Worksheet #6 - Solutions

Worksheet #6 - Solutions - Part C. Part D. qvB F B = , r mv...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Physics 202 SI Session Worksheet 6 Solutions Problem 1. A) F=|q|vBsin60 = 2.8 x 10 -13 N and because v x B is in the positive z direction and the charge is negative, F is in the negative z direction B) v = (8.0 x 10 6 i ) m/s B = (.025cos60 i + .025sin60 j )T F = q v x Bb = -2.8 x 10 -13 N k Problem 2. Part A. Since the force on the electron in this magnetic field is in the downward direction, our Magnetic force must point upwards. F b points up and F E points down. Part B. The magnetic field must go out of the page. Remember the right hand rule applies to positive charges. This is a negative charge, so the force is in the opposite direction that your thumb is pointing with the right hand rule. Part C. E q F B v q F e B ) ) ) ) ) - = - = × = qE qvB = s m x s m C N C N T C N x B E v 3 4 10 133 133 15 . 0 10 00 . 2 = = = = Problem 3. kx F = 2 3 2 10 6 . 19 005 . 0 8 . 9 10 s g x m s m g x F k = = = N s gm m s g x kx F B 0588 . 0 8 . 58 ) 003 . 0 )( 10 6 . 19 ( 2 2 3 = = = = A V R V I 2 12 24 = = = IBh F B = T m A N Ih F B B 588 . 0 050 . 0 2 0588 . 0 = = = Problem 4. Part A. J x keV mv 16 2 10 602 . 1 1 2 - = = , m J x m k v 16 10 602 . 1 ( 2 2 - = =
Background image of page 1

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

View Full DocumentRight Arrow Icon
s m x v e 7 10 875 . 1 = , s m x v p 5 10 376 . 4 = , s m x v n 5 10 374 . 4 = Part B. The magnetic field does no work.
Background image of page 2
Background image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Part C. Part D. qvB F B = , r mv F c 2 = , c B F F = , qB mv r = mm r e 132 . 2 = , mm r p 40 . 91 = , = n r Problem 5. B L I F B ) ) ) = , d L ) ) = B F i IdB j B k Id B d I ) ) ) = =- = ) ( ) ( f i K W K = + 2 2 2 2 2 2 2 4 1 2 1 ) )( 2 1 ( 2 1 2 1 2 1 2 1 cos mv mv R v mR mv I mv Fs + = + = + = + 2 4 3 mv IdBL = s m v 07 . 1 72 . 3 4 45 . 24 . 12 . 48 = Problem 6. Part A. The right most side stays in contact with the table Part B. nILB nILB B L nI F B = = = ) 90 sin( h h h B nIL L F B B 2 = = 2 ) ( L mg g = g B = 2 2 mgL B nIL = A LBn mg I 41 . 2 = = Part C. p n e LBn mg I 2 = If the length is doubled, old I I 2 1 = ....
View Full Document

Page1 / 3

Worksheet #6 - Solutions - Part C. Part D. qvB F B = , r mv...

This preview shows document pages 1 - 3. Sign up to view the full document.

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