PHY_solution-chapter27

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

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

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.

Page1 / 22

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online