# hw11 - Physics 21 Fall 2004 Solution to HW-11 27 P 9 A...

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Physics 21 Fall, 2004 Solution to HW-11 27 P 9 A straight 2.0 mm diameter copper wire can just “ﬂoat” horizontally in air because of the force of the earth’s magnetic field B , which is horizontal, perpendicular to the wire, and of magnitude 5 . 0 × 10 - 5 T. What current does the wire carry? x y z B mg F I The diagram shows a section of wire (of some length L ) carrying current I in the ˆ z direction. The force F on that piece of wire that balances gravity is given by F = IL ˆ k × B = ILB ˆ j We need the mass of the section of wire to get the gravita- tional force acting on it. If the radius of the wire is r , the mass m is m = πr 2 Cu , where ρ Cu is the mass density of copper (8 . 9 × 10 3 kg/m 3 , from Ex. 25-12 in the text or other tables). Balancing the gravitational and magnetic forces, we have ILB = mg = πr 2 Cu g. Solving for I , we see that L drops out and I = ρ Cu πr 2 g B Using r = 1 . 0 × 10 - 3 and the other numbers given, the numerical result is I = (8 . 9 × 10 3 ) π (0 . 001) 2 (9 . 8) 5 × 10 - 5 = 5 . 5 × 10 3 A 27 P 20
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