MIT6_641s09_sol_quiz2006_1

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Unformatted text preview: MIT OpenCourseWare http://ocw.mit.edu 6.641 Electromagnetic Fields, Forces, and Motion Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms . 6.641 — Electromagnetic Fields, Forces, and Motion Spring 2006 Quiz 1 - Solutions Prof. Markus Zahn MIT OpenCourseWare Problem 1 z q ε0 σ=∞ -q image charge z z Figure 1: Charge q of a mass m above a perfectly conducting ground plane and its image charge −q . (Image by MIT OpenCourseWare.) A Question: What is the velo city of the charge as a function of p osition z ? Solution: Force = ma dv = F due to image charge dt dv q (−q ) q 2 dvz ⇒m = =− i ⇒m 2z dt 4π ε0 (2z ) dt 4π ε0 4z 2 dvz dvz q2 dvz dz dvz d ⇒ =− Use of chain rule: = = vz = dt 1 6 π ε 0 mz 2 dt dz dt dz dz ⇒m 2 vz 2 1 Spring 2006 Quiz 1 d dz 2 vz 2 6.641, Spring 2005 q2 ⇒ 1 6 π ε 0 mz 2 2 vz 2 ⇒ =− d = − q2 dz 1 6 π ε 0 mz 2 q2 +C 8 π ε 0 mz Use I.C. vz (z = d) = 0 2 ⇒ vz = + q ⇒ C = − 8πε0 md 2 vz = 2 q2 8 π ε0 m 11 − z d q2 8π ε0 m 1 z ⇒ vz (z ) = − − 1 d since particle is moving towards the conducting ground plane v = vz iz (vz has minus sign) B Question: How long do es it take the charge to reach the z = 0 ground plane? Hint: dz 1 z 1 2 − 1 d = − z d(d − z ) + d 2 tan−1 3 z d−z Solution: vz = dz =− dt dz 1 z q2 8 π ε0 m = − 3 11 − z d q2 dt 8 π ε0 m z = −t d−z q2 + 8 π ε0 m ⇒ − 1 d − z d(d − z ) + d 2 tan−1 C constant We use I.C. to find constant ⇒ at t = 0, z (t = 0) = d. ⇒ d 2 tan−1 π 2 3 π3 d = C ⇒ C = d2 0 2 plugging in z = 0 gives the time (T) it takes for the charge to reach the z = 0 ground plane. ⇒ 0 + d 2 tan−1 0 3 0 =− d q2 π3 T + d2 8 π ε0 m 2 2 Spring 2006 Quiz 1 π3 2 2d q2 8π ε0 m 6.641, Spring 2005 ⇒T = = π 2 3 8 π ε0 m d ⇒ T= 4 q2 2π 3 d3 ε0 m q2 Problem 2 y λ -L I ε0 + + + + + + +++++++ +++++++ a II φ III -a a L x Figure 2: Uniformly distributed line charge λ. (Image by MIT OpenCourseWare.) Φ(r ) = l′ λ(r ′ )dl′ 4π ε0 |r − r ′ | E (r ) = l′ λ(r ′ )ir′ r dl′ 4π ε0 |r − r ′ | 2 A Question: Find the p otential at the p oint (x = 0, y = 0, z = 0) −a Φ(x = 0, y = 0, z = 0) = x =− L −a π λ = − ln x + φ + ln x 4 π ε0 −L 0 L a λ − ln + π + ln = = 4 π ε0 L a λdx + 4π ε0 (−x) π φ=0 λadφ + 4 π ε0 a L a L x =a λdx 4π ε0 (x) λ 4π ε0 (π + 2 ln L ) a B Question: Find the electric field (magnitude and direction) at (x = 0, y = 0, z = 0). 3 Spring 2006 Quiz 1 Solution: −a 6.641, Spring 2005 E (x = 0, y = 0, z = 0) = x =− L λix dx + 4π ε0 x2 −a π φ=0 π λ(−ir )adφ + 4 π ε 0 a2 L x=a λ(−ix )dx 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 = 0) 1 = 4π = = I 4π 0 x =− ∞ 0 x =− ∞ I ix dx × (−xix + aiy ) (x2 + 3 π 2 a2 ) + 3 2 + φ=− π 2 I iφ adφ × (−ir ) + a2 0 0 x =− ∞ I (−ix )dx × (−xix − aiy ) (x2 + a2 ) 2 3 aiz dx (x2 + a2 ) 2 π 2 φ=− π 2 iz dφ + a aiz dx (x2 + a2 ) 2 0 3 x =− ∞ 0 1 π 1 1 x x I 2 iz + φ π iz + 1 2 + a2 ) 2 −∞ 2 + a2 ) 3 4π a (x a − 2 a (x 2 I π π = iz 0 − (−1) + + + 0 − (−1) 4π a 2 2 I 4π a iz −∞ = ( 2 + π ) iz 5 ...
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This note was uploaded on 12/02/2010 for the course ECE 6.641 taught by Professor Zahn during the Spring '09 term at MIT.

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