91Ó°ÊÓ

A circuit consists of a battery, whose emf is K, and five Nichrome wires, three thick and two thin as shown in Figure 19.78. The thicknesses of the wires have been exaggerated in order to give you room to draw inside the wires. The internal resistance of the battery is negligible compared to the resistance of the wires. The voltmeter is not attached until part (e) of the problem. (a) Draw and label appropriately the electric field at the locations marked × inside the wires, paying attention to appropriate relative magnitudes of the vectors that you draw. (b) Show the approximate distribution of charges for this circuit. Make the important aspects of the charge distribution very clear in your drawing, supplementing your diagram if necessary with very brief written descriptions on the diagram. Make sure that parts (a) and (b) of this problem are consistent with each other. (c) Assume that you know the mobile-electron density n and the electron mobility u at room temperature for Nichrome. The lengths $(L_1,L_2,L_3)$and diameters $(d_1,d_2)$of the wires are given on the diagram. Calculate accurately the number of electrons that leave the negative end of the battery every second. Assume that no part of the circuit gets very hot. Express your result in terms of the given quantities $(K,L_1,L_2,L_3,d_1,d_2,n\text{and}u)$ . Explain your work and identify the principles you are using. (d) In the case that ${\mathit{d}}_{\mathbf{2}}\mathbf{≪}{\mathit{d}}_{\mathbf{1}}$, what is the approximate number of electrons that leave the negative end of every second? (e) A voltmeter is attached to the circuit with its + lead connected to location B (halfway along the leftmost thick wire) and its - lead connected to location C (halfway along the leftmost thin wire). In the case that${\mathit{d}}_{\mathbf{2}}\mathbf{≪}{\mathit{d}}_{\mathbf{1}}$ , what is the approximate voltage shown on the voltmeter, including sign? Express your result in terms of the given quantities $(K,L_1,L_2,L_3,d_1,d_2,n\text{and}u)$.

Step by step solution

01

Given data

A circuit is given which has a battery with an emf value K and five Nichrome wires in a number of 5, three out of five are thick, and two are thin. The mobile-electron density is given as n, and the electron mobility is given as u at room temperature for Nichrome. The lengths of the wires are$L_1,L_2,L_3$ , and the diameters of the wires are $d_1,d_2$.

02

Concept

The number of electrons flowing in a given circuit can be written as,

${\mathit{n}}_{\mathbf{e}}\mathbf{=}\mathit{n}\mathit{A}\mathit{u}\mathit{E}\mathit{t}$ (1)

Here n is the charge density of the charge carrier; A is the cross-sectional area of the conductor, E is the electric field in the conductor, u is the mobility of the charge carriers, t is the time.

03

(d) In the case d2≪d1, calculation of the approximate number of electrons that leave the negative end of every second

The number of electrons that leave the negative end of the battery every second is$\frac{nu\mathrm{Ï€}{\left(d_2\right)}^{2}K}{4\left(\frac{2{\left(d_2\right)}^2{L}_1}{{\left(d_1\right)}^2}+\frac{{\left(d_2\right)}^2{L}_3}{{\left(d_1\right)}^2}+2L_2\right)}$ . (Refer to SID:875865-19-59 P-c)

If

$$\begin{gathered} d_2≪d_1 \\ \frac{d_2}{d_1}≪1 \end{gathered}$$

Then in the expression, $n_1=\frac{nu\mathrm{Ï€}{\left(d_2\right)}^{2}K}{4\left(\frac{2{\left(d_2\right)}^2{L}_1}{{\left(d_1\right)}^2}+\frac{{\left(d_2\right)}^2{L}_3}{{\left(d_1\right)}^2}+2L_2\right)}$the term $\frac{2{\left(d_2\right)}^2{L}_1}{{\left(d_1\right)}^2}+\frac{{\left(d_2\right)}^2{L}_3}{{\left(d_1\right)}^2}$ will be written as,

$\frac{2{\left(d_2\right)}^2{L}_1}{{\left(d_1\right)}^2}+\frac{{\left(d_2\right)}^2{L}_3}{{\left(d_1\right)}^2}≈0$

.

Therefore we can write the expression for the number of electrons that leaves the negative end of every second can be written as,

$$\begin{gathered} n_1=\frac{nu\mathrm{Ï€}{\left(d_2\right)}^{2}K}{4\left(2L_2\right)} \\ {n}_1=\frac{nu\mathrm{Ï€}{\left(d_2\right)}^{2}K}{8L_2} \end{gathered}$$

Thus, the approximate number of electrons that leave the negative end of every second when $d_2≪d_1$is $\frac{nu\mathrm{π}{\left(d_2\right)}^{2}K}{8L_2}$.

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