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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 isu=u0+kxwhere 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?

Short Answer

Expert verified

(b) The electric field will decrease and will have two different magnitudes at two locations that are marked as the mobility is increasing, and these two are inversely proportional to each other.

Step by step solution

01

Given data

A long Iron slab with a width w and height h and has temperature difference at both ends. There are n number of iron atoms in a cubic meter.

The electron mobility can be determined by the equation, u=u0+kx where u0is the mobility of the iron at the hot end of the slab, and it is increasing with distance x.

The steady-state current is denoted as I, which runs through the slab. A voltmeter is connected to measure the voltage between distance d.

02

Concept

The number of charge carriers traveling in a conductor can be written as,

ne=nAuEt (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

(b)  Reason for the different magnitude of the electric field at two locations

The above equation tells us that the electric field and mobility are inversely proportional, and mobility is increasing with distance x in this case which will lead to a decrease in the magnitude of the electric field.

Thus, the electric field will decrease in the direction when mobility is increasing in the same direction and will have two different magnitudes at two locations that are marked.

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Most popular questions from this chapter

When a single thin-filament bulb is connected to a 1.5Vbattery, the current through the battery is about80mAIf you add another thin-filament bulb in parallel, the battery current of course increase to160mA. Is the battery ohmic? That is, is the current through the battery proportional to the potential difference across the battery?

The capacitor in Figure 19.67 is initially uncharged, then the circuit is connected. Which graph in Figure 19.66 best describes the current through the bulb as a function of time?

Using thick connecting wires that are very good conductors, a Nichrome wire (“wire 1”) of length L1 and cross-sectional area A1 is connected in series with a battery and an ammeter (this is circuit 1). The reading on the ammeter is I1. Now the Nichrome wire is removed and replaced with a different wire (“wire 2”), which is 2.5 times as long and has 5.5 times the cross-sectional area of the original wire (this is circuit 2). In the following question, a subscript 1 refers to circuit 1, and a subscript 2 refers to circuit 2. It will be helpful to write out your solutions to the following questions algebraically before doing numerical calculations. (Hint: Think about what is the same in these two circuits.)(a) What is the value of I2/ I1, the ratio of the conventional currents in the two circuits? (b) What is the value of R2/ R1, the ratio of the resistances of the wires? (c) What is the value of E2/ E1, the ratio of the electric fields inside the wires in the steady states?

A capacitor with a slab of glass between the plates is connected to a battery by Nichrome wires and allowed to charge completely. Then the slab of glass is removed. Describe and explain what happens. Include diagrams. If you give a direction for a current, state whether you are describing electron current or conventional current.

Suppose that instead of placing an insulating layer between the plates of the capacitor shown in Figure 19.57, you inserted a metal slab of the same thickness, just barely not touching the plates. In the same circuit, would this capacitor keep the current more nearly constant or less so than capacitor 2 in Question Q4? Explain why this is essentially equivalent to making a capacitor with a shorter distance between the plates.

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