821_PartUniversity Physics Solution

821_PartUniversity Physics Solution - 27-28 Chapter 27...

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27-28 Chapter 27 EVALUATE: ! μ is along the z -axis so only and x y B B contribute to the torque. x B produces a y -component of ! τ and y B produces an x -component of . ! Only z B affects U , and U is negative when and z B ! ! are parallel. 27.86. IDENTIFY: q I t Δ = Δ and . IA = SET UP: The direction of ! is given by the right-hand rule that is illustrated in Figure 27.32 in the textbook. I is in the direction of flow of positive charge and opposite to the direction of flow of negative charge. EXECUTE: (a) . 23 u u dq q q v ev I dt t r r π Δ === = Δ (b) 2 . 33 uu ev evr IA r r μπ == = (c) Since there are two down quarks, each of half the charge of the up quark, u . 3 d evr μμ Therefore, total 2 . 3 evr μ = (d) 27 2 7 19 15 ( 9 . 6 6 1 0 A m ) 7.55 10 m s. 22 ( 1 . 6 0 1 0 C ) ( 1 . 2 0 1 0 m ) μ v er −− ×⋅ ×× EVALUATE: The speed calculated in part (d) is 25% of the speed of light. 27.87. IDENTIFY: Eq.(27.8) says that the magnetic field through any closed surface is zero. SET UP: The cylindrical Gaussian surface has its top at zL = and its bottom at 0 z = . The rest of the surface is the curved portion of the cylinder and has radius r and length L . 0 B = at the bottom of the surface, since 0 z = there. EXECUTE: (a) top curved top curved () 0 . zr r dB d AB d A β LdA BdA =+ = + = ∫∫∫ BA ! ! o This gives 2 02 r L rBr L β ππ , and . 2 r β r Br =− (b) The two diagrams in Figure 27.87 show views of the field lines from the top and side of the Gaussian surface. EVALUATE: Only a portion of each field line is shown; the field lines are closed loops. Figure 27.87 27.88. IDENTIFY: U =− ⋅ B ! ! . In part (b) apply conservation of energy. SET UP: The kinetic energy of the rotating ring is 2 1 2 K I ω = . EXECUTE: (a) fi f i 0 && & & & & ( ) ( ( 0 . 8 0 . 6 ) ) ( 1 2 3 4 ) U μ B ⎤⎡ Δ= − ⋅−⋅ = − − ⋅=−−−− + + − ⎦⎣ μ B μ B Bk i j i j k "! !! ! ! . 42 0 [( 0.8)( 12) (0.6)( 3) ( 1)( 4)] (12.5 A)(4.45 10 m )(0.0115 T)( 11.8). UI A B − + + +++− = × 4 7.55 10 J U − × . (b) 2 1 2 KI ω . 4 72 ( 7 . 5 5 1 0 J ) 42.1rad/s. 8.50 10 kg m K ω I Δ× = EVALUATE: The potential energy of the ring decreases and its kinetic energy increases. 27.89. IDENTIFY and SET UP: In the magnetic field, mv R qB = . Once the particle exits the field it travels in a straight line. Throughout the motion the speed of the particle is constant. EXECUTE : (a) 11 5 6 (3.20 10 kg)(1.45 10 m/s) 5.14 m. (2.15 10 C)(0.420 T) mv R qB = ×
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Magnetic Field and Magnetic Forces 27-29 (b) See Figure 27.89. The distance along the curve, , d is given by dR θ = . 0.35 m sin , 5.14 m = so 2.78 0.0486 rad. == ° (5.14 m)(0.0486 rad) 0.25 m. θ = And 6 5 0.25 m 1.72 10 s. 1.45 10 m/s d t v = × × (c) 3 1 tan( /2) (0.25 m)tan (2.79 /2) 6.08 10 m. xd θ Δ= = ° = × (d) 12 x xx Δ=Δ +Δ , where 2 x Δ is the horizontal displacement of the particle from where it exits the field region to where it hits the wall.
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821_PartUniversity Physics Solution - 27-28 Chapter 27...

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