ^, the r resultant magnetic field vector B points out of the paper. Platt, David Homework 18 Due: Oct 21 2005, 4:00 am Inst: Ken Shih 003 (part 3 of 5) 10 points Find the magnitude of the magnetic field at O due to the current segment BC. 1. BBC = 2. BBC = 3. BBC = 4. BBC = 5. BBC = 0 I 24 0 I 12 0 I 24 0 I 3 0 I 3 1 (b - a)2 1 (b - a)2 1 b-a 1 b-a 1 (b - a)2 2 Like Part 2, the magnetic field due to segment DA is also 0. Like Part 1, the magnitude of magnetic field due to segment AB is given by BAB = 0 I 4 a2 B ds = A 0 I 0 I (a ) = 2 4a 4a However, the magnetic field due to segment AB is pointing into the paper, while the magnetic field due to segment CD is pointing out of the paper. Since a < b, the contribution by AB is larger in magnitude than that by CD. Put all the previous results above together, we get the direction of the resultant magnetic field is pointing into the paper. 005 (part 5 of 5) 10 points What is the magnitude of the magnetic field due to the entire loop? 0 I 24 0 I 2. B = 12 0 I 3. B = 3 0 I 4. B = 6 1. B = 5. B = 0 0 I 1 1 - correct 12 a b 1 0 I 1 7. B = - 2 2 6 a b 0 I 1 1 8. B = - 2 2 3 a b 0 I 1 1 9. B = - 2 2 24 a b Explanation: Its magnitude is given by 6. B = B = BAB - BCD = = 0 I 12 1 1 - a b 0 I 4 . 1 1 - a b 1 1 - a b 1 1 - 2 a2 b 1 1 - a b 1 1 - a b 6. BBC = 0 correct 7. BBC = 8. BBC 9. BBC Explanation: For segment BC, ds is from an element on BC to O, so it is antiparallel to the direction of ^. Therefore ds ^ = 0 = BBC = 0 . r r 004 (part 4 of 5) 10 points What is the direction of the resultant magnetic field B at the center of curvature O due to the current loop ABCDA? 1. positive y direction. 2. undetermined. 3. negative y direction. 4. pointing into the paper. correct 5. positive x direction. 6. pointing out of the paper. 7. negative x direction. Explanation: 0 I 1 6 b-a 0 I 1 = 12 b - a 1 0 I = 6 (b - a)2 006 (part 1 of 2) 10 points Platt, David Homework 18 Due: Oct 21 2005, 4:00 am Inst: Ken Shih Consider the two parallel wires shown. They are separated a by distance a. Find the magnitude and the direction of the magnetic field at P due to the two currents, where I1 = I2 = I. The shaded triangle is in the plane perpendicular to the two wires. In the left view, AP = BP . 3 A P BA B BB ^ i ^ j x A P B y A left view z - - - The direction of B = B A + B B is down (or -^). i 007 (part 2 of 2) 10 points What is the magnitude of the field at P ? 1. 2. 3. 0 I 2a 0 I 4a 0 I correct a 0 I a 0 I 2a 0 I 4a a P 4. 5. 6. B As seen from left, what is the direction of the magnetic field at P ? : (Caution: Notice when viewing from the left, +^ is up and + ^ i is to the right.) 1. (^ - ^) i 2 Explanation: From the sketch, the resultant magnetic field is pointing downward. Adding up the vertical projections, the magnitude of the magnetic field at B is B = 2 BA cos =2 0 I a 2 2 0 I = a 4 1 2 2. - ^ (^ - ^) i 3. - 2 (^ + ^) i 4. - 2 5. +^ i 6. + ^ 7. -^ correct i 8. (^ + ^) i 2 Explanation: 008 (part 1 of 2) 10 points Three very long wires are strung parallel to each other as shown in the figure below. Each wire is at a distance 42 cm from the other two, and each wire carries a current of magnitude I = 6.7 A in the directions shown in the figure. Platt, David Homework 18 Due: Oct 21 2005, 4:00 am Inst: Ken Shih I I 3 42 cm 1 42 cm 2 I 1 y x 4 B 3 2 y x z B 1 30 30 B 2 60 3 ^ ^ k ^ i 60 60 z Cross-sectional View Find the magnitude of the net force per unit length exerted on the upper wire (wire 3) by the other two wires. Correct answer: 3.70236 10-5 N/m. Explanation: The magnetic field due to a long straight wire is 0 I B= , 2r 42 cm and the force per unit length between two parallel wires F = 0 I 1 I 2 . 2r 1 1 2 Adding Magnetic Fields F 30 30 Its components will then be There are two ways to solve this problem which are essentially the same. The first way is to find the net magnetic field at the upper wire from the two wires below (Bnet = B1 + B2 ) and then find the force from F = I L B . The crucial step here will be to add the magnetic fields as vectors. The second way would be to use F = I L B to find the net force on the upper wire from the two lower wires Fnet = F1 + F2 , where we must be sure to add the forces as vectors. You should recognize that the two methods are formally identical. Let's do it the first way. The magnitude magnetic field from wire 1 is found from Ampere's law to be B1 = 0 I . 2r B1 = B [sin (30 ) ^ - cos (30 )^] . i Similarly, B2 points down and to the left of wire 3; its components are given by B2 = B [- sin (30 ) ^ - cos (30 )^] . i Notice that by symmetry the ^ (y) compo nent of the magnetic field vanishes. The net magnetic field is therefore Bnet = The force is then F =ILB -2 0 I cos (30 ) =IL 2r -2 0 I cos (30 ) =IL 2r -2 0 I cos (30 ) ^. i 2r F 23 F1 3 60 3 ^ k ^ ^ i 60 60 2 Adding Forces ^ i (-k ^) (-^) Using the right hand rule the direction points up and to the left of wire as shown in figure 2. Platt, David Homework 18 Due: Oct 21 2005, 4:00 am Inst: Ken Shih F L = 0 I 2 cos(30 ) r 5 0 (6.7 A)2 cos(30 ) = (0.42 m) = 3.70236 10-5 N/m . ^ Notice the direction of L is in the -k direction, since that is the direction of the current. After all the negative signs have cancelled, we notice that the force is in the ^ direction. This is 90 from the positive x-axis. 009 (part 2 of 2) 10 points What angle does the net force on the upper wire (wire 3) make with the positive x-axis? Measure your angles in the standard way: counter-clockwise from the positive x-axis. 1. = 300 2. = 60 3. = 270 4. = 240 5. = 180 6. = 210 7. = 90 correct 8. = 0 9. = 120 10. = 30 Explanation: See Part 1.

