17.5 Ex 29-31

17.5 Ex 29-31 - S E C T I O N 17.5 Surface Integrals of...

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SECTION 17.5 Surface Integrals of Vector Fields (ET Section 16.5) 1163 In Exercises 29–30, a varying current i ( t ) fows through a long straight wire in the xy-plane as in Example 5. The current produces a magnetic Feld B whose magnitude at a distance r ±rom the wire is k B k= μ 0 i 2 π r T, where 0 = 4 · 10 7 T - m / A .²urthermore , B points into the page at points P in the xy-plane. 29. Assume that i ( t ) = t ( 12 t ) A( t in seconds). Calculate the fux 8( t ) , at time t ,o F B through a rectangle oF dimensions L × H = 3 × 2 m, whose top and bottom edges are parallel to the wire and whose bottom edge is located d = 0 . 5 m above the wire (similar to ±igure 11). Then use ±araday’s Law to determine the voltage drop around the rectangular loop (the boundary oF the rectangle) at time t . SOLUTION x y Wire 0.5 m H = 2 m L = 3 m Rectangle R Loop C P = ( x , y ) We choose the coordinate system as shown in the ²gure. ThereFore the rectangle R is the region: R = { ( x , y ) : 0 x 3 , 0 . 5 y 2 . 5 } Since the magnetic ²eld points into the page and R is oriented with normal vector pointing out oF the page (as in Example 5) we have B =−k B k k and n = e n = k . Hence: B · n =k B k ( k ) · k B k=− 0 i 2 r The distance From P = ( x , y ) in R to the wire is r = y , hence, B · n =− 0 i 2 y . We now compute the fux t ) oF B through the rectangle R , by evaluating the Following double integral: t ) = ZZ R B · d S = R B · n dydx = Z 3 0 Z 2 . 5 0 . 5 0 i 2 y 0 i 2 Z 3 0 Z 2 . 5 0 . 5 1 y 3 0 i 2 Z 2 . 5 0 . 5 dy y 3 0 i 2 ( ln 2 . 5 ln 0 . 5 ) 3 0 i 2 ln 2 . 5 0 . 5 = 3 · 4 · 10 7 ln 5 2 t ( 12 t ) 9 . 65 × 10 7 t ( 12 t ) T/m 2 We now use ±araday’s Law to determine the voltage drop around the boundary C oF the rectangle. By ±araday’s Law, the voltage drop around C ,when C is oriented according to the orientation oF R and the Right Hand Rule (that is, counterclockwise) is, Z C E · d S d 8
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17.5 Ex 29-31 - S E C T I O N 17.5 Surface Integrals of...

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