HW 22 - benavides(jjb2356 homework22 Turner(59130 This...

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benavides (jjb2356) – homework22 – Turner – (59130) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Two long parallel wires carry the same amount of current and repel each other with a force per unit length of f = F l . If both currents are doubled and the wire separation is tripled, what is the force per unit length? 1. 2 f 9 2. 3 f 4 3. 4 f 3 correct 4. 12 f 5. 3 f 6. 4 f 9 7. 18 f 8. 36 f 9. 6 f 10. 3 f 2 Explanation: f = μ 0 I 1 I 2 2 π r = μ 0 I 2 2 π r f = μ 0 (2 I ) 2 2 π 3 r = 4 3 parenleftbigg μ 0 I 2 2 π r parenrightbigg = 4 3 f 002 (part 1 of 3) 10.0 points Two circular loops of radius a = 0 . 48 cm whose planes are perpendicular to a common axis and whose centers are separated by a distance = a 2 . The currents flow in the directions shown in the figure. C C A x I right = 30 A I left = 30 A Find the magnitude of the magnetic field at point A on the axis midway between the loops. Correct answer: 0 T. Explanation: At point A the B field is zero . Note: The current elements on the two coils that are diagonally opposite through A , gives d vector B contributions that cancel at point A . 003 (part 2 of 3) 10.0 points Find the magnitude of the magnetic field at point C on the axis a distance to the right of the right loop. Correct answer: 0 . 00142152 T. Explanation: Let : I = 30 A , a = 0 . 48 cm = 0 . 0048 m , and μ 0 = 4 π 10 7 . In the text, the magnitude of the magnetic field along an axial point a distance z from the center of a current loop of radius a carrying a current I , is given by B = μ 0 I 0 a 2 2 ( a 2 + x 2 ) 3 / 2 . At point C the field from the right hand loop is in the positive x -direction, and the field from the left hand loop is in the negative x - direction. Adding the fields the two loops, with x = = a/ 2 for the right-hand loop, and x = 2 l = a for the left-hand loop, we obtain B = μ 0 I a 2 2 1 a 2 + a 4 2 3 / 2
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benavides (jjb2356) – homework22 – Turner – (59130) 2 - parenleftbigg 1 a 2 + a 2 parenrightbigg 3 / 2 bracketrightBigg = μ 0 I 2 a bracketleftBigg parenleftbigg 4 5 parenrightbigg 3 / 2 - parenleftbigg 1 2 parenrightbigg 3 / 2 bracketrightBigg = (4 π 10 7 ) (30 A) 2 (0 . 0048 m) (0 . 361988) = 0 . 00142152 T .
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