In the circuit shown in the diagram below, two thick wires connect a 1.7 volt ba

Question

In the circuit shown in the diagram below, two thick wires connect a 1.7 volt battery to a Nichrome (NiCr) wire. Each thick connecting wire is 15 cm long, and has a radius of 9 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 6 cm long, and has a radius of 5 mm. Nichrome has 9e+28 mobile electrons per cubic meter, and an electron mobility of 7e-05 (m/s)/(V/m).

1.What is the magnitude of the electric field in the thick copper wire?

2.What is the magnitude of the electric field in the thin Nichrome wire?

Explanation / Answer

given

V = 1.7 V

thick wire

L = 15 cm

R = 9 mm

N = 8.36*10^28 per m^3

M = 0.00434 m/s / V/m

nichrome wire

l = 6 cm

r = 5 mm

n = 9*10^28

m = 7*10^-5

1. from ohms law

E = J/sigma

now, for thick wire

sigma = eNM = 1.722*10^-8

resistance

R = rho*L/pi*R^2 = 1.105*10^-5 ohm

hence

E1 = V"/R

2. from ohms law

E2 = J/sigma'

now, E = V'/l

also, R' = (1/sigma')l/A = 7.5788*10^-4 ohm

hence Reff = 2R + R' = 7.799*10^-4 ohm

henceE1 = V*R/Reff*L = 0.1605 V/m

E2 = V*R'/Reff*l = 27.5333 V/m

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