811_PartUniversity Physics Solution

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

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27-18 Chapter 27 EXECUTE: (a) The path is sketched in Figure 27.60. (b) Motion is circular: 222 2 2 1 x yR xDy RD +=⇒=⇒= − (path of deflected particle) 2 yR = (equation for tangent to the circle, path of undeflected particle). 22 21 11 1 D D dy yR R D RR R R R =−=− − =− − = −− . If RD >> , 2 1 D D dR RR ⎛⎞ ≈− = ⎢⎥ ⎜⎟ ⎝⎠ ⎣⎦ . For a particle moving in a magnetic field, . mv R qB = But 2 1 2 12 , so . mV mv qV R B q == Thus, the deflection . DB q e d mV mV ≈= (c) 25 1 9 31 ( 0 . 5 0m ) ( 5 . 01 0T ) ( 1 . 61 0 C ) 0.067 m 6.7 cm. 2 2(9.11 10 kg)(750 V) d −− ×× = × 13% of , dD which is fairly significant. EVALUATE: In part (c), 2 3.7 mV D D R DD Be d d = = and 2 14 R D = , so the approximation made in part (b) is valid. Figure 27.60 27.61. IDENTIFY and SET UP: Use Eq.(27.2) to relate ,, a n d . q vB 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 =− = × + + = Bk v i j k ! ! (a) EXECUTE: q F !! ! 6 && && ( 0.120 T)(1.05 10 m/s)( 3 4 12 ) q × −×+ ×+ × F ik jk kk ! && & &&& ,, 0 ×= − ×= ×= ik j jk ikk 55 (1.26 10 N/C)( 3 4 ) (1.26 10 N/C)( 4 3 ) qq × + + × + + Fj ii j ! The magnitude of the vector 43 i s 345 . ++ + = ij Thus 5 (1.26 10 N/C)(5). Fq × 6 1.25 N 1.98 10 C 5(1.26 10 N/C) 5(1.26 10 N/C) F q × (b) so / mm Fa a F 56 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 = = ++ = × × aF i j i j ! ! (c) IDENTIFY and SET UP: 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. EXECUTE: gives ( / ) and / mF m v R R m v F = ! ! 6 2 6 2 1 3 2 2 ( 3.15 10 m/s) ( 4.20 10 m/s) 2.756 10 m /s xy vvv =+= − × + + × = × 2 2 (0.250 N) 4 3 1.25 N FF F = + = 5 1 3 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) IDENTIFY and SET UP: By Eq.(27.12) the cyclotron frequency is /2 . f vR ω ππ = = EXECUTE: The circular motion is in the xy -plane, so 22 6 5.25 10 m/s. xy vv v =+ 6 77 5.25 10 m/s 1.47 10 Hz, and 2 9.23 10 rad/s ( 0 . 0 5 6 9 m ) v ff R ωπ × == = × × (e) IDENTIFY and SET UP 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 zz v t = + EXECUTE: The period of the motion in the xy -plane is given by 8 7 11 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. z v × Thus 68 0 0 (12.6 10 m/s)(2)(6.80 10 s) 1.71 m. z t =+ =+ × × = + The coordinates at 2 a r e , 0 , 1 . 7 1 m . tT xR y z = = + EVALUATE: The circular motion is in the plane perpendicular to . B ! The radius of this motion gets smaller when B increases and it gets larger when v increases. There is no magnetic force in the direction of B ! so the particle moves with constant velocity in that direction. The superposition of circular motion in the xy -plane and constant speed motion in the z -direction is a helical path.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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

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