A composite material is made of alternating thin lamellae of copper (resistivity

Question

A composite material is made of alternating thin lamellae of copper (resistivity 1.69 times 10-6 ohm-cm) and a niobium-titanium alloy (resistivity 7 times 10-5 ohm-cm) of equal thickness. (a) What is the resistivity of this composite measured parallel to the lamellae? If current is passed through the composite in this direction, what fraction of the current will be carried by the copper? How will the electric fields in the two phases compare? (b) What is the resistivity of this composite measured perpendicular to the lamellae? If current is passed through in this direction, what will be the ratio of the electric field in the alloy to that in the copper? (c) At a temperature of 4.2 K, the resistivity of the copper has decreased to 1 times 10-8 ohm-cm, and the niobium-titanium alloy has become superconducting. Answer questions (a) and (b) for this situation.

Explanation / Answer

(R propto ho) i.e. Resistivity linearly depends on Resistance. i.e. Whatever is the formula for R would be for ( ho) (a) The configuration is a parallel combination, ( ho = rac{ ho_{copper} cdot ho_{NbTi}}{ ho_{copper} + ho_{NbTi}}) ( ho = 1.65 mu-Omega-cm) fraction of current passing through copper = (rac{ ho}{ ho_{copper}}) = 0.9764 or 97.64 % From ohm's law (E propto ho cdot I) (E_{copper}/E_{NbTi} = ho{copper} cdot I_{copper}/ ho_{NbTi} cdot I_{NbTi} = 1) or (E_{copper}/E_{NbTi} = 1 or E_{copper} = E_{NbTi}) (b) The configuration is a series combination, ( ho = ho_{copper} + ho_{NbTi} = 71.69 mu-Omega-cm) (E propto ho) as current in this case is same. (E_{copper}/E_{NbTi} = ho_{copper}/ ho_{NbTi} = 41.42 ) (E_{NbTi}/E_{copper} = 0.02414) (c) at 4.2 K from part (a), ( ho = rac{ ho_{copper} cdot ho_{NbTi}}{ ho_{copper} + ho_{NbTi}}) ( ho_{NbTi} = 0) ==> ( ho = 0) but as the potential hasn't changed ==> (E_{copper} = E_{NbTi}) from part (b), ( ho = ho_{copper} + ho_{NbTi}) ( ho_{NbTi} = 0) ==> ( ho = ho_{copper} = 1.69 mu-Omega-cm) and (E_{NbTi} / E_{copper} = ho_{NbTi} / ho_{copper} = 0) (E_{NbTi} = 0)

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