In the circuit shown in the diagram below, two thick wires connect a 1.5 volt ba
ID: 2140524 • Letter: I
Question
In the circuit shown in the diagram below, two thick wires connect a 1.5 volt battery to a Nichrome (NiCr) wire. Each thick connecting wire is 14 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 3 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?
1 V/m
What is the magnitude of the electric field in the thin Nichrome wire?
2 V/m
Explanation / Answer
Total length of copper wire Lc = 2 * 14 cm
= 0.28 m
radius of copper wire rc = 9 mm
= 0.9*10^-3 m
Hence resistance of copper wire RC = ?C * Lc / ? * rc2
= 1.68 * 10-8 * 0.28 / 3.14 * (0.9*10^-3)2
= 1.8494* 10-5 ?
here ?C = resistivity of copper = 1.68 * 10-8 ? - m
Total length of nichrome wire Ln = 6 cm
= 0.06 m
radius of nichrome wire rn = 3 mm
=3 * 10-3 m
Hence resistance of nichrome wire Rn = ?n * Ln / ? * rn2
= 1.10 10-6 * 0.06 / 3.14 * (3 * 10-3)2
=2.335* 10-3 ?
here ?n = resistivity of nichrome = 1.10 * 10-6 ? - m
Net resistance R = Rc + Rn
= 1.8494 * 10-5 + 2.335 * 10-3
= 2.3534* 10-3 ?
Current I = V / R
= 1.5/ 2.3534 * 10-3
= 0.6373757117362114* 103 A
If n be electron concentration, e charge of electron, ? be electron mobility, then
electric field inside wire E = V / L = I * R / L
a. For copper EL = I * Rc / Lc
= 0.6373757117362114* 103 * 1.8494 * 10-5 / 0.28
= 4.2075 * 10-2 V/m
b. For nichrome En = I * Rn / Ln
= 0.6737 * 103 * 2.335 * 10-3 / 0.06
= 26.218 V/m