796_PartUniversity Physics Solution

796_PartUniversity Physics Solution - Magnetic Field and...

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Magnetic Field and Magnetic Forces 27-3 EXECUTE: (a) The smallest possible acceleration is zero, when the motion is parallel to the magnetic field. The greatest acceleration is when the velocity and magnetic field are at right angles: 19 6 2 2 16 31 (1.6 10 C)(2.50 10 m s )(7.4 10 T) 3.25 10 m s . (9.11 10 kg) qvB a m −− ×× × == × (b) If 16 2 1 4 sin (3.25 10 m/s ) , qvB a m φ = then sin 0.25 = and 14.5 . = ° EVALUATE: The force and acceleration decrease as the angle approaches zero. 27.7. IDENTIFY: Apply q × F=v B !! ! . SET UP: & y v v= j ! , with 3 3.80 10 m s y v =− × . 3 7.60 10 N, 0, xy FF = = and 3 5.20 10 N z F × . EXECUTE: (a) () x yz zy Fq v Bv B q v B = . 36 3 (7.60 10 N) ([7.80 10 C)( 3.80 10 m s )] 0.256 T zxy BF q v × × × = 0 , x x z v B = which is consistent with F ! as given in the problem. There is no force component along the direction of the velocity. z yx v B q v B = . 0.175 T xz y q v . (b) y B is not determined. No force due to this component of B ! along v ! ; measurement of the force tells us nothing about . y B (c) 33 ( 0.175 T)(+7.60 10 N) ( 0.256 T)( 5.20 10 N) xx yy zz ⋅= + + = × + − × 0 . B ! and F ! are perpendicular (angle is 90 ) ° . EVALUATE: The force is perpendicular to both v ! and B ! , so vF ! ! is also zero. 27.8. IDENTIFY and SET UP: && & & [ ()()( ) ] [ ( )( ) ] . zx y z y qq B v v v q B vv + × + ×− + F=v B= i k j kk k = j i ! EXECUTE: (a) Set the expression for F ! equal to the given value of F ! to obtain: 7 9 (7.40 10 N) 106 m s ( 5.60 10 C)( 1.25 T) y x z F v qB × = × 7 9 (3.40 10 N) 48.6 m s. ( 5.60 10 C)( 1.25 T) x y z F v qB −× = (b) z v does not contribute to the force, so is not determined by a measurement of F ! . (c) 0; 90 . y x x y F F vF vF vF F F qB qB θ + + = + = ! ! EVALUATE: The force is perpendicular to both v ! and B ! , so B F ! ! is also zero. 27.9. IDENTIFY: Apply q F vB ! to the force on the proton and to the force on the electron. Solve for the components of B ! . SET UP: F ! is perpendicular to both v ! and B ! . Since the force on the proton is in the + y -direction, 0 y B = and x z B B + B =i k ! . For the proton, & (1.50 km/s) i ! . EXECUTE: (a) For the proton, &&& & (1.50 10 m/s) ( ) (1.50 10 m/s) ( ). z qB B q B × ×+ = × F =i i k j ! 16 & (2.25 10 N) , × F = j ! so 16 19 3 2.25 10 N 0.938 T (1.60 10 C)(1.50 10 m/s) z B × . The force on the proton is independent of x B . For the electron, & (4.75 km/s)( ) k ! . & & ( ) ( 4 . 7 51 0 m / s ) ( )( ) ( 4 . 7 0 m / s ) x qe B B e B =×= × −× + + × F k i k= j ! . The magnitude of the force is 3 (4.75 10 m/s) x Fe B . Since 16 8.50 10 N F , 16 19 3 8.50 10 N 1.12 T (1.60 10 C)(4.75 10 m/s) x B × . 1.12 T x B . The sign of x B is not determined by measuring the magnitude of the force on the electron. 22 2 ( 1.12 T) ( 0.938 T) 1.46 T BB B =+ = ± + = . 0.938 T tan 1.12 T z x B B ± . 40 ° . B ! is in the xz -plane and is either at 40 ° from the + x -direction toward the -direction z or 40 ° from the -direction x toward the -direction z .
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27-4 Chapter 27 (b) && x z BB + B= i k ! . & (3.2 km/s)( ) v= j ! . 3 & & & ( )(3.2 km/s)( ) ( ) (3.2 10 m/s)( ( ) ) xz x z qe B B e B B =×= −× + = × −+ F vB j i k k i !! ! . 3 16 16 & & (3.2 10 m/s)( [ 1.12 T] [0.938 T] ) (4.80 10 N) (5.73 10 N) e −− ×− ±− = × ± × F =k i i k !
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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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796_PartUniversity Physics Solution - Magnetic Field and...

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