HOMEWORK-CH 29

HOMEWORK-CH 29 - 2.5 10 T i 1.4 10 N k. F F F qE qv B---= +...

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HOMEWORK ASSIGNMENT 5 18. We consider the point at which it enters the field-filled region, velocity vector pointing downward. The field points out of the page so that v B × r r points leftward, which indeed seems to be the direction of the force; therefore, q > 0 (it is a proton). (a) The particle spends half of a period completing its half-circle path, so 1 2 130 ns. T = The period of a charged particle in a uniform magnetic field is 2 , T m q B = π so ( 29 ( 29 ( 29 27 19 9 2 1.67 10 kg 2 0.252 T. 1.60 10 C 260 10 s m B q T π - - - × = = = × × (b) The period is independent of the speed. It is therefore independent of the kinetic energy. The time is the same: 130 ns. 22. (a) The net force on the proton is given by ( 29 ( 29 ( 29 ( 29 ( 29 net elec mag 19 3 18 ˆ ˆ ˆ 1.6 10 C 4.0V m k 2000m s j
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Unformatted text preview: 2.5 10 T i 1.4 10 N k. F F F qE qv B---= + = + = + - = r r r r r r So the magnitude of the net force is net 18 1.4 10 N. F-= (b) In this case ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 net elec mag 19 3 19 19 1.6 10 C 4.0V m i 2000m s j 2.5 10 T i 6.4 10 N i+ 8.0 10 N k. F F F qE qv B----= + = + = + - = r r r r r r The magnitude of the force is now ( 29 ( 29 2 2 net 19 19 18 6.4 10 N 8.0 10 N 1.0 10 N. F---= + = 28. We use Eq. 29-20 to solve for V : ( 29 ( 29 ( 29 ( 29 ( 29 6 28 3 19 23A 0.65 T 7.4 10 V. 8.47 10 m 150 m 1.60 10 C i B V n e -- = = = l...
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This note was uploaded on 06/01/2008 for the course PHYS 232 taught by Professor Goksu during the Spring '08 term at Millersville.

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