A long Iron slab of width w and height h emerges from a furnace, as shown in Figure 19.79. Because the end of the slab near the furnace is hot and the other end Is cold, the electron mobility increases significantly with the distance x. The electron mobility is $\mathit{u}\mathbf{=}{\mathit{u}}_{\mathbf{0}}\mathbf{+}\mathit{k}\mathit{x}$where u₀is the mobility of the iron at the hot end of the slab. There are n iron atoms per cubic meter, and each atom contributes one electron to the sea of the mobile electron (we can neglect the small thermal expansion of the iron). A steady state conventional current runs through the slab from the hot end towards cold end, and an ammeter (not shown) measures the current to have a magnitude I in amperes. A voltmeter is connected to two locations a distance d apart, as shown. (a) Show the electric field inside the slab at two locations marked with ×. Pay attention to the relative magnitudes of the two vectors that you draw. (b) Explain why the magnitude of the electric field is different at these two locations. (c) At a distance x from the left voltmeter connection, what is the magnitude of the electric field in terms x and the given quantities w,h,d,u₀,k,l, and n ( and fundamental constants)? (d) What is the sign of potential difference displayed on the voltmeter? Explain briefly. (e) In terms of the given quantitiesw,h,d,u₀,k,l, and n and ( and fundamental constants), what is the magnitude of the voltmeter reading? Check your work. (f) What is the resistance of this length of the iron slab?

Step by step solution

01

Given data

The given data can be listed below as,

• A long Iron slab has a width which is w and a height which is h.
• There are n iron atoms per cubic meter.
• One end is hot, and one is cold.
• The electron mobility is$u=u_0+kx$ where u₀ is the mobility of the iron at the hot end of the slab, and it increases with distance x.
• The steady-state current is I.
• A voltmeter is connected to measure the potential drop between d distance.

02

Concept

The steady-state current equation can be written as,

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

Here e is the charge on the electron, A is the cross-sectional area of the conductor, and E is the electric field in the conductor.

03

(c) Calculation of electric field at the distance x from the left end

The cross-sectional area can be calculated as,

$A=wh$

Substitute all the values in equation 1, and we get,

$$\begin{gathered} I=newh\left(u_0+kx\right)E \\ E=\frac{I}{newh\left(u_0+kx\right)} \end{gathered}$$

Thus, the magnitude of the electric field at a distance x from the left end is$\frac{I}{newh\left(u_0+kx\right)}$ .

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