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**Unformatted text preview: **tovar (jdt436) homework22 Turner (59070) 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. 3 f 2. 12 f 3. 18 f 4. 4 f 9 5. 3 f 4 6. 3 f 2 7. 2 f 9 8. 36 f 9. 4 f 3 correct 10. 6 f Explanation: f = μ I 1 I 2 2 π r = μ I 2 2 π r f = μ (2 I ) 2 2 π 3 r = 4 3 parenleftbigg μ I 2 2 π r parenrightbigg = 4 3 f 002 (part 1 of 3) 10.0 points Two circular loops of radius a = 0 . 71 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 . 000961031 T. Explanation: Let : I = 30 A , a = 0 . 71 cm = 0 . 0071 m , and μ = 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 I 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 I a 2 2 1 a 2 + a 4 2 3 / 2 tovar (jdt436) homework22 Turner (59070) 2- parenleftbigg 1 a 2 + a 2 parenrightbigg 3 / 2 bracketrightBigg = μ 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 . 0071 m) (0 . 361988)...

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