91Ó°ÊÓ

Wire diameter$=0.12mm$

Diameter lightbulb filament $=1.5mm$

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

This expression for current density is,

$J≡\frac{I}{A}$

Solve for A.

The radius is$r=0.00075.$

$J=\frac{I}{A}=\frac{I}{r^2\mathrm{Ï€}}$

Find the current,

$I=JA=J{r}^2\mathrm{Ï€}$

Substitute the values,

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

$=0.80A$

Hence, the current is$0.80A$.

Wire diameter$=0.12\mathrm{mm}$

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

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

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.

$J_{\text{filament}}=\frac{I}{A_{\text{filament}}}=\frac{I}{r_{\text{filament}}^2\mathrm{Ï€}}$

Then we will get,

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

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