Chap22Sol

Chap22Sol - SOLUTIONS TO PROBLEMS Easy P22.1 (a) up (b) out...

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161 SOLUTIONS TO PROBLEMS Easy P22.1 (a) up (b) out of the page, since the charge is negative. (c) no deflection (d) into the page FIG. P22.1 *P22.2 At the equator, the Earth’s magnetic field is horizontally north. Because an electron has negative charge, r F = q r v ! r B is opposite in direction to r v ! r B . Figures are drawn looking down. (a) Down × North = East, so the force is directed West . (a) (c) (d) FIG. P22.2 (b) North × North = sin 0 ° = 0 : Zero deflection . (c) West × North = Down, so the force is directed Up . (d) Southeast × North = Up, so the force is Down . P22.3 (a) F B = qvB sin ! = 1.60 " 10 # 19 C ( ) 3.00 " 10 6 m s ( ) 3.00 " 10 # 1 T ( ) sin 37.0 ° F B = 8.67 ! 10 " 14 N (b) a = F m = 8.67 ! 10 " 14 N 1.67 ! 10 " 27 kg = 5.19 ! 10 13 2

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P22.5 Gravitational force: F g = mg = 9.11 ! 10 " 31 kg ( ) 9.80 m s 2 ( ) = 8.93 ! 10 " 30 N down . Electric force: F e = qE = ! 1.60 " 10 ! 19 C ( ) down ( ) = 1.60 " 10 ! 17 N up . Magnetic force: r F B = q r v ! r B = " 1.60 ! 10 " 19 C ( ) 6.00 ! 10 6 m s ˆ E ( ) ! 50.0 ! 10 " 6 N # # m ˆ N ( ) . r F B = ! 4.80 " 10 ! 17 N up = 4.80 " 10 ! 17 N down . P22.15 r F B = I r l ! r B = 2.40 A ( ) 0.750 m ( ) ˆ i ! 1.60 T ( ) ˆ k = " 2.88 ˆ j ( ) N P22.16 (a) F B = ILB sin ! = 5.00 A ( ) 2.80 m ( ) 0.390 T ( ) sin 60.0 ° = 4.73 N (b) F B = 5.00 A ( ) 2.80 m ( ) 0.390 T ( ) sin 90.0 ° = 5.46 N (c) F B = 5.00 A ( ) 2.80 m ( ) 0.390 T ( ) sin120 ° = 4.73 N P22.19 (a) 2 r = 2.00 m so r = 0.318 m μ = IA = 17.0 ! 10 " 3 A ( ) # 0.318 ( ) 2 m 2 \$ % & ' = 5.41 mA ( m 2 (b) r = r " r B so = 5.41 " 10 # 3 A \$ m 2 ( ) 0.800 T ( ) = 4.33 mN \$ m P22.24 B = 0 I 2 R = 4 " 10 # 7 T \$ m A ( ) 1.00 " 10 4 A ( ) 2 100 m ( ) = 2.00 " 10 # 5 T = 20.0 T P22.26 B = 0 I 2 r = 4 " 10 # 7 ( ) 1.00 A ( ) 2 1.00 m ( ) = 2.00 " 10 # 7 T
P22.34 Let both wires carry current in the x direction, the first at y = 0 and the second at y = 10.0 cm . (a) r B = μ 0 I 2 ! r ˆ k = 4 " 10 # 7 T \$ m A ( ) 5.00 A ( ) 2 0.100 m ( ) ˆ k r B = 1.00 ! 10 " 5 T out of the page FIG. P22.34(a) (b) r F B = I 2 r l ! r B = 8.00 A ( ) 1.00 m ( ) ˆ i ! 1.00 ! 10 " 5 T ( ) ˆ k # \$ % & = 8.00 ! 10 " 5 N ( ) " ˆ j ( ) r F B = 8.00 ! 10 " 5 N toward the first wire (c) r B = 0 I 2 r " ˆ k ( ) = 4 # 10 " 7 T \$ m A ( ) 8.00 A ( ) 2 0.100 m ( ) " ˆ k ( ) = 1.60 # 10 " 5 T ( ) " ˆ k ( ) r B = 1.60 ! 10 " 5 T into the page (d) r F B = I 1 r l !

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This note was uploaded on 11/27/2011 for the course PHYS 1A taught by Professor Onuchic during the Fall '07 term at UCSD.

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Chap22Sol - SOLUTIONS TO PROBLEMS Easy P22.1 (a) up (b) out...

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