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28_InstructorSolutions

# 28_InstructorSolutions - SOURCES OF MAGNETIC FIELD 28 28.1...

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28-1 S OURCES OF M AGNETIC F IELD 28.1. IDENTIFY and SET UP: Use Eq.(28.2) to calculate B ! at each point. 00 23 & & , since . 44 qq rr r μμ ππ vr r Br !! ! ! ! ×× == = ( ) 6 & 8.00 10 m/s and × v j r = is the vector from the charge to the point where the field is calculated. EXECUTE: (a) () & 0.500 m , 0.500 m r = ri ! = && & vr vr j ik ×= ×= ( )( ) 66 7 0 2 2 6.00 10 C 8.00 10 m/s 11 0 Tm /A 4 0.500 m qv r μ π −× B kk ! =− = ( ) 5 & 1.92 10 T B k ! = (b) & 0.500 m 0.500 m r −= rj ! = 0 and 0. vr jj B ! (c) & 0.500 m 0.500 m r = rk ! = & vr vr j ki ( )( ) 7 5 2 1 10 T m/A 0.500 m ×⋅ + × B i= i ! = (d) ()() 22 0.500 m 0.500 m 0.500 m 0.500 m 0.7071 m r + = k ! =+ ( ) 62 &&&& & 0.500 m 4.00 10 m /s v × = jjj × ×+× ( )( ) 7 6 3 6.00 10 C 4.00 10 m/s 6.79 10 T 0.7071 m + × B ii ! EVALUATE: At each point B ! is perpendicular to both v ! and . r ! B = 0 along the direction of . v ! 28.2. IDENTIFY: A moving charge creates a magnetic field as well as an electric field. SET UP: The magnetic field caused by a moving charge is 0 2 sin 4 qv B r μφ = , and its electric field is 2 0 1 4 e E r = P since q = e . EXECUTE: Substitute the appropriate numbers into the above equations. 71 9 6 0 21 1 2 sin 4 10 T m/A (1.60 10 C)(2.2 10 m/s)sin90 44( 5 . 3 1 0 m ) qv B r −− × × ° × = 13 T, out of the page. 92 2 1 9 11 1 2 0 1 (9.00 10 N m /C )(1.60 10 C) 5.1 10 N/C, 4( 5 . 3 1 0 m ) e E r × = × × P toward the electron. EVALUATE: There are enormous fields within the atom! 28.3. IDENTIFY: A moving charge creates a magnetic field. SET UP: The magnetic field due to a moving charge is 0 2 sin 4 qv B r = . 28

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28-2 Chapter 28 EXECUTE: Substituting numbers into the above equation gives (a) 71 9 7 0 26 2 sin 4 10 T m/A (1.6 10 C)(3.0 10 m/s)sin30 . 44 ( 2 . 0 0 1 0 m ) qv B r μφ π ππ −− ×⋅ × × ° == × B = 6.00 × 10 &8 T, out of the paper, and it is the same at point B . (b) B = (1.00 × 10 &7 T m/A)(1.60 × 10 &19 C)(3.00 × 10 7 m/s)/(2.00 × 10 &6 m) 2 B = 1.20 × 10 &7 T, out of the page. (c) B = 0 T since sin(180±) = 0. EVALUATE: Even at high speeds, these charges produce magnetic fields much less than the Earth²s magnetic field. 28.4. IDENTIFY: Both moving charges produce magnetic fields, and the net field is the vector sum of the two fields. SET UP: Both fields point out of the paper, so their magnitudes add, giving B = B alpha + B el = () 0 2 sin 40 2 sin140 4 v ee r μ °+ ° EXECUTE: Factoring out an e and putting in the numbers gives 9 5 92 4 10 T m/A (1.60 10 C)(2.50 10 m/s) sin40 2sin140 4 (1.75 10 m) B × × + ° × 3 2.52 10 T 2.52 mT, out of the page. B = EVALUATE: At distances very close to the charges, the magnetic field is strong enough to be important. 28.5. IDENTIFY: Apply 0 3 . 4 μ q π r × vr B= !! ! SET UP: Since the charge is at the origin, ³³³ . x yz ++ r= i j k ! EXECUTE: (a) ³ ,; v= i r= i ! 0, 0 B ×= = . (b) ³³ v= i r= j ³ , 0.500 m. vr r = k 72 2 6 5 6 0 22 (1.0 10 N s /C )(4.80 10 C)(6.80 10 m/s) 1.31 10 T. 4 (0.500 m) qv μ B π r × × ⎛⎞ = × ⎜⎟ ⎝⎠ q is negative, so 6 ³ (1.31 10 T) . −× B =k ! (c) ³ , (0.500 m)( ); v + i j ³ (0.500 m) 0.7071 m. = k 2 6 5 3 0 3 (1.0 10 N s /C )(4.80 10 C)(0.500 m)(6.80 10 m/s) . 4 (0.7071 m) μ Bq r π × × = 7 4.62 10 T. B 7 ³ (4.62 10 T) B . ! (d) v= i r= k ³ , 0.500 m vr r ×− = = j 2 6 5 6 0 4 (0.500 m) μ B π r × × = × ( ) 6 ³ 1.31 10 T .
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28_InstructorSolutions - SOURCES OF MAGNETIC FIELD 28 28.1...

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