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Unformatted text preview: Chapter 9 — OddNumbered 9.1. A point charge, Q = − . 3 µ C and m = 3 × 10 − 16 kg, is moving through the field E = 30 a z V/m. Use Eq. (1) and Newton’s laws to develop the appropriate differential equations and solve them, subject to the initial conditions at t = 0: v = 3 × 10 5 a x m/s at the origin. At t = 3 µ s, find: a) the position P ( x,y,z ) of the charge: The force on the charge is given by F = q E , and Newton’s second law becomes: F = m a = m d 2 z dt 2 = q E = ( − . 3 × 10 − 6 )(30 a z ) describing motion of the charge in the z direction. The initial velocity in x is constant, and so no force is applied in that direction. We integrate once: dz dt = v z = qE m t + C 1 The initial velocity along z , v z (0) is zero, and so C 1 = 0. Integrating a second time yields the z coordinate: z = qE 2 m t 2 + C 2 The charge lies at the origin at t = 0, and so C 2 = 0. Introducing the given values, we find z = ( − . 3 × 10 − 6 )(30) 2 × 3 × 10 − 16 t 2 = − 1 . 5 × 10 10 t 2 m At t = 3 µ s, z = − (1 . 5 × 10 10 )(3 × 10 − 6 ) 2 = − . 135cm. Now, considering the initial constant velocity in x , the charge in 3 µ s attains an x coordinate of x = vt = (3 × 10 5 )(3 × 10 − 6 ) = . 90 m. In summary, at t = 3 µ s we have P ( x,y,z ) = ( . 90 , , − . 135) . b) the velocity, v : After the first integration in part a , we find v z = qE m t = − (3 × 10 10 )(3 × 10 − 6 ) = − 9 × 10 4 m / s Including the intial xdirected velocity, we finally obtain v = 3 × 10 5 a x − 9 × 10 4 a z m / s . c) the kinetic energy of the charge: Have K . E . = 1 2 m  v  2 = 1 2 (3 × 10 − 16 )(1 . 13 × 10 5 ) 2 = 1 . 5 × 10 − 5 J 9.3. A point charge for which Q = 2 × 10 − 16 C and m = 5 × 10 − 26 kg is moving in the combined fields E = 100 a x − 200 a y + 300 a z V/m and B = − 3 a x + 2 a y − a z mT. If the charge velocity at t = 0 is v (0) = (2 a x − 3 a y − 4 a z ) × 10 5 m/s: a) give the unit vector showing the direction in which the charge is accelerating at t = 0: Use F ( t = 0) = q [ E + ( v (0) × B )], where v (0) × B = (2 a x − 3 a y − 4 a z )10 5 × ( − 3 a x + 2 a y − a z )10 − 3 = 1100 a x + 1400 a y − 500 a z 96 9.3a. (continued) So the force in newtons becomes F (0) = (2 × 10 − 16 )[(100+1100) a x +(1400 − 200) a y +(300 − 500) a z ] = 4 × 10 − 14 [6 a x +6 a y − a z ] The unit vector that gives the acceleration direction is found from the force to be a F = 6 a x + 6 a y − a z √ 73 = . 70 a x + . 70 a y − . 12 a z b) find the kinetic energy of the charge at t = 0: K . E . = 1 2 m  v (0)  2 = 1 2 (5 × 10 − 26 kg)(5 . 39 × 10 5 m / s) 2 = 7 . 25 × 10 − 15 J = 7 . 25 fJ 9.5. A rectangular loop of wire in free space joins points A (1 , , 1) to B (3 , , 1) to C (3 , , 4) to D (1 , , 4) to A . The wire carries a current of 6 mA, ﬂowing in the a z direction from B to C ....
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This note was uploaded on 10/02/2011 for the course EE 1 taught by Professor Joshi during the Fall '08 term at UCLA.
 Fall '08
 Joshi
 Electrical Engineering

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