In the circuit shown in the diagram below, two thick wires connect a 1.4 volt ba
ID: 2089918 • Letter: I
Question
In the circuit shown in the diagram below, two thick wires connect a 1.4 volt battery to a Nichrome (NiCr) wire. Each thick connecting wire is 15 cm long, and has a radius of 11 mm. The thick wires are made of copper, which has 8.36e+28 mobile electrons per cubic meter, and an electron mobility of 0.00434 (m/s)/(V/m). The Nichrome wire is 5 cm long, and has a radius of 4 mm. Nichrome has 9e+28 mobile electrons per cubic meter, and an electron mobility of 7e-05 (m/s)/(V/m).
What is the magnitude of the electric field in the thick copper wire?
What is the magnitude of the electric field in the thin Nichrome wire?
What I've tried so far:
Resistence of Copper wire = resistivity * length / area = 1.68e-8 * .3 / 1.9e-4 = 2.65e-5
Resistence of Nichrome wire = resistivity * length / area = 1.1e-6 * .05 / 2.513e-5 = .00109434
Net Resistence = .0011206902
Current = 1.4/.0011206902 = 1249.230124
Density Copper = 1249.23 / 2.65e-5 = 47140759.41
Density of Nichrome = 1249.23 / .00109434 = 1141537.478
Electric field = resistivity * density
Electric filed of Copper = 1.68e-8 * 47140759.41 / .15 = .5.279
Electric field of Nichrome = 1.1e-6 * 1141537.478 / .05 = 25.11
Both of these answers are incorrect.
Explanation / Answer
Look at it this way: assuming your figures for current and resistance are correct, we need the IR drop across each type of wire. For the copper, it's (1249.23 A)(0.0000265 ohm) = 0.0331 V. For the nichrome, it's (1249.23 A)(0.00109434 ohm) = 1.3671 V. Then the electric field in the wires is that voltage divided by the total length of the wire. For copper: (0.0331 V)/(0.30 m) = 0.110 V/m. For the nichrome: (1.3671 V)/(0.05 m) = 27.34 V/m.