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811_PartUniversity Physics Solution

# 811_PartUniversity Physics Solution - 27-18 Chapter 27...

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27-18 Chapter 27 E XECUTE : (a) The path is sketched in Figure 27.60. (b) Motion is circular: 2 2 2 2 2 1 x y R x D y R D + = = = (path of deflected particle) 2 y R = (equation for tangent to the circle, path of undeflected particle). 2 2 2 2 2 1 2 2 1 1 1 D D d y y R R D R R R R R = = = = . If R D >> , 2 2 2 1 1 1 2 2 D D d R R R = . For a particle moving in a magnetic field, . mv R qB = But 2 1 2 1 2 , so . mV mv qV R B q = = Thus, the deflection 2 2 . 2 2 2 2 D B q D B e d mV mV = (c) 2 5 19 31 (0.50 m) (5.0 10 T) (1.6 10 C) 0.067 m 6.7 cm. 2 2(9.11 10 kg)(750 V) d × × = = = × 13% of , d D which is fairly significant. E VALUATE : In part (c), 2 1 2 3.7 2 2 mV D D R D D B e d d = = = = and 2 14 R D = , so the approximation made in part (b) is valid. Figure 27.60 27.61. I DENTIFY and S ET U P : Use Eq.(27.2) to relate , , and . q v B F ! ! ! The force and F a ! ! are related by Newton’s 2nd law. 6 ° ° ° ° (0.120 T) , (1.05 10 m/s)( 3 4 12 ), 1.25 N F = − = × + + = B k v i j k ! ! (a) E XECUTE : q = × F v B ! ! ! 6 ° ° ° ° ° ° ( 0.120 T)(1.05 10 m/s)( 3 4 12 ) q = × × + × + × F i k j k k k ! ° ° ° ° ° ° ° ° , , 0 × = − × = × = i k j j k i k k 5 5 ° ° ° ° (1.26 10 N/C)( 3 4 ) (1.26 10 N/C)( 4 3 ) q q = − × + + = − × + + F j i i j ! The magnitude of the vector 2 2 ° ° 4 3 is 3 4 5. + + + = i j Thus 5 (1.26 10 N/C)(5). F q = − × 6 5 5 1.25 N 1.98 10 C 5(1.26 10 N/C) 5(1.26 10 N/C) F q = − = − = − × × × (b) so / m m = = F a a F ! ! ! ! 5 6 5 ° ° ° ° (1.26 10 N/C)( 4 3 ) ( 1.98 10 C)(1.26 10 N/C)( 4 3 ) q = − × + + = − − × × + + F i j i j ! ° ° 0.250 N(+4 3 ) = + + i j Then 13 2 15 0.250 N ° ° ° ° / ( 4 3 ) (9.69 10 m/s )( 4 3 ) 2.58 10 kg m = = + + = × + + × a F i j i j ! ! (c) I DENTIFY and S ET U P : F ! is in the xy -plane, so in the z -direction the particle moves with constant speed 6 12.6 10 m/s. × In the xy -plane the force F ! causes the particle to move in a circle, with F ! directed in towards the center of the circle. E XECUTE : 2 2 gives ( / ) and / m F m v R R mv F = = = F a ! ! 2 2 2 6 2 6 2 13 2 2 ( 3.15 10 m/s) ( 4.20 10 m/s) 2.756 10 m /s x y v v v = + = − × + + × = × 2 2 2 2 (0.250 N) 4 3 1.25 N x y F F F = + = + = 2 15 13 2 2 (2.58 10 kg)(2.756 10 m /s ) 0.0569 m 5.69 cm 1.25 N mv R F × × = = = =

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Magnetic Field and Magnetic Forces 27-19 (d) I DENTIFY and S ET U P : By Eq.(27.12) the cyclotron frequency is /2 /2 . f v R ω π π = = E XECUTE : The circular motion is in the xy -plane, so 2 2 6 5.25 10 m/s. x y v v v = + = × 6 7 7 5.25 10 m/s 1.47 10 Hz, and 2 9.23 10 rad/s 2 2 (0.0569 m) v f f R ω π π π × = = = × = = × (e) I DENTIFY and S ET U P Compare t to the period T of the circular motion in the xy -plane to find the x and y coordinates at this t . In the z -direction the particle moves with constant speed, so 0 . z z z v t = + E XECUTE : The period of the motion in the xy -plane is given by 8 7 1 1 6.80 10 s 1.47 10 Hz T f = = = × × In t = 2 T the particle has returned to the same x and y coordinates. The z -component of the motion is motion with a constant velocity of 6 12.6 10 m/s.
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811_PartUniversity Physics Solution - 27-18 Chapter 27...

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