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Unformatted text preview: CHAPTER 9 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 ( ) 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 = − . 135 cm. 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 9.2. A point charge, Q = − . 3 µ C and m = 3 × 10 − 16 kg, is moving through the field B = 30 a z mT. Make use of Eq. (2) and Newton’s laws to develop the appropriate differential equations, and solve them, subject to the initial condition at t = 0, v = 3 × 10 5 m/s at the origin. Solve these equations (perhaps with the help of an example given in Section 7.5) to evaluate at t = 3 µ s: a) the position P(x,y,z) of the charge; b) its velocity; c) and its kinetic energy: We begin by visualizing the problem. Using F = q v × B , we find that a positive charge moving along positive a x , would encounter the zdirected B field and be deflected into the negative y direction. 142 9.2 (continued) Motion along negative y through the field would cause further deflection into the negative x direction. We can construct the differential equations for the forces in x and in y as follows: F x a x = m dv x dt a x = qv y a y × B a z = qBv y a x F y a y = m dv y dt a y = qv x a x × B a z = − qBv x a y or dv x dt = qB m v y ( 1 ) and dv y dt = − qB m v x ( 2 ) To solve these equations, we first differentiate (2) with time and substitute (1), obtaining: d 2 v y dt 2 = − qB m dv x dt = − ± qB m ² 2 v y Therefore, v y = A sin (qBt/m) + A cos (qBt/m) . However, at t = 0, v y = 0, and so A = 0, leaving v y = A sin (qBt/m) . Then, using (2), v x = − m qB dv y dt = − A cos ± qBt m ² Now at t = 0, v x = v x = 3 × 10 5 . Therefore A = − v x , and so v x = v x cos (qBt/m) , and v y = − v x sin (qBt/m) . The positions are then found by integrating v x and v y over time: x(t) = Z v x cos ± qBt m ² dt + C = mv x qB sin ± qBt m ² + C where C = 0, since x( ) = 0. Then y(t) = Z − v x sin ± qBt m ² dt + D = mv x qB...
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This note was uploaded on 06/10/2008 for the course EE 325 taught by Professor Brown during the Spring '08 term at University of Texas.
 Spring '08
 Brown
 Electromagnet

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