Question

In the circuit shown in Figure 18.110, the two thick wires and the thin wire are made of Nichrome.

(a) Show the steady-state electric field at indicated locations, including in the thin wire. (b) Carefully draw pluses and minuses on your own diagram to show the approximate surface-charge distribution in the steady state. Make your drawing show the differences between regions of high surface-charge density and regions of low surface-charge density. (c) The emf of the battery is$\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{V}$. In Nichrome, there are $\mathit{n}\mathbf{=}\mathbf{9}\mathbf{\times }\mathbf{10}$²⁸ mobile electrons per m³, and the mobility of mobile electrons is $\mathit{\mu }\mathbf{=}\mathbf{7}\mathbf{\times }\mathbf{10}$^{-5$\raisebox{1ex}{$\mathbf{(}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{)}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{(}\mathbf{V}\mathbf{/}\mathbf{m}\mathbf{)}$}\right.$}. Each thick wire has a length of L₁ $\mathbf{=}\mathbf{}\mathbf{20}\mathbf{cm}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{m}$ and a cross-sectional area of A₁ $\mathbf{=}\mathbf{9}\mathbf{\times }\mathbf{10}$^-8 m². The thin wire has a length of L₂$\mathbf{=}\mathbf{5}\mathbf{cm}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{m}$ and a cross-sectional area of A₂$\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{\times }\mathbf{10}$^-8m². (The total length of the three wires is $\mathbf{45}\mathbf{cm}$)Calculate the number of electrons entering the thin wire every second in the steady state. Do not make any approximations, and do not use Ohm’s law or series-resistance equations. State briefly where each of your equations comes from.

Short answer

The electric field is in a steady state at the indicated locations.

Step-by-step solution

Write the given data from the question.

The emf of the battery, $\mathrm{V}=1.5\mathrm{V}$

Number of the electron in nichrome,$\mathrm{n}=9\times 10$²⁸$\raisebox{1ex}{$\mathrm{e}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{m}$}\right.$³

Mobility of mobile electron, $\mathrm{\mu }=7\times 10$^-5$(\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s})$}\right./(\raisebox{1ex}{$\mathrm{V}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{m})$}\right.$

Length of thick wire, L₁=$20\mathrm{cm}=0.20\mathrm{m}$

Area of thick wire, A₁=$9\times 10$^-8m²

Length of thin wire, L₂=role="math" localid="1668659189411" $5\mathrm{cm}=0.05\mathrm{m}$

Area of thin wire, A₂=$1.5\times 10$^-8m2

The total length of the three wires, L_t= $45\mathrm{cm}$

Determine the formulas to show the steady state electric field at indicated locations.

The electric field is defined as the voltage per unit length.

The expression to calculate the electric field is given as follows.

$\mathbf{E}\mathbf{=}\frac{\mathbf{V}}{\mathbf{L}}$

Here, $\mathbf{V}$the voltage and

$\mathbf{L}$ is the length.

Show the steady state electric field at indicated locations.

The circuit has thick and thin wires connected in the series; therefore, the current is the same. The current in both wires would be the same, and the steady-state electric file drives the steady-state current. Therefore, the electric field at the indicated locations is the same.

Hence the electric field is in a steady state at the indicated locations.