MIT6_641s09_sol_quiz2006_1

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Unformatted text preview: 4π ε0 x2 = 1 1L λ − cos φix − sin φiy − dφ + ix + ix 4 π ε0 x −L a xa φ=0 π π 1� 1� λ 1 1� 1� − � + � ix + − sin φ ix + cos φ iy + � − � ix = 4 π ε0 − − a L� a 0 0 � � a �L 0 −2 λ = 4 π ε0 −2 a iy ⇒ E (x = 0, y = 0, z = 0) = − 2πλ0 a iy ε Problem 3 III y �0 y=a a II -∞ I φ x -∞ I I y = -a 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 xdx 1 [x2 +a2 ] 2 [x +a ] 2 = x2 + a2 1 a 1 2 dx [x2 +a2 ] dx = 3 tan−1 x x a 1 [x2 +a2 ] 2 xdx [x2 +a2 ] 2 = a2 [x2 +a2 ] 2 1 [x2 +a2 ] 2 1 3 =− 4 Spring 2006 Quiz 1 6.641, Spring 2005 Solution: H (r ) = 1 4π I dl′ × ir′ r |r − r ′ | 2 H (x = 0, y = 0, z...
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