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Practice HW 13 solution

# Practice HW 13 solution - song(shs546 – Practice HW 13...

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Unformatted text preview: song (shs546) – Practice HW 13 Solutions – weathers – (22202) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 0.0 points A wire in which there is a current of 5 . 12 A is to be formed into a circular loop of one turn. If the required value of the magnetic field at the center of the loop is 8 . 07 μ T, what is the required radius? The permeability of free space is 1 . 25664 × 10 − 6 T · m / A . Correct answer: 39 . 8636 cm. Explanation: Let : B = 8 . 07 μ T = 8 . 07 × 10 − 6 T , I = 5 . 12 A , and μ = 1 . 25664 × 10 − 6 T · m / A . The magnetic field at the center of circular loop is B = μ I 2 R R = μ I 2 B = (1 . 25664 × 10 − 6 T · m / A) (5 . 12 A) 2 (8 . 07 × 10 − 6 T) × 100 cm 1 m = 39 . 8636 cm . 002 0.0 points The closed loop shown in the figure carries a current of 14 A in the counterclockwise direc- tion. The radius of the outer arc is 60 cm , that of the inner arc is 40 cm . Both arcs ex- tent an angle of 60 ◦ . The permeability of free space is 4 π × 10 − 7 T m / A. 4 c m C B 6 c m D A O 60 ◦ y z Find the component of the magnetic field at point O along the x axis. Correct answer:- 1 . 22173 × 10 − 6 T . Explanation: Let : r i = 40 cm = 0 . 4 m , r o = 60 cm = 0 . 6 m , I = 14 A , and μ = 4 π × 10 − 7 T m / A . The magnetic field due to a circular loop at its center is B = μ I 2 R . Because the loop is only one-sixth of a current loop, the magnetic field due to the inner arc is vector B i =- μ I 12 r i ˆ ı , and the magnetic field due to the outter arc is vector B o = μ I 12 r o ˆ ı . Thus the total magnetic field is vector B = vector B i + vector B o = μ I 12 parenleftbigg 1 r o- 1 r i parenrightbigg ˆ ı = (4 π × 10 − 7 T m / A) (14 A) 12 × parenleftbigg 1 . 6 m- 1 . 4 m parenrightbigg ˆ ı = (- 1 . 22173 × 10 − 6 T ) ˆ ı . 003 0.0 points An infinitely long straight wire is bent as shown in the figure. The circular portion has a radius of 150 cm with its center a distance r from the straight part. song (shs546) – Practice HW 13 Solutions – weathers – (22202) 2 I I 150 cm r Find r so that the magnetic field at the center of the circular portion is zero. Correct answer: 47 . 7465 cm. Explanation: Let : R = 150 cm . For a net magnetic field at the center to be zero, B line + B loop = 0 parenleftbigg μ I 2 π r parenrightbigg ˆ ı + parenleftbigg- μ I 2 R parenrightbigg ˆ ı = 0 μ I 2 parenleftbigg 1 π r- 1 R parenrightbigg = 0 1 π r- 1 R = 0 r = R π = 150 cm π = 47 . 7465 cm ....
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Practice HW 13 solution - song(shs546 – Practice HW 13...

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