Chapter_28_Solutions - 28.1: For a charge with velocity v =...

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28.1: For a charge with velocity G v = (8.00 × 10 6 ms ) ˆ j , the magnetic field produced at a position r away from the particle is G B = µ 0 4 π q G v × ˆ r r 2 . So for the cases below: a) G r = ( + 0.500 m) ˆ i ˆ v × ˆ r =− ˆ k, r 0 2 = 1 4 G B 0 4 qv r 0 2 ˆ k 0 4 (6.0 × 10 6 C)(8.0 × 10 6 ) (0.50 m) 2 ˆ k (1.92 × 10 5 T) ˆ k ≡− B 0 ˆ k . b) G r = ( 0.500 m) ˆ j ˆ v × ˆ r = 0 G B = 0. c) G r = (0.500 m) ˆ k ˆ v × ˆ r =+ ˆ i , r 0 2 = 1 4 . G B 0 4 qv r 0 2 ˆ i = B 0 ˆ i . d) G r (0.500 m) ˆ j + (0.500 m) ˆ k ˆ v × ˆ r ˆ i , r 2 = 1 2 = 2r 0 G B 0 4 qv r 2 ˆ i 2 B 0 2 ˆ i 2 B 0 ˆ i 22
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28.2: B total = B +′ B = µ 0 4 π qv d 2 + q v d 2 B = 0 4 (8.0 × 10 6 C)(4.5 × 10 6 ms ) (0.120 m) 2 + (3.0 × 10 6 C)(9.0 × 10 6 ) (0.120 m) 2 B = 4.38 × 10 4 T, into the page.
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28.3: G B = µ 0 4 π q G v × G r r 3 a) G v = v ˆ i , G r = r ˆ i ; G v × G r = 0, B = 0 b) G v = v ˆ i , G r = r ˆ j ; G v × G r = vr ˆ k, r = 0.500 m B = 0 4 q v r 2 = 1 × 10 7 N s 2 C 2 )(4.80 × 10 6 C)(6.80 × 10 5 ms ) (0.500 m) 2 = 1.31 × 10 6 T q is negative, so G B = − (1.31 × 0 6 T) ˆ k c) G v = v ˆ i , G r = (0.500 m)( ˆ i + ˆ j ); G v × G r = (0.500 m)v ˆ k, r = 0.7071 m B = 0 4 q G v × G r r 3 () = (1 × 10 7 N s 2 C 2 )(4.80 × 10 6 C)(0.500 m)(6.80 × 10 5 ) (0.7071 m) 3 B = 4.62 × 10 7 T; G B =− (4.62 × 10 7 ˆ k d) G G v = v ˆ i , G r = r ˆ k ; v × G r = − vr ˆ j , r = 0.500 m B = 0 4 q v r 2 = (1 × 10 7 N s 2 C 2 )(4.80 × 10 6 C)(6.80 × 10 5 ) (0.500 m) 2 B = 1.31 × 10 6 G B = (1.31 × 10 6 T) ˆ j
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28.4: a) Following Example 28.1 we can find the magnetic force between the charges: F B = µ 0 4 π q q v v r 2 = (10 7 T mA ) (8.00 × 10 6 C)(3.00 × 10 6 C)(4.50 × 10 6 ms )(9.00 × 10 6 ) (0.240 m) 2 = 1.69 × 10 3 N (the force on the upper charge points up and the force on the lower charge points down). The Coulomb force between the charges is F = k q 1 q 2 r 2 = (8.99 × 10 9 N m 2 C 2 ) (8.00)(3.00) × 10 12 C 2 (0.240 m) 2 = 3.75 N (the force on the upper charge points up and the force on the lower charge points down). The ratio of the Coulomb force to the magnetic force is 3.75N 1.69 × 10 3 N = 2.22 × 10 3 = c 2 v 1 v 2 . b) The magnetic forces are reversed when the direction of only one velocity is reversed but the magnitude of the force is unchanged.
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28.5: The magnetic field is into the page at the origin, and the magnitude is B = B + B = µ 0 4 π qv r 2 + q v r 2 B = 0 4 (4.0 × 10 6 C)(2.0 × 10 5 ms ) (0.300 m) 2 + (1.5 × 10 6 C)(8.0 × 10 5 ) (0.400 m) 2 B = 1.64 × 10 6 T, into the page.
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28.6: a) q =− q; B q = µ 0 qv 4 π d 2 into the page; B q = 0 q v 4 d 2 out of the page. (i) v = v 2 B = 0 qv 4 (2d 2 ) into the page. (ii) v = v B = 0. (iii) v = 2v B = 0 qv 4 d 2 out of the page. b) G F =′ q G v × G B q 0 q 2 v v 4 (2d) 2 and is attractive. c) F B = 0 q 2 v v 4 2 ,F C = q 2 4 πε 0 2 F B F C = 0 ε 0 v v = 0 0 (3.00 × 10 5 ms ) 2 = 1.00 × 10 6 .
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28.7: a) ˆ r = cos θ ˆ i + sin ˆ j = cos(150 ° ) ˆ i + sin(150 ° ) ˆ j =− (0.866) ˆ i + (0.500) ˆ j .
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This note was uploaded on 04/01/2008 for the course PHYSICS 240 taught by Professor Davewinn during the Fall '08 term at University of Michigan.

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Chapter_28_Solutions - 28.1: For a charge with velocity v =...

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