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Since there is an electric field inside a wire in a circuit, why don鈥檛 the mobile electrons in the wire accelerate continuously?

Short Answer

Expert verified

Electrons in a wire don't accelerate continuously because the backward force due to resistance cancels out the forward force from the external electric field.

Step by step solution

01

Given data

Electrons in a wire don't accelerate continuously.

02

Concept of resistance in a wire

In the presence of an external electric field, there is a steady flow of electrons inside a wire in a circuit. These wires continuously collide with each other which hinders the flow. Thus there is a net opposing force against the flow proportional to the velocity. This is called resistance.

03

Determination of the reason why electrons in a wire don't accelerate continuously

Electrons in a wire face a net forward force from the external electric field and a backward force due to resistance. These forces cancel each other and thus the electrons move with a constant velocity called the drift velocity.

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Most popular questions from this chapter

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?

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鈥檛 matter.

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?

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.

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 8.41028mobile electrons per cubic meter and electron mobility. The Nichrome wire is 8 cm long and has a radius of 3 mm. Nichrome has 91028mobile electrons per cubic meter and electron mobility of 710-5(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?

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