The conductivity of tungsten at room temperature,$\mathbf{1}\mathbf{.}\mathbf{8}\times {\mathbf{10}}^{\mathbf{7}}\left(\mathbf{A}\mathbf{/}{\mathbf{m}}^{\mathbf{2}}\right)\mathbf{/}\left(\mathbf{V}\mathbf{/}\mathbf{m}\right)$ , is significantly smaller than that of copper. At the very high temperature of a glowing light-bulb filament (nearly 3000 kelvins), the conductivity of tungsten is 18 times smaller than it is at room temperature. The tungsten filament of a thick-filament bulb has a radius of about 0.015 mm. Calculate the electric field required to drive 0.20 A of current through the glowing bulb and show that it is very large compared to the field in the connecting copper wires.

The data can be listed as follows,

• Current is, $I=\text{0.20}\text{A}$.
• The radius of the filament is, $r=0.015\text{mm}=0.015\times {10}^{-3}\text{m}$.
• The conductivity of the tungsten filament at room temp is,$5.95\times {10}^7\left({\text{A/m}}^2\right)\text{/}\left(\text{V/m}\right)$ .

The required electric field can be determined using the formula,

$\mathit{E}\mathbf{=}\frac{\mathbf{I}}{\mathbf{\sigma }\mathbf{A}}$

Here A is the cross-sectional area of the wire and $\mathit{\sigma }$is the conductivity of the wire.

The amount of flowing current and conductivity is not constant. As the temperature of the wire changes, these quantities will also change.

The cross-sectional area can be calculated as,

$A=\pi \times r^2$

Substitute the value in the above expression, and we get,

$\begin{array}{rcl}A& =& \pi \times {\left(0.015\times {10}^{-3}\text{m}\right)}^2\\ & =& 7.065\times {10}^{-10}{\text{m}}^2\\ & & \end{array}$

The required electric field for tungsten filament can be calculated as,

$E_W=\frac{I}{{\sigma }_{W}A}$

Here ${\sigma }_W$is the conductivity of the tungsten wire, whose value at 3000 kelvins is, 18 times smaller than it is at room temperature that means $5.95\times {10}^7\left({\text{A/m}}^2\right)\text{/}\left(\text{V/m}\right)/18=3.30\times {10}^6\left({\text{A/m}}^2\right)\text{/}\left(\text{V/m}\right)$.

Substitute the value in the above expression, and we get,

$\begin{array}{rcl}{E}_W& =& \frac{\left(\text{0.20}\text{A}\right)}{\left(3.30\times {10}^6\left({\text{A/m}}^2\right)\text{/}\left(\text{V/m}\right)\right)\left(7.065\times {10}^{-10}{\text{m}}^2\right)}\\ & =& 85.78\text{V/m}\\ & & \end{array}$

The required electric field for copper wire can be calculated as,

$E_C=\frac{I}{{\sigma }_{C}A}$

Here ${\sigma }_C$is the conductivity of the copper wire whose value is$\text{1.8}\times {\text{10}}^{\text{7}}\left({\text{A/m}}^2\right)\text{/}\left(\text{V/m}\right)$ .

Substitute the value in the above expression, and we get,

$\begin{array}{rcl}{E}_C& =& \frac{\left(\text{0.20}\text{A}\right)}{\left(\text{1.8}\times {\text{10}}^{\text{7}}\left({\text{A/m}}^2\right)\text{/}\left(\text{V/m}\right)\right)\left(7.065\times {10}^{-10}{\text{m}}^2\right)}\\ & =& 15.726\text{V/m}\\ & & \end{array}$

The required electric field for copper wire is $15.726\text{V/m}$. The required electric field for tungsten filament is $85.78\text{V/m}$.

The ratio can be calculated as,

$$\begin{gathered} =\frac{85.78\text{V/m}}{15.726\text{V/m}} \\ =5.45 \end{gathered}$$

Thus, The required electric field for tungsten filament is$85.78\text{V/m}$ , and it is very large compared to the field in the connecting copper wires by a factor 5.45.