- #1
moenste
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Homework Statement
ABCD is a plane rectangular strip of conducting material of uniform thickness, with a steady current flowing uniformly from AD to BC. The potential difference between E and F, the mid-points respectively of AB and CD, is zero, but when a magnetic field is set up at right angles to ABCD, into the plane of the diagram, a small, steady potential difference appears between E and F. Explain these observations and briefly describe a practical application of the phenomenon.
When the rectangle is made of copper, of density 9 * 103 kg m-3, relative atomic mass 63, with thickness 1 mm, breadth 20 mm, carrying a current of 10 A and with an applied field of flux density 1.67 T, the potential difference between E and F is found to be 1 μV, F being positive with respect to E. Taking the electronic charge to be 1.6 * 10-19 C and the Avogadro constant to be 6 * 1023 mol-1, find:
(a) the drift velocity of the current carriers in the copper;
(b) the number of charge carriers per unit volume in the copper and the sign of their charge;
(c) the mobility (drift velocity per unit electric field) of these carriers, given the resistivity of copper to be 1.7 * 10-8 Ω m;
(d) the ratio of the number of atoms per unit volume to the number of charge carriers per unit volume in the copper.
Answers: (a) 3.0 * 10-5 m s-1, (b) 1.0(4) * 1029 m-3, (c) 3.5 * 10-3 m2 s-1 V-1, (d) 0.82.
2. The attempt at a solution
(a) I = n A v e → v = I / n A e where: I = 10 A, n = 9 * 103 kg m-3, A = (1 / 10 / 100) * (20 / 10 / 100) = 2 * 10-5 m2, e = 1.6 * 10-19 C.
v = 10 / (9 * 103) * (2 * 10-5) * (1.6 * 10-19) = 3.47 * 1020 m s-1.
Doesn't fit the answer. What am I missing?