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There are very roughly the same number of iron atoms per m3 as there are copper atoms per m3 , but copper is a much better conductor than iron. How does uiron compare with ucopper?

Short Answer

Expert verified

The ucopper is greater than uiron

Step by step solution

01

Given data

Copper is a better conductor than iron.

02

Define the concept for the mobility

The mobility in a conductor of cross sectional area A and n number of atoms per unit volume, through which a current i passes on application of electric field E, is defined as follows:

u=inAE ..... (i)

03

Determine the mobility of copper to that of iron

From equation (i) it is evident that for equal number density, area and applied electric field, the conductor which allows more current has better mobility that is the material which is more conducting has larger mobility. Thus, copper has larger mobility that is given as:

ucopper>uiron

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Most popular questions from this chapter

What would be the potential difference VC-VBacross the thin resistor in Figure 18.103 if the battery emf is3.5V ? Assume that the electric field in the thick wires is very small (so that the potential differences along the thick wires are negligible). Do you have enough information to determine the current in the circuit?

Since there is an electric field inside a wire in a circuit, why donโ€™t the mobile electrons in the wire accelerate continuously?

The drift speed in a copper wire is 7ร—10-5msfor a typical electron current. Calculate the magnitude of the electric field inside the copper wire. The mobility of mobile electrons in copper is 4.5ร—10-3ms/NC. (Note that though the electric field in the wire is very small, it is adequate to push a sizable electron current through the copper wire.)

Why donโ€™t all mobile electrons in a metal have exactly the same speed?

Question: Three identical light bulbs are connected to two batteries as shown in Figure 18.106. (a) To start the analysis of this circuit you must write energy conservation (loop) equations. Each equation must involve a round-trip path that begins and ends at the same location. Each segment of the path should go through a wire, a bulb, or a battery (not through the air). How many valid energy conservation (loop) equations is it possible to write for this circuit? (b) Which of the following equations are valid energy conservation (loop) equations for this circuit? E1refers to the electric field in bulb 1; L refers to the length of a bulb filament. Assume that the electric field in the connecting wires is small enough to neglect.

(1) +E2L-E3L=0, (2) E1L-E3L=0, (3)+2emf-E2L-E3L=0, (4)E1L-E2L=0, (5)+2emf-E1L-E2L=0, (6)+2emf-E1L-E3L=0, (7)+2emf-E1L-E2L-E3L=0. (c) It is also necessary to write charge conservation equations (node) equations. Each such equation must relate electron current flowing into a node to electron current flowing out of a node. Which of the following are valid charge conservation equations for this circuit? (1)i1=i3, (2)i1=i2, (3)i1=i2+i3. Each battery has an emf of 1.5V. The length of the tungsten filament in each bulb is 0.008m. The radius of the filament is5ร—10-6m(it is very thin!). The electron mobility of tungsten is localid="1668588909714" 1.8ร—10-3(m/s)/(V/m). Tungsten has localid="1668588927161" 6ร—1028mobile electrons per cubic meter. Since there are three unknown quantities, we need three equations relating these quantities. Use any two valid energy conservation equations and one valid charge conservation equation to solve for localid="1668588943223" E1,E2,i1and localid="1668588965567" i2.

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