35 volume atom mole m problem parallel g k and

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Unformatted text preview: gt; 0). so they agree. (d), where (a)is = charge = charge. atoms. moles. grams = (e)(N) A P 5.35 volume atom mole M Problem parallel G K, and depends only on volume 9.6 (15 points) to riffiths Problem 5.gram z, so A = A(z) x. 39 - = -~~ I J ~ I I x y z aA I (~ ) K = 1.6 X 10-19 C, e For a ring, m = I1I"r2. Here I=--+charge=of electron m = foR 1I"r2aUJr = 11I"aUJR4 av aUJrdr, so dr /4.1 B=VxA=1 a/ax a/ay =a/az dr =8y=xfL~ y. 6.0 X io23 mole, = N0 Avogadro's number Problem 5.36 A(z) 0 z M ring is dq mass of scopper '7 The total charge on the shaded = atomic= a {211"Rin O)R dO. 64gm/mole, 9.0gm/cm3. The time for one revolutiond is = = 211" of SO the current dt density /UJ. copper JA -¥Izi = x I y \" ~ill do the job-or this plus any constant. x in the ring is I = ~~ = aUJR2 sin 0 dO. The area of the ring is = (1.6 x 5.27 so the xmagnetic moment of X 104ring is dm Problem 0)2, 11.4 the C/cm3.1 p 1I"(Rsin 10-19)(6.0 1023)(~.~) / z = = RsinO {aUJR2...
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This note was uploaded on 04/23/2013 for the course PHSX 531 taught by Professor Chiu during the Fall '12 term at Kansas.

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