Chapter_27_Solutions - -8 4 j i 27.1: a) F = qv B = (-1.24...

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27.1: a) G F = q G v × G B = ( 1.24 × 10 8 C)( 3.85 × 10 4 ms )(1.40T)( ˆ j × ˆ i ) G F =− (6.68 × 10 4 N) ˆ k . b) G F = q G v × G B G F = ( 1.24 × 10 8 C)(1.40 T)[( 3.85 × 10 4 )( ˆ j × ˆ k ) + (4.19 × 10 4 )( ˆ i × ˆ k )] G F = (6.68 × 10 4 N) ˆ i + (7.27 × 10 4 N) ˆ j .
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27.2: Need a force from the magnetic field to balance the downward gravitational force. Its magnitude is: qvB = mg B = mg qv = (1.95 × 10 4 kg)(9.80 m s 2 ) (2.50 × 10 8 C)(4.00 × 10 4 ms ) = 1.91 T. The right-hand rule requires the magnetic field to be to the east, since the velocity is northward, the charge is negative, and the force is upwards.
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27.3: By the right-hand rule, the charge is positive.
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27.4: G F = m G a = q G v × G B G a = q G v × G B m G a = (1.22 × 10 8 C)(3.0 × 10 4 ms )(1.63 T)( ˆ j × ˆ i ) 1.81 × 10 3 kg =− (0.330 m s 2 ) ˆ k.
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27.5: See figure on next page. Let F 0 = qvB, then: F a = F 0 in the ˆ k direction F b = F 0 in the direction + ˆ j F c = 0, since and velocity are parallel B in the direction F d = F 0 sin 45 D ˆ j F e = F 0 in the direction ( ˆ j + ˆ k )
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27.6: a) The smallest possible acceleration is zero, when the motion is parallel to the magnetic field. The greatest acceleration is when the velocity and magnetic field are at right angles: a = qvB m = (1.6 × 10 19 C)(2.50 × 10 6 ms )(7.4 × 10 2 T) (9.11 × 10 31 kg) = 3.25 × 10 16 2 . b) If a = 1 4 (3.25 × 10 16 2 ) = qvBsin φ m sin = 0.25 = 14.5 °
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27.7: F = q v B sin φ v = F q B sin = 4.60 × 10 15 N (1.6 × 10 19 C)(3.5 × 10 3 T) sin 60 ° = 9.49 × 10 6 ms .
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27.8: a) G F = q G v × G B = qB z [v x ( ˆ i × ˆ k) + v y ( ˆ j × ˆ k) + v z ( ˆ k × ˆ k )] = qB z x ( ˆ j ) + v y ( ˆ i )]. Set this equal to the given value of G F to obtain: v x = F y qB z = (7.40 × 10 7 N) ( 5.60 × 10 9 C)( 1.25 T) =− 106 m s v y = F x qB z = (3.40 × 10 7 N) ( 5.60 × 10 9 C)( 1.25 T) 48.6 m s . b) The value of is indeterminate. v z c) G v G F = v x F x + v y F y + v z F z = F y qB z F x + F x qB z F y = 0; θ = 90 ° .
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27.9: G F = q G v × G B , G v = v y ˆ j with v y =− 3.80 × 10 3 ms and F x =+ 7.60 × 10 3 N, F y = 0, F z = − 5.20 × 10 3 N F x = q(v y B z v z B y ) = qv y B z B z = F x qv y = (7.60 × 10 3 N) [(7.80 × 10 6 C)( 3.80 × 10 3 )] 0.256 T which is consistent with F y = q(v z B x v x B z ) = G F as given in the problem. No force component along the direction of the velocity. F z = x B y v y B x ) qv y B x B x F z qv y 0.175 T b) B is not determined. No force due to this component of y G B along measurement of the force tells us nothing about G v ; B y . c) G B G F = B x F x + B y F y + B z F z = ( 0.175 T)( + 7.60 × 10 3 N) + ( 0.256 T) ( 5.20 × 10 3 N) and are perpendicular (angle is 90 ) G B G F = 0; G B G F °
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27.10: a) The total flux must be zero, so the flux through the remaining surfaces must be Wb. 0.120 b) The shape of the surface is unimportant, just that it is closed. c)
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27.11: a) Φ B = G B G A = (0.230 T) π (0.065 m) 2 = 3.05 × 10 3 Wb. b) Φ B = G B G A = (0.230 T) (0.065 m) 2 cos 53.1 °= 1.83 × 10 3 Wb.
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Chapter_27_Solutions - -8 4 j i 27.1: a) F = qv B = (-1.24...

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