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 8.4×1028mobile electrons per cubic meter and electron mobility. The Nichrome wire is 8 cm long and has a radius of 3 mm. Nichrome has 9×1028mobile electrons per cubic meter and electron mobility of 7×10-5(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?

Short Answer

Expert verified

The magnitude of the electric field inside copper wire is8.82V/m.

Step by step solution

01

Identification of the given data

The emf of the battery is ε=1.5V.

The charge density for copper wire is nc=8.4×1028electrons/m3

The length of the copper wire is role="math" localid="1668668353251" lc=17cm

The radius of copper wire is rc=9mm

The electron mobility of copper wire is μc=4.4×10-3ms/Vm

The magnitude of the electric field inside the copper wire is found by dividing the emf of the battery by the length of the copper wire.

02

Determination of magnitude of the electric field inside copper wire.

The magnitude of the electric field inside copper wire is given as:

Ec=Vlc

Substitute all the values in the above equation.

Ec=1.5V17cm1m100cmEc=8.82Vm

Therefore, the magnitude of the electric field inside copper wire is 8.82V/m.

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Most popular questions from this chapter

Question: The following questions refer to the circuit shown in Figure 18.114, consisting of two flashlight batteries and two Nichrome wires of different lengths and different thicknesses as shown (corresponding roughly to your own thick and thin Nichrome wires).

The thin wire is 50 cm long, and its diameter is 0.25 mm. The thick wire is 15 cm long, and its diameter is 0.35 mm. (a) The emf of each flashlight battery is 1.5 V. Determine the steady-state electric field inside each Nichrome wire. Remember that in the steady state you must satisfy both the current node rule and energy conservation. These two principles give you two equations for the two unknown fields. (b) The electron mobility

in room-temperature Nichrome is about . Show that it takes an electron 36 min to drift through the two Nichrome wires from location B to location A. (c) On the other hand, about how long did it take to establish the steady state when the circuit was first assembled? Give a very approximate numerical answer, not a precise one. (d) There are about mobile electrons per cubic meter in Nichrome. How many electrons cross the junction between the two wires every second?

How can there be a nonzero electric field inside a wire in a circuit? Isn’t the electric field inside a metal always zero?

In the circuit shown in Figure 18.110, the two thick wires and the thin wire are made of Nichrome.

(a) Show the steady-state electric field at indicated locations, including in the thin wire. (b) Carefully draw pluses and minuses on your own diagram to show the approximate surface-charge distribution in the steady state. Make your drawing show the differences between regions of high surface-charge density and regions of low surface-charge density. (c) The emf of the battery is1.5V. In Nichrome, there are n=9×1028 mobile electrons per m3, and the mobility of mobile electrons is μ=7×10-5(m/s)(V/m). Each thick wire has a length of L1 =20cm=0.2m and a cross-sectional area of A1 =9×10-8 m2. The thin wire has a length of L2=5cm=0.05m and a cross-sectional area of A2=1.5×10-8m2. (The total length of the three wires is 45cm)Calculate the number of electrons entering the thin wire every second in the steady state. Do not make any approximations, and do not use Ohm’s law or series-resistance equations. State briefly where each of your equations comes from.

When a single thick-filament bulb of a particular kind and two batteries are connected in series, 3×1018 electrons pass through the bulb every second. When two batteries in series are connected to a single thin-filament bulb, with a filament made of the same material and length as the thick-filament bulb but a smaller cross-section, only 1.5×1018 electrons pass through the bulb every second. (a) In the circuit shown in Figure 18.109, how many electrons per second flow through the thin-filament bulb? (b) What approximations or simplifying assumptions did you make? (c) Show approximately the surface charge on a diagram of the circuit.

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?

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