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 number of electrons crossing the junction is $\begin{array}{rcl}& & 1.72\times {10}^{18}\frac{\mathrm{electrons}}{\mathrm{s}}\end{array}$.

Step-by-step solution

Write the given data

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

The flashlight battery is1.5 V.

Determine the formula

Consider the equation for the current in the wire is derived as:

$\mathit{i}\mathbf{=}\mathit{n}\mathit{A}\mathit{u}\mathit{E}$

Determine the number of electrons that cross the junction as:

Solve for the number of electrons that are crossing the junction as:

$$\begin{gathered} i=nAv \\ =\left(9\times {10}^{28}\frac{\mathrm{electrons}}{{\mathrm{m}}^3}\right)\left(0.04908\times {10}^6\right)\left(3.9\times {10}^4\frac{\mathrm{m}}{\mathrm{s}}\right) \\ =1.72\times {10}^{28}\frac{\mathrm{electrons}}{\mathrm{s}} \end{gathered}$$