20_InstSolManual_PDF_Part24

20_InstSolManual_PDF_Part24 - 3 10 4 m / s 2 1 1.60 3 10 2...

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20.84. Set Up: Segments bc and cd each have length Draw each segment in the plane of the page. The direction of is given by the right-hand rule. Solve: (a) segment ab (Figure 20.84a): is in the direction. so segment bc (Figure 20.84b): is in the direction. segment cd (Figure 20.84c): is parallel to the yz plane, midway between the and axes. segment de (Figure 20.84d): is in the direction. segment ef (Figure 20.84e): so (b) The force on segment cd has a component of 4.24 N in the direction and a component of 4.24 N in the direction. Adding the forces on the segments therefore gives a net force of 4.24 N, in the direction. Figure 20.84 20.85. Set Up: The ion has charge After being accelerated through a potential difference V the ion has kinetic energy qV. The acceleration in the circular path is Solve: and gives R 5 m v 0 q 0 B 5 1 1.16 3 10 2 26 kg 21 7.79
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Unformatted text preview: 3 10 4 m / s 2 1 1.60 3 10 2 19 C 21 0.723 T 2 5 7.81 3 10 2 3 m 5 7.81 mm. q v B 5 m v 2 R . F S 5 m a S f 5 90. F B 5 q v B sin f . v 5 2 eV m 5 2 1 1.60 3 10 2 19 C 21 220 V 2 1.16 3 10 2 26 kg 5 7.79 3 10 4 m / s. 1 2 m v 2 5 eV K 5 qV 5 1 eV . v 2 / R . q 5 1 e . y z x I F ( a ) x B z y F I ( b ) y B x z I F ( c ) x y B I ( e ) y z x B I F ( d ) B f 2 y 1 y 1 z F 5 0. f 5 180 F 5 IlB sin f 5 1 6.58 A 21 0.750 m 21 0.860 T 2 sin 90 5 4.24 N. 2 y F S F 5 IlB sin f 5 1 6.58 A 21 1.06 m 21 0.860 T 2 sin 90 5 6.00 N. 1 z 1 y F S F 5 IlB sin f 5 1 6.58 A 21 1.06 m 21 0.860 T 2 sin 45 5 4.24 N. 2 y F S F 5 IlB 5 1 6.58 A 21 0.750 m 21 0.860 T 2 5 4.24 N. f 5 90 2 z F S F S F 5 IlB sin f . l 5 " 2 1 0.750 m 2 2 5 1.06 m. 20-24 Chapter 20...
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This note was uploaded on 03/06/2009 for the course PHYS 114 taught by Professor Shoberg during the Spring '07 term at Pittsburg State Uiversity.

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