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Chapter 27: Q. 31 (page 763)

The terminals of a 0.70V watch battery are connected by a 100m long gold wire with a diameter of 0.10mm. What is the current in the wire?

Short Answer

Expert verified

The current in the wire is2.29mA.

Step by step solution

01

Given Information

Voltage=0.70V

Gold wire length=100m

Diameter=0.10mm

02

Explanation 

If we can calculate the potential difference across the wire and the resistance, we can find the current according to Ohm's law.

I=UR→(1)

The resistance of the conductor appears in the following expression:

R=ρLA→(2)

Substituting the second expression into the first one,

I=AUρL.

The cross-section area is a circle, we can substitute A=πD24to find

I=πUD24ρL

Substitute the expression,

role="math" localid="1649055691425" I=π×0.7V×0.00012mm4×2.4×10-8Ωm×100m

I=2.29mA

03

Final Answer 

Hence, the current in the wire is2.29mA.

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Most popular questions from this chapter

74. FIGURE CP27.74 shows a wire that is made of two equal diameter segments with conductivities σ1and σ2. When current I passes through the wire, a thin layer of charge appears at the boundary between the segments.
a. Find an expression for the surface charge density ηon the boundary. Give your result in terms of I,σ1,σ2, and the wire's cross-section area A.
b. A 1.0-mm-diameter wire made of copper and iron segments carries a 5.0A current. How much charge accumulates at the boundary between the segments?

For what electric field strength would the current in a 2.0mmdiameter nichrome wire be the same as the current in a 1.0mmdiameter aluminum wire in which the electric field strength is0.0080V/m?

15. A hollow copper wire with an inner diameter of 1.0mmand an outer diameter of 2.0mm carries a current of 10A. What is the current density in the wire?

75. A 300μFcapacitor is charged to 9.0V, then connected in parallel with a 5000Ωresistor. The capacitor will discharge because the resistor provides a conducting pathway between the capacitor plates, but much more slowly than if the plates were connected by a wire. Let t=0s be the instant the fully charged capacitor is first connected to the resistor. At what time has the capacitor voltage decreased by half, to 4.5V?
Hint: The current through the resistor is related to the rate at which charge is leaving the capacitor. Consequently, you'll need a minus sign that you might not have expected.

An ideal battery would produce an extraordinarily large current if "shorted" by connecting the positive and negative terminals with a short wire of very low resistance. Real batteries do not. The current of a real battery is limited by the fact that the battery itself has resistance. What is the resistance of a 9.0V battery that produces a 21A current when shorted by a wire of negligible resistance?
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