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Chapter 18: Q37 P (page 759)

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.)

Short Answer

Expert verified

The electric field inside the wire is 5V/m and it remains same for new wire diameter.

Step by step solution

01

Identification of given data

The emf of the flashlight battery is 蔚=1.5V

The length of Nichrome wire is l=30cm

The diameter of wire is d=0.25mm

The new diameter of wire is D=0.35mm
The amount of charge at the positive end of battery plate is found by the product of the capacitance and emf of the battery.

02

Determination of electric field inside the wire

The electric field inside the wire is given as:

E=Vl

Substitute all the values in above equation.

E=1.5V30cm1m100cmE=5V/m

The electric field does not change with the diameter of the wire because the length of wire and emf of flashlight battery is constant.

Therefore, the electric field inside the wire is 5V/m and it remains same for new wire diameter.

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Most popular questions from this chapter

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.

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?

Suppose that wire A and wire B are made of different metals and are subjected to the same electric field in two different circuits. Wire B has the 6 times the cross sectional area, 1.3 times as many mobile electrons per cubic centimetre and 4 times the mobility of wire A. In the steady state \({\bf{2 \times 1}}{{\bf{0}}^{{\bf{18}}}}\) electrons enters wire A every second. How many electrons enter wire B every second?

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.

Describe the following attributes of a metal wire in steady

state vs. equilibrium:

Metal Wire

Steady-state

Equilibrium

Location of excess charge

Motion of mobile electrons

inside the metal wire
See all solutions

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