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.

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Most popular questions from this chapter

Compare the direction of the average electric field inside a battery to the direction of the electric field in the wires and resistors of a circuit.

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 1.4×10-6m2and is 25cmlong. The thin wire has a cross-sectional area 5.9×10-6m2and is 6.1cmlong. In this metal, the electron mobility is 5×10-4(ms)(Vm), and there are 4×1028mobile 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.

Why does the brightness of a bulb not change noticeably when you use longer copper wires to connect it to the battery? (1) Very little energy is dissipated in the thick connecting wires. (2) The electric field in connecting wires is very small, so emfEbulbLbulb. (3) Electric field in the connecting wires is zero, so emfEbulbLbulb. (4) Current in the connecting wires is smaller than current in the bulb. (5) All the current is used up in the bulb, so the connecting wires don’t matter.

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?

Question:In figure 18.102 suppose that VC-VF=8 V and VD-VE=4.5 V.

(a) What is the potential difference VC-VD?

(b) If the element between the battery C and D is a battery, is the + end of the battery at C or D?

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