Chapter 28 - 28.1: For a charge with velocity , ˆ ) s m 10...

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Unformatted text preview: 28.1: For a charge with velocity , ˆ ) s m 10 8.00 ( 6 j v × = the magnetic field produced at a position r away from the particle is . ˆ 4 2 r q μ r v B × = π So for the cases below: a) 4 1 2 , ˆ ˆ ˆ ˆ ) m 0.500 ( =- = × ⇒ + = r k r v i r . ˆ ˆ ) T 10 1.92 ( ˆ ) m 0.50 ( ) s m 10 8.0 )( C 10 (6.0 4 ˆ 4 5 2 6 6 2 k k k k B B π μ r qv π μ- ≡ ×- = × ×- =- = ⇒-- b) . ˆ ˆ ˆ ) m 0.500 ( = ⇒ = × ⇒- = B r v j r c) . 4 1 , ˆ ˆ ˆ ˆ ) m 0.500 ( 2 = + = × ⇒ = r i r v k r . ˆ ˆ 4 i i B 2 B r qv = + = ⇒ π μ d) 2 2 2 1 , ˆ ˆ ˆ ˆ ) m 0.500 ( ˆ ) m 0.500 ( r r = =- = × ⇒ +- = i r v k j r 2 2 ˆ 2 ˆ 2 2 ˆ 4 i B B r qv + = + = + = ⇒ i i B 2 π μ 28.2: ′ ′ + = ′ + = 2 2 total 4 d v q d qv B B B π μ page. the into T, 10 38 . 4 ) m 120 . ( ) s m 10 . 9 )( C 10 . 3 ( ) m 120 . ( ) s m 10 5 . 4 )( C 10 . 8 ( 4 4 2 6 6 2 6 6--- × = ⇒ × × + × × = ⇒ B μ B π 28.3: 3 4 r q π μ r v B × = a) ; ˆ , i r i v r v = = , = = × B r v b) ; ˆ , ˆ j r i v r v = = m 0.500 , ˆ = = × r vr k r v k B ˆ ) T 1.31 ( so negative, is T 10 1.31 ) m 0.500 ( ) s m 10 6.80 )( C 10 4.80 )( C s N 10 1 4 6 6 2 5 6 2 2 7 2---- ×- = × = × × ⋅ × = = q r v q π μ B c) ); ˆ ˆ )( m 0.500 ( , ˆ j i r i v + = = v m 0.7071 , ˆ ) m 0.500 ( = = × r v k r v ( 29 k B ˆ ) T 10 4.62 ( ; T 10 4.62 ) m 0.7071 ( ) s m 10 6.80 )( m 0.500 ( C) 10 4.80 )( C s N 10 1 4 7 7 3 5 6 2 2 7 3---- ×- = × = × × ⋅ × = × = B r r v q π μ B d) ; ˆ , ˆ k r i v r v = = m 0.500 , ˆ =- = × r vr j r v j B ˆ T) 10 1.31 ( ; T 10 1.31 m) 0.500 ( ) s m 10 6.80 )( C 10 4.80 )( C s N 10 1 4 6 6 2 5 6 2 2 7 2---- × = × = × × ⋅ × = = B r v q π μ B 28.4: a) Following Example 28.1 we can find the magnetic force between the charges: down). points char lower the on force the and up points charge upper the on force (the N 10 1.69 ) m 0.240 ( m 10 9.00 )( s m 10 4.50 )( C 10 3.00 )( C 10 8.00 ( ) A m T 10 ( 4 3 2 6 6 6 6 7 2---- × = × × × × ⋅ = ′ ′ = r v v q q π μ F B The Coulomb force between the charges is N 3.75 ) C m N 10 8.99 ( 2 2 12 m) (0.240 C 10 0) (8.00)(3.0 2 2 9 = ⋅ × = =- × 2 2 1 r q q k F (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 2 1 2 v v c = × =- × 3 N 10 1.69 N 3.75 10 2.22 3 . b) The magnetic forces are reversed when the direction of only one velocity is reversed but the magnitude of the force is unchanged. 28.5: The magnetic field is into the page at the origin, and the magnitude is page. the into T, 10 1.64 m) 0.400 ( ) s m 10 8.0 )( C 10 1.5 ( ) m 0.300 ( ) s m 10 2.0 )( C 10 4.0 ( 4 4 6 2 5 6 2 5 6 2 2--- × = ⇒...
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This homework help was uploaded on 02/26/2008 for the course PHYS 11 and 21 taught by Professor Hickman during the Spring '08 term at Lehigh University .

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Chapter 28 - 28.1: For a charge with velocity , ˆ ) s m 10...

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