chap5o hayt

chap5o hayt - Chapter 5 Odd-Numbered 5.1. Given the current...

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Unformatted text preview: Chapter 5 Odd-Numbered 5.1. Given the current density J = 10 4 [sin(2 x ) e 2 y a x + cos(2 x ) e 2 y a y ]kA / m 2 : a) Find the total current crossing the plane y = 1 in the a y direction in the region 0 < x < 1, < z < 2: This is found through I = Z Z S J n S da = Z 2 Z 1 J a y y =1 dxdz = Z 2 Z 1 10 4 cos(2 x ) e 2 dxdz = 10 4 (2) 1 2 sin(2 x ) 1 e 2 = 1 . 23MA b) Find the total current leaving the region 0 < x,x < 1, 2 < z < 3 by integrating J dS over the surface of the cube: Note first that current through the top and bottom surfaces will not exist, since J has no z component. Also note that there will be no current through the x = 0 plane, since J x = 0 there. Current will pass through the three remaining surfaces, and will be found through I = Z 3 2 Z 1 J ( a y ) y =0 dxdz + Z 3 2 Z 1 J ( a y ) y =1 dxdz + Z 3 2 Z 1 J ( a x ) x =1 dy dz = 10 4 Z 3 2 Z 1 cos(2 x ) e cos(2 x ) e 2 dxdz 10 4 Z 3 2 Z 1 sin(2) e 2 y dy dz = 10 4 1 2 sin(2 x ) 1 (3 2) 1 e 2 + 10 4 1 2 sin(2) e 2 y 1 (3 2) = 0 c) Repeat part b , but use the divergence theorem: We find the net outward current through the surface of the cube by integrating the divergence of J over the cube volume. We have J = J x x + J y y = 10 4 2cos(2 x ) e 2 y 2cos(2 x ) e 2 y = 0 as expected 5.3. Let J = 400sin r 2 + 4 a r A / m 2 a) Find the total current owing through that portion of the spherical surface r = 0 . 8, bounded by 0 . 1 < < . 3 , 0 < < 2 : This will be I = Z Z J n S da = Z 2 Z . 3 . 1 400sin ( . 8) 2 + 4 ( . 8) 2 sin d d = 400( . 8) 2 2 4 . 64 Z . 3 . 1 sin 2 d = 346 . 5 Z . 3 . 1 1 2 [1 cos(2 )] d = 77 . 4A b) Find the average value of J over the defined area. The area is Area = Z 2 Z . 3 . 1 ( . 8) 2 sin d d = 1 . 46m 2 The average current density is thus J avg = (77 . 4 / 1 . 46) a r = 53 . a r A / m 2 . 54 5.5. Let J = 25 a 20 2 + 0 . 01 a z A / m 2 a) Find the total current crossing the plane z = 0 . 2 in the a z direction for < . 4: Use I = Z Z S J n z = . 2 da = Z 2 Z . 4 20 2 + . 01 dd = 1 2 20ln( . 01 + 2 ) . 4 (2 ) = 20 ln(17) = 178 . 0A b) Calculate v /t : This is found using the equation of continuity: v t = J = 1 ( J ) + J z z = 1 (25) + z 20 2 + . 01 = 0 c) Find the outward current crossing the closed surface defined by = 0 . 01, = 0 . 4, z = 0, and z = 0 . 2: This will be I = Z . 2 Z 2 25 . 01 a ( a )( . 01) ddz + Z . 2 Z 2 25 . 4 a ( a )( . 4) ddz + Z 2 Z . 4 20 2 + . 01 a z ( a z ) dd + Z 2 Z . 4 20 2 + .....
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chap5o hayt - Chapter 5 Odd-Numbered 5.1. Given the current...

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