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Question: A circuit is constructed from two batteries and two wires, as shown in Figure 18.104. Each battery has an emf of 1.3V. Each wire is26cmlong and has a diameter of 710-4m. The wires are made of a metal that has71028mobile electrons per cubic meter; the electron mobility is 510-5(m/s)/(V/m). A steady current runs through the circuit. The locations marked by and labeled by a letter are in the interior of the wire. (a) Which of these statements about the electric field in the interior of the wires, at the locations marked by 's, are true? List all that apply. (1) The magnitude of the electric field at location G is larger than the magnitude of the electric field at location F. (2) At every marked location the magnitude of the electric field is the same. (3) At location B the electric field points to the left. (b) Write a correct energy conservation (round-trip potential difference) equation for this circuit, along a round-trip path starting at the negative end of battery 1 and traveling counterclockwise through the circuit (that is, traveling to the left through the battery, and continuing on around the circuit in the same direction). (c) What is the magnitude of the electric field at location B? (d) How many electrons per second enter the positive end of battery 2? (e)If the cross-sectional area of both wires were increased by a factor of 2, what would be the magnitude of the electric field at location B? (f) Which of the diagrams in Figure 18.105 best shows the approximate distribution of excess charge on the surface of the circuit?

Short Answer

Expert verified

The correct statement is (2): At every marked location the magnitude of the electric field is the same.

Step by step solution

01

Write the given data from the question.

Emf of the battery, V=1.3V

Length of wire, L=26cm

Diameter of wire,d=710-4m

Number of mobile electrons,role="math" localid="1668583532869" N=710281/m3

Electron mobility,=510-5(m/s)/(V/m)

02

Determine the equation to find out the correct statement.

The electric field is defined as the ratio of the voltage and length of the wire.

The expression to calculate the electric field is given as follows.

E=VL 鈥︹ (i)

Here, isL the length of the wire.

03

Find out the correct statement.

The circuit is connected in the series connection, and the current in the series is always the same. The direction of the electric field produce inside the wire is always towards the positive terminal and away from the negative terminal.

From equation (i), it is clear that the electric field depends on the voltage and length of the wire. Therefore, the electric field in the battery's wire is the same at all locations.

Hence the correct statement is (2): At every marked location, the magnitude of the electric field is the same.

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Most popular questions from this chapter

When a single thick-filament bulb of a particular kind and two batteries are connected in series, 31018 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 1.51018 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.

Inside a chemical battery it is not actually individual electrons that are transported from the + end to the 鈥 end. At the + end of the battery an 鈥渁cceptor鈥 molecule picks up an electron entering the battery, and at the 鈥 end a different 鈥渄onor鈥 molecule gives up an electron, which leaves the battery. Ions rather than electrons move between the two ends to support the charge inside the battery.

When the supplies of acceptor and donor molecules are used up in a chemical battery, the battery is dead because it can no longer accept or electron. The electron current in electron per second times the number of seconds of battery life, is equal to the number of donor molecules in the battery.

A flashlight battery contains approximately half a mole of donor molecules. The electron current through a thick filament bulb powered by two flashlight batteries in series is about 0.3 A. About how many hours will the batteries keep this bulb lit?

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.

The circuit shown in Figure 18.107 consists of a single battery, whose emf is 1.8V, and three wires made of the same material but having different cross-sectional areas. Each thick wire has a cross-sectional area 1.410-6m2and is 25cmlong. The thin wire has a cross-sectional area 5.910-6m2and is 6.1cmlong. In this metal, the electron mobility is 510-4(ms)(Vm), and there are 41028mobile electrons/m3.

(a) Which of the following statements about the circuit in the steady state are true? (1) At location B, the electric field points toward the top of the page. (2) The magnitude of the electric field at locations F and C is the same. (3) The magnitude of the electric field at locations D and F is the same. (4) The electron current at location D is the same as the electron current at location F . (b) Write a correct energy conservation (loop) equation for this circuit, following a path that starts at the negative end of the battery and goes counterclockwise. (c) Write this circuit's correct charge conservation (node) equation. (d) Use the appropriate equation(s), plus the equation relating electron current to electric field, to solve for the magnitudes EDand EF of the electric field at locations D and F . (e) Use the appropriate equation(s) to calculate the electron current at location D in the steady state.

A Nichrome wire 30 cm long and 0.25 mm in diameter is connected to a 1.5 V flashlight battery. What is the electric field inside the wire? Why you don鈥檛 have to know how the wire is bent? How would your answer change if the wire diameter change were 0.35 mm? (Not that the electric field in the wire is quiet small compared to the electric field near a charged tape.)

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