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University of Texas - PHY - 303L
Platt, David Homework 19 Due: Oct 24 2005, 4:00 am Inst: Ken Shih This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time.L I1 I2
University of Texas - PHY - 303L
Platt, David Homework 19 Due: Oct 24 2005, 4:00 am Inst: Ken Shih This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (par
University of Texas - PHY - 303L
Platt, David Homework 20 Due: Oct 26 2005, 4:00 am Inst: Ken Shih This print-out should have 5 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part
University of Texas - PHY - 303L
Platt, David Homework 20 Due: Oct 26 2005, 4:00 am Inst: Ken Shih This print-out should have 5 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part
University of Texas - PHY - 303L
Platt, David Homework 21 Due: Oct 28 2005, 4:00 am Inst: Ken Shih This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part
University of Texas - PHY - 303L
Platt, David Homework 21 Due: Oct 28 2005, 4:00 am Inst: Ken Shih This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part
University of Texas - PHY - 303L
Platt, David Homework 22 Due: Oct 31 2005, 4:00 am Inst: Ken Shih This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part
University of Texas - PHY - 303L
Platt, David Homework 22 Due: Oct 31 2005, 4:00 am Inst: Ken Shih This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part
University of Texas - PHY - 303L
Platt, David Homework 23 Due: Nov 2 2005, 4:00 am Inst: Ken Shih This print-out should have 7 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part
University of Texas - PHY - 303L
Platt, David Homework 23 Due: Nov 2 2005, 4:00 am Inst: Ken Shih This print-out should have 7 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part
University of Texas - PHY - 303L
Platt, David Homework 24 Due: Nov 4 2005, 4:00 am Inst: Ken Shih This print-out should have 5 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part
University of Texas - PHY - 303L
Platt, David Homework 24 Due: Nov 4 2005, 4:00 am Inst: Ken Shih This print-out should have 5 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part
University of Texas - PHY - 303L
Platt, David Homework 25 Due: Nov 7 2005, 4:00 am Inst: Ken Shih This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part
University of Texas - PHY - 303L
Platt, David Homework 25 Due: Nov 7 2005, 4:00 am Inst: Ken Shih This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part
University of Texas - PHY - 303L
Platt, David Homework 26 Due: Nov 9 2005, 4:00 am Inst: Ken Shih This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part
University of Texas - PHY - 303L
Platt, David Homework 26 Due: Nov 9 2005, 4:00 am Inst: Ken Shih This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part
University of Texas - PHY - 303L
Platt, David Homework 27 Due: Nov 11 2005, 4:00 am Inst: Ken Shih This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part
University of Texas - PHY - 303L
Platt, David Homework 27 Due: Nov 11 2005, 4:00 am Inst: Ken Shih This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part
University of Texas - PHY - 303L
Platt, David Homework 28 Due: Nov 14 2005, 4:00 am Inst: Ken Shih This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part
University of Texas - PHY - 303L
Platt, David Homework 28 Due: Nov 14 2005, 4:00 am Inst: Ken Shih This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part
University of Texas - PHY - 303L
Platt, David Homework 29 Due: Nov 18 2005, 4:00 am Inst: Ken Shih This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part
University of Texas - PHY - 303L
Platt, David Homework 29 Due: Nov 18 2005, 4:00 am Inst: Ken Shih This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part
University of Texas - PHY - 303L
Platt, David Homework 30 Due: Nov 21 2005, 4:00 am Inst: Ken Shih This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part
University of Texas - PHY - 303L
Platt, David Homework 30 Due: Nov 21 2005, 4:00 am Inst: Ken Shih This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part
University of Texas - PHY - 303L
Platt, David Oldquiz 1 Due: Sep 18 2005, 4:00 am Inst: Ken Shih This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part
University of Texas - PHY - 303L
Platt, David Oldquiz 2 Due: Oct 16 2005, 4:00 am Inst: Ken Shih This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part
University of Texas - PHY - 303L
Platt, David Oldquiz 3 Due: Nov 13 2005, 4:00 am Inst: Ken Shih This print-out should have 32 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part
University of Texas - PHY - 303L
Platt, David Oldquiz 4 Due: Dec 4 2005, 4:00 am Inst: Ken Shih This print-out should have 28 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1
University of Texas - PHY - 303L
E &amp; M - Basic Physical ConceptsElectric force and electric fieldElectric force between 2 point charges: |F | = k 1r2 2 k = 8.987551787 109 N m2 /C2 1 -12 C2 /N m2 0 = 4 k = 8.854187817 10 qp = -qe = 1.60217733 (49) 10-19 C mp = 1.672623 (10) 1
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