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Current Balance

# Current Balance - Current Balance Performed Due Theory A...

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Current Balance Performed: 3/22/07, Due 3/29/07 Theory : A current carrying wire produces a magnetic field around it. The field is represented by B. The differential magnetic field produced a distance r from a wire segment with length l , where l points in the same direction as the current, and the current is I, is given by the equation dB = (μ 0 /4π)((I d l X r)/r 2 ). In this equation r is the unit vector that points from the wire element to the field point and μ 0 is a constant called the vacuum permeability. The permeability of air can be taken to be μ 0 for almost all purposes. This constant is defined so that μ 0 /4π is exactly equal to 10 -7 . This equation can be used to determine the magnetic field produced by one segment of wire at all points in another wire. As we can see in the equation, the magnetic field will be proportional to I. When two current carrying wires are close to one another, they experience forces due to each other’s magnetic fields. If a wire element d l with current I is in a magnetic field, B, it will experience the force dF, given by dF = I d X B. This equation can be integrated to give the total force on the wire. Using this equation it can be shown that the magnitude of the force F between a length L of two thin infinitely long parallel wires separated by the distance D is given by F = (2μ 0 /4π)((I A I B L)/D), where I A is the current in one wire, and I B is the current in the other wire. We can expect that the error in using this equation gets smaller as the distance D between the wires becomes smaller and the length of our finite wires becomes longer. When the currents of two wires are both in the same directions the wires will experience forces that attract them to the other, and with current traveling in opposite direction in the two wires, they will experience repulsive forces away from the each other. We assumed that the force on the single wire is due entirely to the current in the long side of the rectangular coil that is nearest the single wire. The experiment is wired so that the current through the single wire also passes through each turn of the coil. The coil has N turns so the previous equation can be rewritten as F = (2μ 0 /4π)((NI 2 L)/D), where I is the common current. We break down the distance D into two parts such that D = d + b. The distance d will be known and have a value of a=1.0mm or 2a=2.0mm. the distance b will be obtained from the data and (a+b) or (2a+b) will be approximately equal to the distance between the single wire and the longer side of the coil nearest the single wire.

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