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Chapter 18: Q46 P (page 761)

In the circuit shown figure 18.108, two thick copper wires connect a 1.5 V battery to a Nichrome wire. Each thick connecting wire is 17 cm long and has a radius of 9 mm. Copper has 8.4脳1028mobile electrons per cubic meter and electron mobility. The Nichrome wire is 8 cm long and has a radius of 3 mm. Nichrome has 9脳1028mobile electrons per cubic meter and electron mobility of 7脳10-5(ms)(Vm).

(a) What is the magnitude of the electric field in the thick copper wire?

(b) What is the magnitude of the electric field in the thin Nichrome wire?

Short Answer

Expert verified

The magnitude of the electric field inside copper wire is8.82V/m.

Step by step solution

01

Identification of the given data

The emf of the battery is 蔚=1.5V.

The charge density for copper wire is nc=8.4脳1028electrons/m3

The length of the copper wire is role="math" localid="1668668353251" lc=17cm

The radius of copper wire is rc=9mm

The electron mobility of copper wire is 渭c=4.4脳10-3ms/Vm

The magnitude of the electric field inside the copper wire is found by dividing the emf of the battery by the length of the copper wire.

02

Determination of magnitude of the electric field inside copper wire.

The magnitude of the electric field inside copper wire is given as:

Ec=Vlc

Substitute all the values in the above equation.

Ec=1.5V17cm1m100cmEc=8.82Vm

Therefore, the magnitude of the electric field inside copper wire is 8.82V/m.

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Most popular questions from this chapter

The emf of a particular flashlight battery is 1.7 V. If the battery is 4.5 cm long and radius of cylindrical battery is 1 cm, estimate roughly the amount of charge on the positive end plate of the battery.

When a single thick-filament bulb of a particular kind and two batteries are connected in series, 3脳1018 electrons pass through the bulb every second. When two batteries in series are connected to a single thin-filament bulb, with a filament made of the same material and length as the thick-filament bulb but a smaller cross-section, only 1.5脳1018 electrons pass through the bulb every second. (a) In the circuit shown in Figure 18.109, how many electrons per second flow through the thin-filament bulb? (b) What approximations or simplifying assumptions did you make? (c) Show approximately the surface charge on a diagram of the circuit.

During the initial transient leading to the steady state, the electron current going into a bulb may be greater than the electron current leaving the bulb. Explain why and how these two currents come to be equal in the steady state.

Suppose that a wire leads into another, thinner wire of the same material that has only a third the cross-sectional area. In the steady state, the number of electrons per second flowing through the thick wire must be equal to the number of electrons per second flowing through the thin wire. If the drift speedV1炉in the thick wire is 4脳10-5ms, what is the drift speed V炉2in the thinner wire?

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 is5脳10-6m(it is very thin!). The electron mobility of tungsten is localid="1668588909714" 1.8脳10-3(m/s)/(V/m). Tungsten has localid="1668588927161" 6脳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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