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**Unformatted text preview: **Chapter 19 Magnetism Problem Solutions 19.1 The direction in parts (a) through (d) is found by use of the right hand rule. You must remember that the electron is negatively charged and thus experiences a force in the direction exactly opposite that predicted by the right hand rule for a positively charged particle.  (a)  horizontal and due east  (b)  horizontal and 30 N of E °  (c)  horizontal and due east  (d)  zero force , ( ) sin sin 180 F qvB qvB θ = = ° = 19.3 Since the particle is positively charged, use the right hand rule. In this case, start with the fingers of the right hand in the direction of v and the thumb pointing in the direction of F . As you start closing the hand, the fingers point in the direction of B after they have moved 90°. The results are  (a)  into the page  (b)  toward the right  (c)  toward bottom of page 19.8 The speed attained by the electron is found from ( ) 2 1 2 mv q V = Δ , or ( ) ( ) ( ) 19 7 31 2 1.60 10 C 2 400 V 2 2.90 10 m s 9.11 10 kg e V v m-- × Δ = = = × × (a) Maximum force occurs when the electron enters the region perpendicular to the field. ( ) ( ) ( ) max 19 7 12 sin90 1.60 10 C 2.90 10 m s 1.70 T 7.90 10 N F q vB-- = ° = × × = × 141 142       CHAPTER 19 (b) Minimum force occurs when the electron enters the region parallel to the field. min sin0 F q vB = ° = 19.9 ( ) ( ) ( ) ( ) 27 13 2 19 7 1.67 10 kg 2.0 10 m s 0.021 T 1.60 10 C 1.0 10 m s ma F B qv qv-- × × = = = = × × The right hand rule shows that B must be in the direction y- to yield a force in the + x direction when v is in the + z direction. 19.11 From sin F BIL θ = , the magnetic field is ( ) 3 0.12 N m 8.0 10 T sin 15 A sin90 F L B I θ- = = = × ° The direction of B must be the direction z + to have F in the – y direction when I is in the + x direction. 19.15 ( )( )( ) ( ) sin 0.300 T 10.0 A 5.00 m sin 30.0 7.50 N F BIL θ = = ° = 19.19 For the wire to move upward at constant speed, the net force acting on it must be zero....

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