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Combination of fields from a bundle of wires.

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A thin electrically insulated wire is unwound from a reel and cut into {nwires} sections by a student, each {length} cm long. A single wire is analysed and found to have a diameter of {diametermm} mm and an electrical resistance of {resistance} Ω.  The {nwires} pieces of wire are all straightened, tighly bundled together and laid on the benchtop in the lab from left to right in front of the student. The left-hand ends of each of the {nwires} wires are electrically connected together, and this end of the bundle is connected to the positive electrode of a power supply. The right-hand ends of the wires are also connected together electrically and this end of the bundle is attached to the negative electrode of the power supply.

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When the power supply is switched on, an electrical current flows through the bundle of wires.

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When providing numerical answers you may express them using scientific notation.  Express values to four significant figures and use the values of physical constants as provided in the course notes.  You may need to consider the concepts of Ohm's Law and how resistors combine.

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This is a question that draws upon the ideas of the field from a long straight wire, and from the notions of current and voltage.  The bundle of wires can be thought of as a single current, but we have to be careful about determining how much current we have passing through the wires.

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First, we note that if there is a current $j_1$ in each wire, the total current is $Nj_1$, where $N$ is the number of wires.  The fields are then 

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$\\displaystyle \\left|\\vec{H}\\right|={j\\over 2\\pi R}={\\left|\\vec{B}\\right|\\over\\mu}$

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In the first case we have a constant potential difference across the bundle.  Each wire has the same potential difference (voltage) so that by Ohms-law we can determine the current in one wire to be 

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$j_1=V/R_1$

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where $R_1$ is the resistance of a single wire.  We could determine the total resistance of the bundle to be $R_1/N$, since they are in parallel.  Alternatively we could note that the same current, $j_1$, is flowing through each wire, so the total current is the sum of the currents over all the wires.  Either way 

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$j=NV/R_1$

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For the second case we have a constant current source attached to the wires.  This specifies the current through all the wires together, so that in this case the input current is split between the {nwires} wires, i.e. 

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$j_1=j/N$

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We're not really interested in $j_1$, so we simply use $j$ for the determination of the field. 

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The assumptions we need to make is that the bundle of wires is tightly packed so that the distance we're determining the field further from the bundle that the overall cross-sectional area of the bundle.  

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Number of wires in the bundle

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Permeability of free space in Tm/A

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Diameter of each wire in mm.

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Distance from the (centre of) the bundle to the point at which the field is being determined.

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Resistance of each wire in Ohms.

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Length of wires - this is not relevant to the calculations.

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Write down an expression for the magnetising field strength as a function of distance from the bundle. (You will have to make some assumptions.)  Express your answer in terms of the current in each wire, $j$, the number of wires, $N$ and the distance of the bundle to the point at which the field is being determined, $R$.

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$\\displaystyle\\left|\\vec{H}\\right|=$[[0]]

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The power source is initially set so that the potential difference across the bundle of wires is held at a constant value of {voltage} V. What is the magnitude of the magnetizing field strength at a distance of {distancemm} mm from the wire bundle?

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$\\displaystyle\\left|\\vec{H}\\right|=$[[0]] A/m

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The power source is next set to a current source of {current} A. What is the magnitude of the magnetic flux density also at a distance of {distancemm} mm from the wire bundle in this different case where the current through the bundle of wires is held constant?

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$\\displaystyle\\left|\\vec{B}\\right|=$[[0]] μT

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