Chapter_29_08 - Chapter 29 29.5 In Fig. 29-36, two circular...

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Chapter 29 29.5 In Fig. 29-36, two circular arcs have radii a = 13.5 cm and b = 10.7 cm , subtend angle θ = 74 ° , carry current i = 0.411 A , and share the same center of curvature P. What are (a) the magnitude and (b direction of the net field at P. This is exactly the same kind of problem as 29.6. See the full derivation below. Only the radii and angles are different. We begin near the end of the previous problem, changing the limits of integration B = μ 0 i 4 π a d ˆ k /2 /2 = 0 i 4 a ˆ k = 0 i 4 a ˆ k = 74 ° 180 ° = 0.4111 B = 0 i 4 a 0.4111 ˆ k = 0.4111 0 i 4 a ˆ k To get the full field, we can add the contributions from each arc, and again taking into the paper as positive. B = ( 0.4111 0 i 4 b 0.4111 0 i 4 a ) ˆ k = 1.028 × 10 7 T Field points out of paper. 29.6 In Fig. 29-38, two semicircular arcs have radii R 2 = 7.8 cm and R 1 = 3.15 cm and carry current i = 0.281 A , and share the same center of curvature C. What are the (a) magnitude and (b) direction (into or out of the page) of the net magnetic field at C? To do this problem, we begin by writing the Biot-Savart Law and describe what the field at (0,0,0) will look like. B = 0 4 id l × ˆ r r 2 Carefully write r , r , ˆ r , and d l for the tiny current element given. Hint. You can find d l by taking the derivative of r , Our observation point is taken to be the origin (0,0,0). Note that we are finding the field due to a generic arc with radius a that is centered on the origin.
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l = x ˆ i + y ˆ j = a cos θ ˆ i + a sin ˆ j d l d = a sin ˆ i + a cos ˆ j d l = ( a sin ˆ i + a cos ˆ j ) d r = (0 a cos ) ˆ i + a sin ) ˆ j + 0) ˆ k = a cos ˆ i a sin ˆ j r = a cos ) 2 + ( a sin ) 2 = 2 = a ˆ r = a cos ˆ i a sin ˆ j a = cos ˆ i sin ˆ j Set up to compute the field at (0,0,0). Be sure to show that the field that results will only point in the z direction. B = μ 0 4 π id l × ˆ r r 2 0 = = 0 i 4 ( a sin ˆ i + a cos ˆ j ) × ( cos ˆ i sin ˆ j ) a 2 0 d = 0 i 4 a 2 ( a sin ˆ i + a cos ˆ j ) × ( cos ˆ i sin ˆ j ) 0 d = 0 i 4 a (sin 2 + cos 2 ) ˆ k 0 d B = 0 i 4 a (sin 2 + cos 2 ) ˆ k 0 2 d = 0 i 4 a d ˆ k 0 = 0 i 4 a ˆ k = 0 i 4 a ˆ k To get the total field, we sum contributions from each half circle.
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This note was uploaded on 10/03/2008 for the course PHYS 1322 taught by Professor Ndili during the Spring '07 term at University of Houston.

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Chapter_29_08 - Chapter 29 29.5 In Fig. 29-36, two circular...

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