Question

The wires leading to and from a $0.12mm$diameter lightbulb filament are $1.5\mathrm{mm}$ in diameter. The wire to the filament carries a current with a current density of role="math" localid="1648887651623" $4.5\times {10}^5\mathrm{A}/{\mathrm{m}}^2$. What are (a) the current and (b) the current density in the filament?

Short answer

a. The current is $0.80A$

b. The current density in the filament is$7.0·{10}^7\frac{\mathrm{A}}{{\mathrm{m}}^2}.$

Step-by-step solution

Given Information (Part a) 

Wire diameter$=0.12mm$

Diameter lightbulb filament $=1.5mm$

Current density$=4.5\times {10}^5\mathrm{A}/{\mathrm{m}}^2$

Explanation (Part a) 

This expression for current density is,

$J\equiv \frac{I}{A}$

Solve for A.

The radius is$r=0.00075.$

$J=\frac{I}{A}=\frac{I}{r^2\pi }$

Find the current,

$I=JA=J{r}^2\pi$

Substitute the values,

localid="1648888370524" $=4.5\times {10}^{5}A/m^2\times {\left(0.00075m\right)}^2\times 3.14$

$=0.80A$

Final Answer

Hence, the current is$0.80A$.

Given Information (Part b)  

Wire diameter$=0.12\mathrm{mm}$

Diameter lightbulb filament$=1.5\mathrm{mm}$

Current density$=4.5\times {10}^5\mathrm{A}/{\mathrm{m}}^2$

Explanation

The same amount of current goes into the filament as into the wire, which is due to the conservation of current. There can, however, be differences in density between wires. Therefore, we keep using (31.13), but our info in the equation requires us to move to a new area.

To convert $0.12mm$into a radius of $0.00006m$, the reader must multiply $0.12mm$by 0.00006m.

$J_{\text{filament}}=\frac{I}{A_{\text{filament}}}=\frac{I}{r_{\text{filament}}^2\pi }$

Then we will get,

$J_{\text{filament}}=7\cdot {10}^7\mathrm{A}/{\mathrm{m}}^2\text{.}$

Final Answer

Hence, the current density is$7.0\cdot {10}^7\frac{\mathrm{A}}{{\mathrm{m}}^2}.$