Answer to Question #138588 in Electric Circuits for Roy Bruno

Question

A copper wire AB is fused at one end B to an iron wire BC. The diameter of the copper wire is 0.50 mm and that of the iron wire is 2.50 mm. The compound wire is stretched so that its length increases by a total amount of 0.1 cm. (AB=9.00 x 10^2mm and BC = 4.50 x 10^2mm) Calculate; (a) The ratio of the extension of the copper wire to that of the iron wire (b) The tension applied to the compound wire (NB: Young modulus of copper = 1.3 x 10^11 Pa while that of iron = 2.0 x 10^11 Pa)

Answer 1

A) Since the young's modulus for a copper wire of equal $E_{cu}=\frac{\frac{F}{S_{cu}}}{\frac{\Delta l_{cu}}{l_{cu}}}$ , where $F$ is the force applied to the body, $S_{cu}$ is a cross – sectional area of copper wire, $\Delta l_{cu}$ elongation of copper wire, $l_{cu}$ length of copper wire, then

$\frac{E_{fr}}{E_{cu}}$$= \frac{S_{fr}\times l_{fr}}{\Delta l_{fr} \times S_{fr}}\times \frac{\Delta l_{cu} \times S_{cu} }{S_{cu} \times l_{cu}}$ ;

$\frac{\Delta l_{cu}}{\Delta l_{fr}}=\frac{E_{fr}\times S_{fr} \times l_{cu}}{ E_{cu}\times S_{cu} \times l_{fr} }=$ $\frac{2\times 10^{11} \times \pi \times (\frac{2.5}{2})^2\times 9 \times 10^2 \times 10^{-3}}{1.3\times 10^{11} \times \pi \times (\frac{0.5}{2})^2\times 4.5 \times 10^2 \times 10^{-3}} \approx76. 92$ .

B) The total young's modulus is $E_{total}=E_{cu}+E_{fr}=\frac{\sigma}{\varepsilon}$ , where $\sigma$ is the total stress, $\varepsilon=\frac {\Delta l}{l_{cu}+l_{fr}}$ is the total elongation, then

$\sigma=\frac{\Delta l \times (E_{cu}+E_{fr})}{l_{cu}+l_{fr}}$ $=\frac{0.1\times 10^{-2} \times 10^{11}\times(1.3+2)}{(9+4.5)\times 10^2\times 10^{-3}}\approx2.44\times 10^8$