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Question Number 99744 by Rio Michael last updated on 23/Jun/20

$$\mathrm{A}\mathrm{wire}\mathrm{that}\mathrm{is}\mathrm{highly}\mathrm{insulated}\mathrm{has}\mathrm{a}\mathrm{radius}\mathrm{of}2.1\mathrm{mm}\mathrm{and}\mathrm{a}\mathrm{current}$$$$\mathrm{of}6\mathrm{Agoes}\mathrm{through}\mathrm{it}.\mathrm{The}\mathrm{material}\mathrm{used}\mathrm{in}\mathrm{insulation}\mathrm{has}\mathrm{thickness}$$$$\mathrm{of}2.1\mathrm{mm}\mathrm{with}\mathrm{a}\mathrm{thermal}\mathrm{conductivity}\mathrm{of}0.2\mathrm{W}/\mathrm{Km}.\mathrm{the}\mathrm{material}\mathrm{used}$$$$\mathrm{in}\mathrm{constructing}\mathrm{the}\mathrm{wire}\mathrm{has}\mathrm{a}\mathrm{resistivity}\mathrm{of}4.2\times {10}^{-7}\mathrm{\Omega }\mathrm{m}.\mathrm{assume}\mathrm{the}$$$$\mathrm{the}\mathrm{materials}\mathrm{reach}\mathrm{steady}\mathrm{state}.\mathrm{Find}\mathrm{the}\mathrm{difference}\mathrm{in}\mathrm{temperature}$$$$\mathrm{between}\mathrm{the}\mathrm{outer}\mathrm{suface}\mathrm{and}\mathrm{inner}\mathrm{surface}.$$

Answered by ajfour last updated on 23/Jun/20

$$P=I^{2}R=\frac{I^2\rho l}{\pi a^2}$$$$R_T=\int \frac{dr}{k\left(2\pi rl\right)}=\frac{1}{2k\pi l}\ln \frac{b}{a}$$$$\mathrm{\Delta }T={PR}_T=\frac{I^2\rho l}{\pi a^2}\left(\frac{1}{2k\pi l}\right)\ln \frac{b}{a}$$$$=\frac{\left(36A^2\right)(4.2\times {10}^{-7}\mathrm{\Omega }m)}{2{\pi }^2{(2.1\times {10}^{-3}m)}^2(0.2W/Km)}\ln 2$$$$\mathrm{\Delta }T=\frac{36\times 4.2\left(\ln 2\right)}{2{\pi }^2\times 4.41\times 2}K$$$$=\frac{18\left(\ln 2\right)}{2.1{\pi }^2}{C}^{°}\approx 0.602C°.$$

Commented by Rio Michael last updated on 23/Jun/20

$$\mathrm{thanks}\mathrm{sir},\mathrm{please}\mathrm{you}\mathrm{got}\mathrm{any}\mathrm{other}$$$$\mathrm{procedure}\mathrm{than}\mathrm{this}??$$

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