91Ó°ÊÓ

Aluminum wire thickness$=2.0\mathrm{mm}×2.0\mathrm{mm}$

Current$=2.5A$

In electrical circuits, the current density is equal to the current divided by the cross-section area of the wire.

This wire has a square cross-section of area,

$A=2.0\mathrm{mm}·2.0\mathrm{mm}=4.0·{10}^{-6}{\mathrm{m}}^2$

The expression will be,

$j=\frac{I}{A}$

role="math" localid="1648898970401" $=\left(\frac{2.5\mathrm{A}}{4.0×{10}^{−6}{\mathrm{m}}^2}\right)$

role="math" localid="1648898954139" $=6.25×{10}^5\mathrm{A}/{\mathrm{m}}^2$

Hence, the current density is$6.25×{10}^5\mathrm{A}/{\mathrm{m}}^2.$

Aluminum wire thickness

Current $=2.5\mathrm{A}$

The expression for the current density is related to the electron drift speed $v_{\mathrm{D}}$is,

$j=e{n}_e{v}_{\mathrm{D}},$

Where $n_e$is free electron density in aluminum.

From Table 27.1 the value for aluminum is $n_e=6.0·{10}^{28}{\mathrm{m}}^{-3}$

The expression will be,
$v_{\mathrm{D}}=\frac{j}{e{n}_e}$

Substitute the values,

$=\left(\frac{6.25×{10}^{25}A/m^2}{1.6×{10}^{-19}C×6.0×{10}^{28}{m}^{-3}}\right)$

localid="1648899774223" $v_D=6.5×{10}^{−5}\mathrm{m}/\mathrm{s}$

Hence, the electron drift speed is$6.5×{10}^{−5}\mathrm{m}/\mathrm{s}.$