A wire 4.00mlong and 6.00mmin diameter has a resistance of$\mathbf{15}\mathbf{.}\mathbf{0}\mathbf{m\Omega }$. A potential difference of 23.0 Vis applied between the ends. (a) What is the current in the wire? (b) What is the magnitude of the current density? (c) Calculate the resistivity of the wire material. (d) Using Table, identify the material.

1. Length of the wire is L = 4.0 m
2. Diameter of the wire is$d=6.0\mathrm{mm}\mathrm{or}6.0\times {10}^{-3}\mathrm{m}$
3. Resistance of the wire,$R=15.0\mathrm{m\Omega }\mathrm{or}15\times {10}^{-3}\mathrm{\Omega }$
4. Potential difference, V = 23.0 V

The term "current density" refers to the quantity of electric current moving across a certain cross-section. We can find the current in the wire by substituting the values of voltage and resistance in Ohm’s law. We can find the magnitude of current density by using the values of area and current through the wire. The resistivity of the wire can be found from resistance, area, and length of wire using the formula for resistance. Based on the value of resistivity, we can identify the material.

Formulae:

The voltage equation using Ohm’s law, V = lR …(i)

The current density of a current flowing through an area,$J=\frac{l}{A}$ …(ii)

The resistance of a material related to its resistivity,$R=\frac{pL}{A}$ …(iii)

Where,

p is the resistivity

l is current in the wire

A is the area of wire

Substituting the given values in equation (i), we can get the value of the current in the wire as follows:

$$\begin{gathered} l=\frac{23V}{15\times {10}^{-3}\Omega } \\ =1.53\times {10}^{3}A \end{gathered}$$

Hence, the value of the current is $1.53\times {10}^{3}A$.

Substituting the given values in equation (ii), we can get the magnitude of the current density through the wire as follows:

$$\begin{gathered} J=\frac{l}{{\mathrm{\pi r}}^2}(\because areaofwire={\mathrm{\pi r}}^2) \\ =\frac{\mathrm{l}}{{\displaystyle \frac{{\mathrm{\pi d}}^2}{4}}}(\because \mathrm{raduis}=\mathrm{diameter}/2) \\ =\frac{1.53\times {10}^3\mathrm{A}}{{\displaystyle \frac{3.142{(6.0\times {10}^{-3}\mathrm{m})}^2}{4}}} \\ =5.41\times {10}^7\mathrm{A}/{\mathrm{m}}^2 \end{gathered}$$

Thus, the magnitude of the current density is $5.41\times {10}^7\mathrm{A}/{\mathrm{m}}^2$.

We can get the resistivity of the material using the given data and the area formula from part (b) calculations in equation (iii) as follows:

$$\begin{gathered} \mathrm{p}=\frac{\mathrm{R}\left({\displaystyle \frac{{\mathrm{\pi d}}^2}{4}}\right)}{\mathrm{L}} \\ =\frac{(15\times {10}^{-3}\mathrm{\Omega })\left({\displaystyle \frac{3.142{(6.0\times {10}^{-3}\mathrm{m})}^2}{4}}\right)}{4\mathrm{m}} \\ =10.6\times {10}^{-8}\mathrm{\Omega m} \end{gathered}$$

Thus, resistivity of the wire material is$10.6\times {10}^{-8}\mathrm{\Omega m}$

As, $\mathrm{p}=10.6\times {10}^{-8}\mathrm{\Omega m}$, from table 26 - 1, we can conclude that the material is Platinum.