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796_PartUniversity Physics Solution

# 796_PartUniversity Physics Solution - Magnetic Field and...

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Magnetic Field and Magnetic Forces 27-3 E XECUTE : (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 φ = and 14.5 . φ = ° E VALUATE : The force and acceleration decrease as the angle φ approaches zero. 27.7. I DENTIFY : Apply q × F = v B ! ! ! . S ET U P : ° y v v = j ! , with 3 3.80 10 m s y v = − × . 3 7.60 10 N, 0, x y F F = + × = and 3 5.20 10 N z F = − × . E XECUTE : (a) ( ) x y z z y y z F q v B v B qv B = = . 3 6 3 (7.60 10 N) ([7.80 10 C)( 3.80 10 m s )] 0.256 T z x y B F qv = = × × × = − ( ) 0, y z x x z F q v B v B = = which is consistent with F ! as given in the problem. There is no force component along the direction of the velocity. ( ) z x y y x y x F q v B v B qv B = = − . 0.175 T x z y B F qv = − = − . (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) 3 3 ( 0.175 T)(+7.60 10 N) ( 0.256 T)( 5.20 10 N) x x y y z z B F B F B F = + + = − × + − × B F ! ! 0 = B F ! ! . B ! and F ! are perpendicular (angle is 90 ) ° . E VALUATE : The force is perpendicular to both v ! and B ! , so v F ! ! is also zero. 27.8. I DENTIFY and S ET U P : ° ° ° ° ° ° ° ° [ ( ) ( ) ( )] [ ( ) ( )]. z x y z z x y q qB v v v qB v v × × + × + × + F = v B = i k j k k k = j i ! ! ! E XECUTE : (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 x y y z z x y z z F F v F v F v F F F qB qB θ = + + = + = = ° v F ! ! E VALUATE : The force is perpendicular to both v ! and B ! , so B F ! ! is also zero. 27.9. I DENTIFY : Apply q = × F v B ! ! ! to the force on the proton and to the force on the electron. Solve for the components of B ! . S ET U P : 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) v = i ! . E XECUTE : (a) For the proton, 3 3 ° ° ° ° (1.50 10 m/s) ( ) (1.50 10 m/s) ( ). x z z q B 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)( ) v = k ! . 3 3 ° ° ° ° ( )(4.75 10 m/s)( ) ( ) (4.75 10 m/s) x z x q e B B e B = × = − × × + + × F v B k i k = j ! ! ! . The magnitude of the force is 3 (4.75 10 m/s) x F e 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. 2 2 2 ( 1.12 T) ( 0.938 T) 1.46 T x z B B B = + = ± + − = . 0.938 T tan 1.12 T z x B B θ = = ± . 40 θ = ± ° . B !

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796_PartUniversity Physics Solution - Magnetic Field and...

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