A pia c i r q r i r aa a a pia c i r sin sin x

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Unformatted text preview: ¢ a PQ i  C  x C  C  y ” ” €lugging this into equ—tion @IA yieldsX r r< P ¢ i Y l @; 'A C Y @; 'A¡ x C Y l @; 'A C Y @; 'A¡ y£ I ~r ” ” A@~A a  !a P@IA C I r> Q r< I ~r A@~A a  !a ” ” P@IA C I r> ‘ sin  sin 'x C sin  ™os 'y“ I r< sin  ‘ sin 'x C ™os 'y“ ~r A@~A a  !a ”{z ”} | Qr 1 2 + 2 2 3 2 =0 0 0 cos =1 1 1 1 1 0 1 1 1 1 (1)+1 1 1 3 (1)+1 3 2 > 1 1 1 1 1 1 3 1 1 1 1 @ A a  !a I r< sin ' ” Q r> ~r A~ 1 1 1 1 0 1 1 2 3 ' ^ 3 2 I 1 1 1 1 1 =a snside the sphereX r I ” @ A a Q  !a a sin ' ” a I  !ar sin ' Q ~r Ain ~ 3 0 2 0 yutside the sphereX I @ A a Q  !a ra sin ' ” I ” a Q  ! a sin ' r ~ Aout ~ r 3 0 2 4 0 ~ B ~ ar¢A I@ @ ” a r sin  @ @sin A'A” I @r @rA'A r r   ~ Bin 2   I@ ” ” r@ Q a r sin  @ I  !ar sin  r I @r I  !ar sin   Q I ” a r sin  I  !arP sin  ™os r I I  !aPr sin  ” rQ Q  P ” ” a Q  !a ™os r sin  2 0 2 0 0 0 0  ~ Bout    I@ ” a r sin  @ I  ! a sin  r I @r I  ! a sin   ” r@ Q Q r r   I I  ! a P sin  ™os r I I  ! a sin  ” a r sin  Q ” rQ r r I  ! a P ™os r C sin  ” aQ ” r 4 0 2 0 2 4 0 4 0 2 4 0 4 3 P 2 2 Problem 5.15 2.1 ~ pirstD we use empre9s v—w to determine B for — single wire ™—rrying ™urrent I z X e ” s ¡ a I ~ B d~ l 0 end P a I I ~ aA B a Pr ' ” B r 0 0 ~ ~ ~ where the dire™tion of B ™omes from —pplying the rightEh—nd ruleF ƒin™e H a B= X 0 ~ H I ” a Pr...
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This note was uploaded on 12/08/2011 for the course PHYSICS 505 taught by Professor Liu during the Winter '08 term at University of Michigan.

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