Question

15. A hollow copper wire with an inner diameter of $1.0\mathrm{mm}$and an outer diameter of $2.0\mathrm{mm}$ carries a current of $10\mathrm{A}$. What is the current density in the wire?

Short answer

The current density in the wire is $4.2\times {10}^6\mathrm{A}/\mathrm{m}$.

Step-by-step solution

Given information

A hollow copper wire's inner diameter is $d=1.0\mathrm{mm}$and an outer diameter is $2.0\mathrm{mm}$carries a current of $10\mathrm{A}$.

Explanation

The area of the wire is calculated by the area of the outer circle with diameter$D=2.0\mathrm{mm}$minus the area of the inner circle $d=1.0\mathrm{mm}$. Hence,
$A=\frac{1}{4}\pi D^2-\frac{1}{4}\pi d^2$

$=\frac{1}{4}\pi \left(D^2-d^2\right)$

The current density $j$is given by the current $I=10\mathrm{A}$divided by area of circle localid="1648796043058" $A=\frac{1}{4}\pi \left(D^2-d^2\right)$

So,

$j=\frac{I}{A}=\frac{4I}{\pi \left(D^2-d^2\right)}$

localid="1648796295327" $=4.2\times {10}^6\mathrm{A}/{\mathrm{m}}^2$

Final Answer

The current density in the wire is $4.2\times {10}^6\mathrm{A}/\mathrm{m}$.