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# chap27 - 27.1 a F b F qv B 1.24 10 8 C 3.85 104 m s(1.40 T...

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27.1: a) ) ˆ ˆ T)( )(1.40 s m 10 3.85 C)( 10 24 . 1 ( 4 8 i j B v F q . ˆ N) 10 68 . 6 ( 4 k F b) B v F q ˆ )( s m 10 19 . 4 ( ) ˆ ˆ )( s m 10 3.85 T)[( C)(1.40 10 24 . 1 ( 4 4 8 i k j F . ˆ N) 10 27 . 7 ( ˆ N) 10 68 . 6 ( 4 4 j i F 27.2: Need a force from the magnetic field to balance the downward gravitational force. Its magnitude is: T. 91 . 1 ) s m 10 C)(4.00 10 50 . 2 ( ) s m kg)(9.80 10 95 . 1 ( 4 8 2 4 qv mg B mg qvB 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. 27.3: By the right-hand rule, the charge is positive. 27.4: m q q m B v a B v a F . ˆ ) s m 330 . 0 ( kg 10 81 . 1 ) ˆ ˆ T)( )(1.63 s m 10 C)(3.0 10 22 . 1 ( 2 3 4 8 k i j a

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27.5: See figure on next page. Let , 0 qvB F then: 0 F F a in the k ˆ direction 0 F F b in the j ˆ direction , 0 c F since B and velocity are parallel o 45 sin 0 F F d in the j ˆ direction 0 F F e in the ) ˆ ˆ ( k j direction 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: . s m 10 25 . 3 kg) 10 (9.11 T) 10 )(7.4 s m 10 C)(2.50 10 6 . 1 ( 2 16 31 2 6 19 m qvB a b) If . 5 . 14 25 . 0 sin sin ) s m 10 25 . 3 ( 4 1 2 16 m qvB a 27.7: 60 sin ) T 10 C)(3.5 10 (1.6 N 10 60 . 4 sin sin 3 19 - 15 B q F v B v q F . s m 10 49 . 9 6
27.8: a) )]. ˆ ( ) ˆ ( [ )] ˆ ˆ ( ) ˆ ˆ ( ) ˆ ˆ ( [ i j k k k j k i B v F y x z z y x z v v qB v v v qB q Set this equal to the given value of F to obtain: s m 106 T) 1.25 C)( 10 5.60 ( N) 10 40 . 7 ( 9 7 z y x qB F v . s m 6 . 48 T) 1.25 C)( 10 5.60 ( N) 10 40 . 3 ( 9 7 z x y qB F v b) The value of z v is indeterminate. c) . 90 ; 0 y z x x z y z z y y x x F qB F F qB F F v F v F v F v 27.9: s m 10 80 . 3 ˆ , 3 y y v with v q j v B v F , 0 N, 10 60 . 7 3 y x F F and N 10 20 . 5 3 z F z y y z z y x B qv B v B v q F ) ( T 0.256 )] s m 10 3.80 C)( 10 ([7.80 N) 10 60 . 7 ( 3 6 3 y x z qv F B , 0 ) ( z x x z y B v B v q F which is consistent with F as given in the problem. No force component along the direction of the velocity. x y x y y x z B qv B v B v q F ) ( T 175 . 0 y z x qv F B b) y B is not determined. No force due to this component of B along ; v measurement of the force tells us nothing about . y B c) N) 10 7.60 T)( 175 . 0 ( 3 z z y y x x F B F B F B F B N) 3 10 5.20 T)( 0.256 ( B F B ; 0 and F are perpendicular (angle is ) 90 o

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27.10: a) The total flux must be zero, so the flux through the remaining surfaces must be 120 . 0 Wb.
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