Question

Question: The following questions refer to the circuit shown in Figure 18.114, consisting of two flashlight batteries and two Nichrome wires of different lengths and different thicknesses as shown (corresponding roughly to your own thick and thin Nichrome wires).

The thin wire is 50 cm long, and its diameter is 0.25 mm. The thick wire is 15 cm long, and its diameter is 0.35 mm. (a) The emf of each flashlight battery is 1.5 V. Determine the steady-state electric field inside each Nichrome wire. Remember that in the steady state you must satisfy both the current node rule and energy conservation. These two principles give you two equations for the two unknown fields. (b) The electron mobility

in room-temperature Nichrome is about $\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{}\raisebox{1ex}{$\left({\displaystyle \frac{m}{s}}\right)$}\!\left/ \!\raisebox{-1ex}{$\left({\displaystyle \frac{N}{s}}\right)$}\right.$. Show that it takes an electron 36 min to drift through the two Nichrome wires from location B to location A. (c) On the other hand, about how long did it take to establish the steady state when the circuit was first assembled? Give a very approximate numerical answer, not a precise one. (d) There are about $\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{28}}$mobile electrons per cubic meter in Nichrome. How many electrons cross the junction between the two wires every second?

Short answer

The time taken by the electron is 36 min.

Step-by-step solution

Write the given data from the question.

The thin wire is 50 cm long and the diameter is 0.25 mm.

The flashlight battery is 1.5 V.

Determine the formula

Consider the formula for the average drift speed of the electrons in the wire is as follows:

$\mathit{v}\mathbf{=}\mathit{u}\mathit{E}$

Determine the time taken by the electrons to pass through the nichrome wire.

Determine the value of the speed of the electron through the thin wire as:

$\begin{array}{rcl}{v}_1& =& \left(7.5\times {10}^{-5}\frac{{\displaystyle \raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.}}{{\displaystyle \raisebox{1ex}{$\mathrm{N}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{C}$}\right.}}\right)\left(5.2\frac{\mathrm{V}}{\mathrm{m}}\right)\\ & =& 3.9\times {10}^{-4}\frac{\mathrm{m}}{\mathrm{s}}\end{array}$

Solve for the average drift speed for the electron in thin wire as:

$\begin{array}{rcl}{v}_2& =& \left(7.5\times {10}^{-5}\frac{{\displaystyle \raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.}}{{\displaystyle \raisebox{1ex}{$\mathrm{N}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{C}$}\right.}}\right)\left(2.65\frac{\mathrm{V}}{\mathrm{m}}\right)\\ & =& 1.98\times {10}^{-4}\frac{\mathrm{m}}{\mathrm{s}}\end{array}$

Solve for the total time taken by the electron as:

$\begin{array}{rcl}t& =& \frac{L_1}{v_1}+\frac{L_2}{v_2}\\ & =& \left(\frac{50\times {10}^2\mathrm{m}}{3.90\times {10}^{-4}\frac{\mathrm{m}}{\mathrm{s}}}\right)\left(\frac{1.50\times {10}^2}{1.98\times {10}^{-4}\frac{\mathrm{m}}{\mathrm{s}}}\right)\\ & =& 2140\mathrm{s}\\ & =& 36\min \end{array}$