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Chapter 19: Q57P (page 799)

Consider a copper wire with a cross-sectional area of 1 mm2 (similar to your connecting wires ) and carrying 0.3 A of current, which is about what you get in a circuit with a thick-filament bulb and two batteries in series. Calculate the strength of the very small electric field required to drive this current through the wire.

Short Answer

Expert verified

The strength of the very small electric field required to drive this current through the wire is 5脳10-3V/m.

Step by step solution

01

Given data

The data can be listed as,

  • The cross-sectional area of copper wire is, A=1mm2=1脳10-6m2.
  • Current is, I=0.3A.
02

Concept

If a current flows in the circuit, it is influenced by many factors, such as the length of the wire, the cross-sectional area of the wire, and the applied voltage on the wire.

03

Determination of the required electric field

The wire is made of copper metal only.

The required electric field can be determined using the formula as,

E=蚁IA

Here 蚁is the resistivity of the copper wire whose value is 16.78脳10-9惟路m.

Substitute the values in the above expression, and we get,

E=16.78脳10-9惟路m0.3A1脳10-6m2=5.03脳10-3惟路m-1路A~5脳10-3V/m

Thus, the strength of the very small electric field required to drive this current through the wire is 5脳10-3V/m.

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Most popular questions from this chapter

Consider two capacitors whose only difference is that the plates of capacitor number 2 are closer together than those of capacitor number 1 (Figure 19.56). Neither, capacitors has an insulating layer between the plates. They are placed in two different circuits having similar batteries and bulbs in series with the capacitor.

Show that in the first fraction of a second, the current stays nearly constant (decreases less rapidly) in the circuit with capacitor number 2. Explain your reasoning in detail.

Hint: Show charges on metal plates, and consider the electric fields they produce in the nearby wires. Remember that the fringe field near a plate outside a circular capacitor is approximately-

(QA蔚o)(s2R)

More extensive analysis shows that this trend holds true for the entire charging process: the capacitor with the narrower gap ends up with more charge on the plates.

Using thick connecting wires that are very good conductors, a Nichrome wire (鈥渨ire 1鈥�) of length L1 and cross-sectional area A1 is connected in series with a battery and an ammeter (this is circuit 1). The reading on the ammeter is I1. Now the Nichrome wire is removed and replaced with a different wire (鈥渨ire 2鈥�), which is 2.5 times as long and has 5.5 times the cross-sectional area of the original wire (this is circuit 2). In the following question, a subscript 1 refers to circuit 1, and a subscript 2 refers to circuit 2. It will be helpful to write out your solutions to the following questions algebraically before doing numerical calculations. (Hint: Think about what is the same in these two circuits.)(a) What is the value of I2/ I1, the ratio of the conventional currents in the two circuits? (b) What is the value of R2/ R1, the ratio of the resistances of the wires? (c) What is the value of E2/ E1, the ratio of the electric fields inside the wires in the steady states?

A battery with negligible internal resistance is connected to a resistor. The power produced in the battery and power dissipated in the resistor are both P1. Another resistor of same kind is added, so circuit consists of a battery and two resistors in series. (a) in terms of P1 how much power is dissipated in the first resistor ? . (a) in terms of P1 how much power is produced in the battery ? (c ) The circuit is rearranged so that the two resistors are in parallel rather than in series. In terms of P1, now how much power is produced in the battery?

You are marooned on a desert island full of all kinds of standard electrical apparatus including a sensitive voltmeter, but you don鈥檛 have an ammeter. Explain how you could use the voltmeter to measure currents.

A particular capacitor is initially charged. Then a high-resistance Nichrome wire is connected between the plates of the capacitor, as shown in Figure 19.69. The needle of a compass placed under the wire deflects 20掳to the east as soon as the connection is made. After 60sthe compass needle no longer deflects.

(a)Which of the diagrams in Figure 19.69 best indicates the electron current at three locations in this circuit? (1) 0.01safter the circuit is connected, (2) 15s after the circuit is connected, (3) 120s after the circuit is connected.

(b)Which of the diagrams in Figure 19.70 best indicates the net electric field inside the wire at three locations in this circuit? (1) 0.01s after the circuit is connected, (2) 15safter the circuit is connected, (3) 120s after the circuit is connected.
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