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 10 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 8 cm long, and has a radius of 2 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?
       V/m

What is the magnitude of the electric field in the thin Nichrome wire?
        V/m

Explanation / Answer

Electric field in the thick copper wire is E = rho*J

rho is the resistivity of copper wire

j is the current density j = i/A

but drift speed is vd = i/(n*e*A)

i/A = vd*n*e = 0.00434*8.36*10^28*1.6*10^-19 = 5.8*10^7 A/m^2

then E_copper = 1.7*10^-8*5.8*10^7 = 0.986 V/m or N/C

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nichrome wire
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i/A = vd*n*e = 7*10^-5*9*10^28*1.6*10^-19 = 1.00*10^6

rho is the resistivity of nichrome = 100*10^-8 ohm-m

then E= rho*(i/A) = 100*10^-8*1*10^6 = 1 V/m

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