91Ó°ÊÓ

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

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

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 = $36×{10}^6$S/m

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

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

Bottom will be same as top as dimensions are same.