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50P-a

Expert-verifiedFound in: Page 45

Book edition
4th edition

Author(s)
Ruth W. Chabay, Bruce A. Sherwood

Pages
1135 pages

ISBN
9781118875865

**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(\frac{m}{s}\right)$}\!\left/ \!\raisebox{-1ex}{$\left(\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?**

The value of the electric field in the thin wire is ${E}_{1}=5.2\frac{\mathrm{V}}{\mathrm{m}}$ and electric field in the thick wire is ${E}_{1}=2.65\frac{\mathrm{V}}{\mathrm{m}}$.

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

The flashlight battery is 1.5 V.

Consider the expression for the current in the metal wire as:

${\mathit{i}}{\mathbf{=}}{\mathit{n}}{\mathit{a}}{\mathit{v}}$ …... (1)

Consider the formula for the drift sped of the mobile electrons is proportional to the magnitude of the electric field as:

${\mathit{v}}{\mathbf{=}}{\mathit{u}}{\mathit{E}}$ ….. (2)

Here, u is the proportionality constant and the is called the electrons mobility.

Consider from equation (1) and (2), the equation for the current in the wire is derived as:

${\mathit{i}}{\mathbf{=}}{\mathit{n}}{\mathit{A}}{\mathit{u}}{\mathit{E}}$ ….. (3)

Determine the area of the thin nichrome wire as:

$\begin{array}{rcl}{A}_{1}& =& \mathrm{\pi}{\left(\frac{{\mathrm{D}}_{1}}{2}\right)}^{3}\\ & =& \left(3.14\right){\left(\frac{0.25\mathrm{mm}}{2}\right)}^{2}\\ & =& 0.04908\times {10}^{-6}{\mathrm{m}}^{2}\end{array}$

Determine the area of thick nichrome wire as:

$\begin{array}{rcl}{A}_{1}& =& \mathrm{\pi}{\left(\frac{{\mathrm{D}}_{1}}{2}\right)}^{3}\\ & =& \left(3.14\right){\left(\frac{0.35\mathrm{mm}}{2}\right)}^{2}\\ & =& 0.09621\times {10}^{-6}{\mathrm{m}}^{2}\end{array}$

Consider the emf of each battery is 1.5 V and since there are two batteries then the total emf is:

$\begin{array}{rcl}E& =& 1.5\mathrm{V}+1.5\mathrm{V}\\ & =& 3\mathrm{V}\end{array}$

Consider the KCL equation for the steady state. In this case the current in the thick and the thin wire must be equal.

$\begin{array}{rcl}n{A}_{1}u{E}_{1}& =& n{A}_{2}u{E}_{2}\\ {E}_{2}& =& \left(\frac{{A}_{1}}{{A}_{2}}\right){E}_{1}\end{array}$

Consider the loop equation as follows:

role="math" localid="1662199994922" $3\mathrm{V}={E}_{1}\left({L}_{1}+\left(\frac{{A}_{1}}{{A}_{2}}\right){L}_{2}\right)\phantom{\rule{0ex}{0ex}}{E}_{1}=\frac{3\mathrm{V}}{\left({L}_{1}+\left(\frac{{A}_{1}}{{A}_{2}}\right){L}_{2}\right)}$

Substitute the values and solve as:

role="math" localid="1662200569351" $\begin{array}{rcl}{E}_{1}& =& \frac{3\mathrm{V}}{\left(50\times {10}^{2}\mathrm{m}+\left(\frac{0.04908\times {10}^{-6}{\mathrm{m}}^{2}}{0.9062\times {10}^{-6}{\mathrm{m}}^{2}}\right)15\times {10}^{-2}\right)}\\ & =& 5.2\frac{\mathrm{V}}{\mathrm{m}}\end{array}$

Also,

role="math" localid="1662200616486" $\begin{array}{rcl}{E}_{2}& =& \left(\frac{0.04908\times {10}^{-6}{\mathrm{m}}^{2}}{0.9062\times {10}^{-6}{\mathrm{m}}^{2}}\right)\left(5.2\frac{\mathrm{V}}{\mathrm{m}}\right)\\ & =& 2.65\frac{\mathrm{V}}{\mathrm{m}}\end{array}$

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