quiz91sprng99 - Bd = 5.6 x10-6 T coming out of the paper...

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01/15/12 P. Kumar PHY 2054 QUIZ IX/3455 Name:______________________________________________________ 1. The drawing shows two perpendicular wires which lie in the plane of the paper. They also define the Cartesian axes from which the distances are measured. One wire is along the x-axis and the other along the y-axis. The point of their crossing is the origin (0,0). Each wire carries a current of 5.6A. Determine the magnitude and direction of the net magnetic field at points (0.4m, 0.2m) and (-0.4m, -0.2m). Let's consider point A = (0.4, 0.2). Suppose the vertical wire is C and the Horizontal wire is D. Then the magnetic field at A due to C is Bc= μ I 2 π R = 2x10 -7 x5.6/(0.4) = 2.8 x10 -6 T . The magnetic field due to wire D is
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Unformatted text preview: Bd = 5.6 x10-6 T, coming out of the paper. The net field at A is therefore the difference between the two, B = 2.8 x10-6 T, coming out of the paper . At the other point B = (-0.4m, -0.2m), the field due to wire C is coming out but smaller than the field due to wire D, which is pointed in to the paper (the numbers are otherwise the same). The net field at B is therefore B = 2.8 x10-6 T , going in to the paper. 2. A straight wire in a magnetic field experiences a force of 0.03N when the current in the wire is 2.7A. What is the current in the wire when it experiences a force of .047N? since F = IBl, F 1 /F 2 = I 1 /I 2 . Thus I 2 = I 1 × F 2 / F 1 = 2.7 .047/.03 = 4.23 A C D...
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