P29_025 - 8 kg / s) 3 . 92 10 25 kg = 2 . 27 10 2 A . (c)...

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25. (a) We solve for B from m = B 2 qx 2 / 8 V (see Sample Problem 29-3): B = s 8 Vm qx 2 . We evaluate this expression using x =2 . 00 m: B = s 8(100 × 10 3 V)(3 . 92 × 10 25 kg) (3 . 20 × 10 19 C)(2 . 00 m) 2 =0 . 495 T . (b) Let N be the number of ions that are separated by the machine per unit time. The current is i = qN and the mass that is separated per unit time is M = mN ,where m is the mass of a single ion. M has the value M = 100 × 10 6 kg 3600 s =2 . 78 × 10 8 kg / s . Since N = M/m we have i = qM m = (3 . 20 × 10 19 C)(2 . 78 × 10
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Unformatted text preview: 8 kg / s) 3 . 92 10 25 kg = 2 . 27 10 2 A . (c) Each ion deposits energy qV in the cup, so the energy deposited in time t is given by E = NqV t = iqV q t = iV t . For t = 1 . 0h, E = (2 . 27 10 2 A)(100 10 3 V)(3600 s) = 8 . 17 10 6 J . To obtain the second expression, i/q is substituted for N ....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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