hw17-solutions

# hw17-solutions - howard (cah3459) hw17 turner (56705) This...

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howard (cah3459) – hw17 – turner – (56705) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 of 2) 10.0 points A circular coil oF 693 turns and area 0 . 62 m 2 is in a uniForm magnetic feld oF 0 . 89 T. The maximum torque exerted on the coil by the feld is 0 . 0027 N · m. Calculate the current in the coil. Correct answer: 7 . 06072 × 10 - 6 A. Explanation: Let : τ =0 . 0027 N · m , N = 693 , A . 62 m 2 , and B . 89 T . The maximum torque is τ = N I A B , so the current in the coil is I = τ N A B = (0 . 0027 N · m) (693) (0 . 62 m 2 ) (0 . 89 T) = 7 . 06072 × 10 - 6 A . 002 (part 2 of 2) 10.0 points Assume the 693 turns oF wire are used to Form a single-turn coil with the same shape but much larger area. What is the current iF the maximum torque exerted on the coil by the feld is 0 . 0027 N · m? Correct answer: 1 . 01886 × 10 - 8 A. Explanation: Let : N = 693 . The radius oF the coil is r = ± A π . and the total length oF the coil is C = N 2 π r . The new radius oF the single-turn coil is R = C 2 π = N r , so the new area oF the coil is A new = π R 2 = N 2 A. Using the equation For the current, we fnd the new value oF the current: I new = τ N 2 AB = I N = 7 . 06072 × 10 - 6 A 693 = 1 . 01886 × 10 - 8 A . 003 (part 1 of 3) 10.0 points A circular current loop oF radius R is placed in a horizontal plane and maintains a current I . There is a constant magnetic feld \$ B in the xy -plane, with the angle α ( α < 90 ) defned with respect to y -axis. The current in the loop ±ows counterclockwise as seen From above. I z ˆ k x ˆ ı y B α What is the direction oF the magnetic mo- ment ? 1. - ˆ ı 2. ı

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howard (cah3459) – hw17 – turner – (56705) 2 3. correct 4. - sin α ˆ - cos α ˆ ı 5. - ˆ 6. + ˆ k 7. - ˆ k 8. cos α ˆ - sin α ˆ ı 9. sin α ˆ + cos α ˆ ı 10. - cos α ˆ + sin α ˆ ı Explanation: I z ˆ k τ x ˆ ı μ loop B α According to the right-hand rule, the mag- netic moment is upward. 004 (part 2 of 3) 10.0 points If the current in the loop is 0 . 202 A and its radius is 3 . 29 cm, what is the magnitude of the magnetic moment of the loop? Correct answer: 0 . 000686899 A · m 2 . Explanation: Let : I =0 . 202 A , and R =3 . 29 cm . The magnetic dipole moment is μ = IA = I π R 2 = (0 . 202 A) (3 . 15) (3 . 29 cm) 2 = 0 . 000686899 A · m 2 005 (part 3 of 3) 10.0 points What is the direction of the torque vector ? 1. - sin α ˆ - cos α ˆ ı 2. - ˆ k 3. sin α ˆ + cos α ˆ ı 4. - ˆ ı 5. - ˆ 6. - cos α ˆ + sin α ˆ ı 7. 8. cos α ˆ - sin α ˆ ı 9. + ˆ k correct 10. ı Explanation: The torque is = × \$ B = μ ) × [ B x ( - ˆ ı )+ B y (+ˆ )] and ˆ × ˆ = 0 and +ˆ × ( - ˆ ı )= ˆ k , so = μB x ˆ k so the direction of the torque is + ˆ k , which agrees with the answer based on the right- hand rule.
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## This note was uploaded on 11/28/2010 for the course PHY 56705 taught by Professor Turner during the Spring '10 term at University of Texas at Austin.

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hw17-solutions - howard (cah3459) hw17 turner (56705) This...

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