CH-26 WEB - Chapter-26 The Magnetic Field: (Additional...

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Chapter-26 The Magnetic Field: (Additional Questions) 1-) An electron that has velocity ( 29 ( 29 6 6 ˆ ˆ 2 10 m/s 3 10 m/s v i j = × + × r moves through the magnetic field ( 29 ( 29 ˆ ˆ 0.03 T 0.15 T B i j = - r . (a) Find the force on the electron. (b) Repeat your calculation for a proton having the same velocity. Solution: (a) The force on the electron is ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 19 6 6 14 ˆ ˆ ˆ ˆ ˆ ˆ 1.6 10 C 2 10 m/s 0.15 T 3 10 m/s 0.03 T ˆ 6.2 10 N B x y x y x y y x F qv B q v i v j B i B j q v B v B k k k - - = × = + × + = - = - × × - - × = × r r r Thus, the magnitude of B F r is 14 6.2 10 N - × , and B F r points in the positive z direction. (b) This amount to repeating the above computation with a change in the sign in the change. Thus, B F r has the same magnitude but points in the negative z direction. 2-) An electron has a velocity of 1.20 × 10 4 m/s (in the positive x direction), and an acceleration of 2.00 × 10 12 m/s 2 (in the positive z direction) in uniform electric and magnetic fields. If the electric field has a magnitude of 20.0 N/C (in the positive z direction), what can you determine about the magnetic field in the region? What can you not determine? Solution:
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3-) An alpha particle ( 2 , 4 u q e m = + = ) travels in a circular path of radius 4.5 cm in a uniform magnetic field with 1.2 B = T. Calculate (a) its speed, (b) its period of revolution, (c)
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This note was uploaded on 04/03/2012 for the course ENGINEERIN 12 taught by Professor Who during the Spring '09 term at Kadir Has Üniversitesi.

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CH-26 WEB - Chapter-26 The Magnetic Field: (Additional...

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