An aluminum wire consists of the three segments shown in FIGURE P27.64. The current in the top segment is 10 A. For each of these three segments, find the a. Current I. b. Current density J. c. Electric field E. d. Drift velocity vd. e. Electron current i. Place your results in a table for easy viewing.

Current in top section of wire: 10A

Radius of top section: 2/2mm = 0.001m

Diameter of mid section: 1/2mm = 0.0005m

Diameter of bottom section: 2/2mm = 0.001m

Current will remain same throughout the wire, i.e. 10A

To find current density, we need to find cross sectional area of all three regions.

For top: $$\begin{gathered} A=\mathrm{\pi }\times 0.{001}^2=0.00000314{\mathrm{m}}^2 \\ \mathrm{J}=\frac{\mathrm{I}}{\mathrm{A}}=\frac{10}{0.00000314}=3184713.37\mathrm{A}{\mathrm{m}}^{-2} \end{gathered}$$

For mid: $$\begin{gathered} A=\mathrm{\pi }\times 0.{0005}^2=0.000000785{\mathrm{m}}^2 \\ \mathrm{J}=\frac{\mathrm{I}}{\mathrm{A}}=\frac{10}{0.000000785}=12738853{\mathrm{Am}}^{-2} \end{gathered}$$

Bottom will be same as top as its dimensions are same.

We know, relation between current density and electric field can be expressed as

E = $\frac{CurrentDensity}{Conductivity}$

Conductivity of Aluminum = $36\times {10}^6$S/m

For top: E = $\frac{3184713.37}{36\times {10}^6}=0.088V/m$

For mid : E = $\frac{12738853}{36\times {10}^6}=0.353V/m$

Bottom will be same as top as dimensions are same.