Question

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.

Short answer

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 of 5.45.

Step-by-step solution

Given data

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,$5.95\times {10}^7\left({\text{A/m}}^2\right)\text{/}\left(\text{V/m}\right)$ .

Concept

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.

 Calculation of the required electric field

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,

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

Comparision between two fields

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.