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# ass19 - SOLUTIONS ASSIGNMENT 19 27.28 IDENTIFY:For no...

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SOLUTIONS, ASSIGNMENT 19 27.28. I DENTIFY : For no deflection the magnetic and electric forces must be equal in magnitude and opposite in direction. S ET U P : / v E B = for no deflection. With only the magnetic force, 2 / q vB mv R = E XECUTE : (a) 4 3 6 (1.56 10 V m ) (4.62 10 T) 3.38 10 m s. v E B - = = × × = × (b) The directions of the three vectors , v r E r and B r are sketched in Figure 27.28. (c) 31 6 3 19 3 (9.11 10 kg)(3.38 10 m/s) 4.17 10 m. (1.60 10 C)(4.62 10 T) mv R q B - - - - × × = = = × × × 3 9 6 2 2 2 (4.17 10 m) 7.74 10 s. (3.38 10 m s ) m R T q B v π π π - - × = = = = × × E VALUATE : For the field directions shown in Figure 27.28, the electric force is toward the top of the page and the magnetic force is toward the bottom of the page. Figure 27.28 27.30. I DENTIFY : For no deflection the magnetic and electric forces must be equal in magnitude and opposite in direction. S ET U P : / v E B = for no deflection. E XECUTE : To pass undeflected in both cases, 3 (5.85 10 m s)(1.35 T) 7898 N C. E vB = = × = (a) If 9 0.640 10 C, q - = × the electric field direction is given by ˆ ˆ ˆ ( ( )) , - × - j k = i since it must point in the opposite direction to the magnetic force. (b) If 9 0.320 10 C, q - = - × the electric field direction is given by ˆ ˆ ˆ (( ) ( )) , - × - = j k i since the electric force must point in the opposite direction as the magnetic force. Since the particle has negative charge,

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