91Ó°ÊÓ

The data can be listed as follows,

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

The required electric field can be determined using the formula,

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

Here A is the cross-sectional area of the wire and $\mathit{σ}$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,

$A=\mathrm{π}×r^2$

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

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

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

$E_W=\frac{I}{{\mathrm{σ}}_{W}A}$

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

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

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

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

$E_C=\frac{I}{{\mathrm{σ}}_{C}A}$

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

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

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

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.