Exam_03-solutions

# Exam_03-solutions - Version 065 – Exam 03 – Chiu...

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Unformatted text preview: Version 065 – Exam 03 – Chiu – (58295) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Two long straight wires in the xy plane are parallel to the x axis. One wire is at y = − 8 . 5 cm and the other wire is at y = 8 . 5 cm. The current in each wire is 16 A and the currents in figure are in the negative x direction. The permeability of free space is 4 π × 10 − 7 T · m / A. ˆ ı ˆ ˆ k Find the magnitude of the force per unit length exerted by one wire on the other. 1. 470.588 2. 4320.0 3. 1364.21 4. 301.176 5. 1425.45 6. 677.647 7. 4000.0 8. 576.0 9. 2560.0 10. 1137.78 Correct answer: 301 . 176 μ N / m. Explanation: Let : d = y 2 − y 1 = 17 cm = 0 . 17 m , I = 16 A , and μ = 4 π × 10 − 7 T · m / A . The magnetic field at either location due to the current in the wires at the other location is B = μ 4 π 2 I d , so the force exerted on either wire is F = I ℓ B F ℓ = I B = μ 4 π 2 I 2 d = ( 1 × 10 − 7 T · m / A ) 2 (16 A) 2 . 17 m × 10 6 μ N N = 301 . 176 μ N / m . 002 10.0 points A rectangular loop consists of 319 closely wrapped turns of wire and has dimensions . 5 m by 0 . 32 m. The loop is hinged along the y-axis, and its plane makes an angle of 53 ◦ with the x-axis. A uniform magnetic field of 0 . 24 T is di- rected along the x-axis and the current in the loop is 6 . 6 A in the direction shown. x y z 5 3 ◦ B = 0 . 24 T B = 0 . 24 T . 5m . 3 2 m i = 6 . 6 A What is the magnitude of the torque ex- erted on the loop? 1. 48.6552 2. 26.1997 3. 46.7161 4. 269.85 5. 1247.81 6. 76.9161 7. 25.1487 8. 3.32034 9. 426.088 10. 84.5026 Correct answer: 48 . 6552 N · m. Explanation: Version 065 – Exam 03 – Chiu – (58295) 2 Let : n = 319 , ℓ = 0 . 5 m , w = 0 . 32 m , θ = 53 ◦ , B = 0 . 24 T , and I = 6 . 6 A . The field makes an angle of α = 90 ◦ − θ with a line perpendicular to the plane of the loop, so the torque acting on the loop is τ = μ loop B sin α = n B I A sin α = n B I ℓ w sin α (1) = (319)(0 . 24 T)(6 . 6 A)(0 . 5 m)(0 . 32 m) · sin(90 ◦ − 53 ◦ ) = 48 . 6552 N · m . Notice: If the axis of rotation is parallel to y-axis, the torque is the same — no matter where the axis of rotation is place; e.g. , if the axis of rotation were in the middle of the coil, we have τ = 2 n B I ℓ 2 w sin α = n B I ℓ w sin α, which is the same as Eq. 1. keywords: 003 10.0 points A solenoid with circular cross section pro- duces a steadily increasing magnetic flux through its cross section. There is an octago- nally shaped circuit surrounding the solenoid as shown. B B B B X Y i i Figure 1: The increasing magnetic flux gives rise to a counterclockwise induced emf E . The circuit consists of two identical light bulbs of equal resistance, R , connected in series, leading to a loop equation E − 2 iR = 0....
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## This note was uploaded on 04/26/2009 for the course PHY 58310 taught by Professor Chiu during the Spring '09 term at University of Texas.

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Exam_03-solutions - Version 065 – Exam 03 – Chiu...

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