Question

When 115 Vis applied across a wire that is 10 mlong and has a 0.33mmradius, the magnitude of the current density is$\mathbf{1}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}\mathbf{A}\mathbf{/}{\mathbf{m}}^{\mathbf{2}}$. Find the resistivity of the wire.

Short answer

The resistivity of the wire is$8.2\times {10}^{-8}\mathrm{\Omega m}$ .

Step-by-step solution

Identification of given data

1. Voltage across the wire, V = 115V
2. Length of the wire, L = 10m
3. Radius of the wire, r = 0.30mm
4. Magnitude of the current density, J =$1.4\times {10}^8\mathrm{A}/{\mathrm{m}}^2$

Significance of the resistivity

The resistivity is the opposition to the flow of current. It is the characteristic of the material for the given temperature. We have to use the formula for resistivity in terms of voltage, current density, and length of the wire to find the resistivity of the wire

Formulae:

The current density in terms of resistivity, $\mathrm{J}=\frac{\mathrm{E}}{\mathrm{p}}$ …(i)

Here J is the current density, E is the electric field, and p is resistivity.

The voltage of the wire in terms of the electric field, V = EL …(ii)

Here, V is the voltage, E is the electric field, L is the length of the wire.

Calculation of the resistivity of the wire

Substituting the given data and electric field from equation (ii) in equation (i) and rearranging the equation to get the resistivity, we can get the magnitude of the resistivity of the wire as follows:

$$\begin{gathered} \mathrm{p}=\frac{\mathrm{V}}{\mathrm{LJ}} \\ =\frac{115\mathrm{V}}{10\mathrm{m}\times 1.4\times {10}^8\mathrm{A}/{\mathrm{m}}^2} \\ =8.2\times {10}^{-8}\mathrm{\Omega }.\mathrm{m} \end{gathered}$$

Hence, the value of the resistivity of the wire is$8.2\times {10}^{-8}\mathrm{\Omega }.\mathrm{m}$ .