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Chapter 18: 50P-c (page 762)

Question: The following questions refer to the circuit shown in Figure 18.114, consisting of two flashlight batteries and two Nichrome wires of different lengths and different thicknesses as shown (corresponding roughly to your own thick and thin Nichrome wires).

The thin wire is 50 cm long, and its diameter is 0.25 mm. The thick wire is 15 cm long, and its diameter is 0.35 mm. (a) The emf of each flashlight battery is 1.5 V. Determine the steady-state electric field inside each Nichrome wire. Remember that in the steady state you must satisfy both the current node rule and energy conservation. These two principles give you two equations for the two unknown fields. (b) The electron mobility

in room-temperature Nichrome is about . Show that it takes an electron 36 min to drift through the two Nichrome wires from location B to location A. (c) On the other hand, about how long did it take to establish the steady state when the circuit was first assembled? Give a very approximate numerical answer, not a precise one. (d) There are about mobile electrons per cubic meter in Nichrome. How many electrons cross the junction between the two wires every second?

Short Answer

Expert verified

The time taken by electron is 1.5 ns.

Step by step solution

01

Write the given data from the question.

The thin wire is 50 cm long and the diameter is 0.25 mm.

The flashlight battery is 1.5 V.

02

Determine the formula

Write the formula for the drift speed of the electron.

03

Determine the time taken to obtain the steady state.

Consider the electrical field is propagating the speed at 0.3ms.

Here, the shift of the electron sea takes very little time as the electron moves a very little distance. So the time taken to make a change of 0.5 m is 1.5 ns.

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Most popular questions from this chapter

The drift speed in a copper wire is 7×10-5msfor a typical electron current. Calculate the magnitude of the electric field inside the copper wire. The mobility of mobile electrons in copper is 4.5×10-3ms/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.)

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 is5×10-6m(it is very thin!). The electron mobility of tungsten is localid="1668588909714" 1.8×10-3(m/s)/(V/m). Tungsten has localid="1668588927161" 6×1028mobile 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.

What is the most important general difference between a system in steady state and a system in equilibrium?

In a table like the one shown, write an inequality comparing each quantity in the steady state for a narrow resistor and thick connecting wires, which are made of the same material as the resistor.

Electron current in resistor

<,=, or >

Electron current in Thick Wires

nR

nw

AR

Aw

uR

uw

ER

Ew

vR

vw

In the circuit shown in Figure 18.91, all of the wire is made of Nichrome, but one segment has a much smaller cross-sectional area. On a copy of this diagram, using the same scale for magnitude that you used in the previous question for Figure 18.90, show the steady-state electric field at the locations indicated, including in the thinner segment. Before attempting to answer these questions, draw a copy of this diagram. All of the locations indicated by letters are inside the wire.

(a)On your diagram, show the electric field at the locations indicated, paying attention to relative magnitude. Use the same scale for magnitude as you did in the previous question.

(b)Carefully draw pluses and minuses on your diagram to show the approximate surface charge distribution that produces the electric field you drew. Make your drawing show clearly the differences between regions of high surface charge density and regions of low surface-charge density. Use your diagram to determine which of the following statements about this circuit are true.

(1) There is a large gradient of surface charge on the wire between locations Cand E. (2) The electron current is the same at every location in this circuit.

(3) Fewer electrons per second pass location Ethan location C.

(4) The magnitude of the electric field at location Gis smaller in this circuit than it

was in the previous circuit (Figure 18.90).

(5) The magnitude of the electric field is the same at every location in this circuit.

(6) The magnitude of the electric field at location D is larger than the magnitude of the electric field at location G.

(7) There is no surface charge at all on the wire near location G.

(8) The electron current in this circuit is less than the electron current in the previous circuit (Figure 18.90).
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