(a) What is the magnitude of the(perpendicular) magnetic field in the separator?
ID: 1671038 • Letter: #
Question
(a) What is the magnitude of the(perpendicular) magnetic field in the separator?(b) If the machine is used to separate out 100 mg of material perhour, calculate the current of the desired ions in the machine.
(c) Considering the same separation rate, calculate the thermalenergy produced in the cup in 1.00 h
Explanation / Answer
(a) According to conservation of energy we have 0.5 mv2 = qV ==> v = 2qV/m = 2 * 3.2 x 10-19 * 95 * 1000 / 3.92 x10-25 ] = 39.38 x 104 m/s Therefore B =mv / qr = 3.92 x 10-25 * 39.38 x 104 / 3.2 x10-19 * 1.00 = 0.482 T (b) The number of ions per sec in 100mg material is 1.00 x 10-6 = n (ion/s) 3600 s* 3.92 x10-25kg /ions Therefore n = 7.09 x1016 ions/s then i = n q = 7.09 x 1016 * 3.2 x 10-19 = 0.0277A c) energy U = q V = 3.2 x 10-19 * 95 * 1000 = 3.04 x 10-14 J /ion = 7.09 x 1016 ion /s * 3600s /hr * 3.04 x10-14 J /ion = 7759296J
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