Chapter 35 - 35.1. Model: Apply the Galilean transformation...

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35.1. Model: Apply the Galilean transformation of velocity. Solve: (a) In the laboratory frame S, the speed of the proton is () 22 66 6 1.41 10 m/s 2.0 10 m/s v + × = × The angle the velocity vector makes with the positive y -axis is 6 1 6 tan 45 θ ⎛⎞ × = ⎜⎟ × ⎝⎠ (b) In the rocket frame S , we need to first determine the vector v G . Equation 34.1 yields: ( ) ( ) 6 ˆˆ ˆ 1.41 10 1.41 10 m/s 1.00 10 m/s 0.41 10 1.41 10 m/s vvV i j i i j =−= × + × × = × + × G GG The speed of the proton is 666 0.41 10 m/s 1.47 10 m/s v + × = × The angle the velocity vector makes with the positive y -axis is 6 1 6 tan 16.2 × ′ = ×
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35.2. Model: Apply the Galilean transformation of fields. Visualize: Please refer to Figure EX35.2. Solve: (a) Equation 35.11 gives the Galilean field transformation equation for magnetic fields: 2 1 B BV E c ′ = −× G GG G B G is in the positive ˆ k direction, ˆ B Bk = G . For B > B , VE × G G must be in the negative ˆ k direction. Since ˆ , E Ej = G V G must be in the negative ˆ i direction, so that ( ) ( ) ˆ ˆˆ . V E Vi Ej VEk ×= − × = G G The rocket scientist will measure B > B , if the rocket moves along the – x -axis. (b) For B = B , × must be zero. The rocket scientist will measure B = B if the rocket moves along either the + y -axis or the – y -axis. (c) For B < B , × must be in the positive ˆ k direction. The rocket scientist will measure B < B , if the rocket moves along the + x -axis.
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35.3. Model: Use the Galilean transformation of fields. Solve: Equation 35.11 gives the Galilean transformation equations for the electric and magnetic fields in S and S frames: 2 1 E EVB B B VE c ′′ =+× =− × GGGG GG GG In a region of space where 0 B = G G , 6 ˆ 1.0 10 V/m. EE j == − × GG The magnetic field is () ( ) 12 66 5 2 2 8 11 . 0 1 0 ˆˆ 0 1.0 10 m/s 1.0 10 V/m T 1.11 10 T 3.0 10 B ij k k c × × ×− × = = × × G G
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35.4. Model: Use the Galilean transformation of fields. Visualize: Please refer to Figure EX35.4. We are given 6 ˆ 2.0 10 m/s, Vi G ˆ 1.0 T, B j ′ = and 6 ˆ 1.0 10 V/m. Ek G Solve: Equation 35.11 gives the Galilean transformation equations for the electric and magnetic fields in S and S frames: 2 1 E EVB BB VE c ′′ =−× =+ × GGGG GG The electric and magnetic fields viewed from earth are ( ) ( ) ( ) 66 6 ˆˆ 1.0 10 V/m 2.0 10 m/s 1.0 T 1.0 10 V/m E ki j k − × × = G () 12 2 2 8 1 2.0 10 V/m ˆ ˆ ˆ ˆ 1.0 T 1.0 10 V/m 0.99998 T 3.0 10 m/s B ji k j j j c × =+ × × × = = × G Assess: Although , B B < you need 5 significant figures of accuracy to tell the difference between them.
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35.5. Model: Use the Galilean transformation of fields. Visualize: Please refer to Figure EX35.5. We are given 6 ˆ 1.0 10 m/s, Vi G ˆ 0.50 T, B k = G and () 6 11 22 ˆˆ 10 V/m. Eij =+× G Solve: Equation 35.11 gives the Galilean transformation equation for the electric field in the S and S frames: E EVB GGGG . The electric field from the moving rocket is ( ) ( ) ( ) ( ) 66 6 6 ˆ ˆ ˆ ˆ 0.707 10 V/m 1.0 10 m/s 0.50 T 0.707 10 0.207 10 V/m Ei j i k i j =+ × + × × = × + × G 6 1 6 0.207 10 V/m tan 16.3 0.707 10 V/m θ ⎛⎞ × == ° ⎜⎟ × ⎝⎠ above the x -axis
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35.6.
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Chapter 35 - 35.1. Model: Apply the Galilean transformation...

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