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 7×10-4m. The wires are made of a metal that has7×1028mobile electrons per cubic meter; the electron mobility is 5×10-5(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=7×10-4m

Number of mobile electrons,role="math" localid="1668583532869" N=7×10281/m3

Electron mobility,μ=5×10-5(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 …… (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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Inside a chemical battery it is not actually individual electrons that are transported from the + end to the – end. At the + end of the battery an “acceptor” molecule picks up an electron entering the battery, and at the – end a different “donor” 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?

In the circuit shown in Figure 18.87, bulbs 1 and 2 are identical in mechanical construction (the filaments have the same length and the same cross-sectional area), but the filaments are made of different metals. The electron mobility in the metal used in bulb 2 is three times as large as the electron mobility in the metal used in bulb 1, but both metals have the same number of mobile electrons per cubic meter. The two bulbs are connected in series to two batteries with thick copper wires (like your connecting wires).

(a)In bulb 1, the electron current is i1and the electric field is E1. In terms of these quantities, determine the corresponding quantities i2and E2for bulb 2, and explain your reasoning.

(b)When bulb 2 is replaced by a wire, the electron current through bulb 1 is i0and the electric field in bulb 1 is E0. How big is i1 in terms of i0? Explain your answer, including explicit mention of any approximations you must make. Do not use ohms or series-resistance equations in your explanation, unless you can show in detail how these concepts follow from the microscopic analysis introduced in this chapter.

(c)Explain why the electric field inside the thick copper wires is very small. Also explain why this very small electric field is the same in all of the copper wires, if they all have the same cross-sectional area.

(d)Figure 18.88 is a graph of the magnitude of the electric field at each location around the circuit when bulb 2 is replaced by a wire. Copy this graph and add to it, on the same scale, a graph of the magnitude of the electric field at each location around the circuit when both bulbs are in the circuit. The very small field in the copper wires has been shown much larger than it really is in order to give you room to show how that small field differs in the two circuits.

How can there be a nonzero electric field inside a wire in a circuit? Isn’t the electric field inside a metal always zero?

A steady-state current flows through the Nichrome wire in the circuit shown in Figure 18.90. Before attempting to answer the following 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.

(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 some excess negative charge on the surface of the wire near location B.

(2) Inside the metal wire the magnitude of the electric field is zero.

(3) The magnitude of the electric field is the same at locations Gand C.

(4) The electric field points to the left at location G.

(5) There is no excess charge on the surface of the wire.

(6) There is excess charge on the surface of the wire near the batteries but nowhere else.

(7) The magnitude of the electric field inside the wire is larger at location Gthan at location C.

(8) The electric field at location Dpoints to the left.

(9) Because the current is not changing, the circuit is in static equilibrium.

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?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free