Dx x2 a2 1 ln x 2 2 x a 2 xdx 1 x2 a2 2 dx x2

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Unformatted text preview: = 0 −2 a λ = 4 π ε0 −2 iy ⇒ E (x = 0, y = 0, z = 0) = − 2πλ0 a iy ε Problem 3 y III I -∞ �0 y=a a φ x II -∞ y = -a I I Figure 3: Current. (Image by MIT OpenCourseWare.) Question: What is the magnetic field H at the p oint (x = 0, y = 0, z = 0)? Hint: a. b. c. d. e. One or more of the following indefinite integrals may b e useful. √ dx x2 + a2 1 = ln x + 2 2 [x +a ] 2 xdx 1 [x2 +a2 ] 2 dx [x2 +a2 ] = dx 1 a tan−1 = 3 =− xdx [x2 +a2 ] 2 x a x 3 [x2 +a2 ] 2 1 2 = x2 + a2 1 a2 [x2 +a2 ] 2 1 1 [x2 +a2 ] 2 4 Spring 2006 Quiz 1 6.641, Spring 2005 Solution: H (r ) = 1 4π I dl′ × ir′ r 2 |r − r ′ | H (x = 0, y = 0, z = 0) 1 = 4π = I 4π 0 I ix dx × (−xix + aiy ) (x2 x =...
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