Question

One end of an aluminium wire whose diameter is 2.5 mm is welded to one end of a copper diameter is 1.8 mm. The composite wire carries a steady current i of I .3 A. What is the current density in each wire ?

• A. $j_{Cu}=26A/c{m}^2,j_{Al}=51A/c{m}^2$
  
• B. $j_{Cu}=51A/c{m}^2,j_{Al}=26A/c{m}^2$
• C. $j_{Cu}=40A/c{m}^2,j_{Al}=60A/c{m}^2$
• D. $j_{Cu}=60A/c{m}^2,j_{Al}=40A/c{m}^2$

Answer 1

Answer: (B)

$j_{Cu}=51A/c{m}^2,j_{Al}=26A/c{m}^2$

$\begin{array}{c}\text{Given,}\\ \text{diameter of Aluminium wire}=2.5mm=2.5\times {10}^{-3}m\\ \text{diameter of copper wire}=1.8mm=18\times {10}^{-3}m\\ \text{Steady curvent}=1.3A\end{array}$

$\begin{array}{c}J\text{(current density)}=\frac{I}{\text{Area of cross section}}\\ J_{aluminium}=\frac{4i}{\pi d^2}\\ J_{A1}=0.26\times {10}^{6}A/m^2\end{array}$

$\begin{array}{cc}{J}_{\text{copper}}& =\frac{4i}{\pi d^2}\\ & =\frac{4\times 1.3}{3.14\times {(1.8\times {10}^{-3})}^2}\\ J_{cu}& =0.51\times {10}^{+6}A/m^2\end{array}$