PHY_solution-chapter27

# PHY_solution-chapter27 - Page 1 of 2227.1: a) ) ˆ ˆ T)(...

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Unformatted text preview: Page 1 of 2227.1: 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 ˆ )( s m 10 19 . 4 ( ) ˆ ˆ )( s m 10 3.85 T)[( C)(1.40 10 24 . 1 ( 4 4 8 k 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 . ( 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- = × × × × = ⇒-- 27.5: See figure on next page. Let , qvB F = then: F F a = in the k ˆ- direction F F b = in the j ˆ + direction , = c F since B and velocity are parallel o 45 sin F F d = in the j ˆ- direction F F e = in the ) ˆ ˆ ( k j +- direction 19dbb73ef466c110619aa14a26965a5399d55140.doc Page 1 of 22 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 . sin sin ) s m 10 25 . 3 ( 4 1 2 16 = ⇒ = ⇒ = × = φ φ φ m qvB a 27.7: 60 sin ) T 10 C)(3.5 10 (1.6 N 10 60 . 4 sin sin 3 19- 15-- × × × = = ⇒ = φ φ 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 ; = = +- = + + = ⋅ θ 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 , 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 , ) ( =- = 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. 19dbb73ef466c110619aa14a26965a5399d55140.doc Page 2 of 22 x y x y y x z B qv B v B v q F- =- = ) ( T 175 .- =- = 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 ....
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## This note was uploaded on 02/25/2009 for the course PHY 127 taught by Professor Mr.yang during the Spring '08 term at SUNY Stony Brook.

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PHY_solution-chapter27 - Page 1 of 2227.1: a) ) ˆ ˆ T)(...

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