In the circuit shown in the diagram below, two thick wires connect a 1.5 volt ba
ID: 2140525 • 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
asssumotions:
1. There is no internal resistance in battery.
2. Electric Field in Copper wire Ec& Current Density = Jc
3. Electric Field in NiCr wire En& Current Density = Jn
4. I am assuming 9 e+28 = 9*10^28(where e is exponential)
5. e' = charge on one electron
Now
Ec* 2 *(14/1000) + En* (8/1000) = 1.5 [ Since Voltage = Ed]
=> 28 Ec + 8En = 1500
=>14Ec + 4 En = 750 --------------------- eqn 1
Since both wires are in series current will be same
Jn= nn* e' *? n* En
= 9 * e28* 7 * e-5* e' *En= 6.14 * 1011*e' * En
In=Jn*? * (3/1000)2? 1.74 * 107*e' * En
Also
Jc= nc* e' *?c*Ec= 8.36 * e28* e' * 0.00434 * Ec= 5.25 * 1010*e' *Ec
Ic= Jc*? * (10/1000)2? 1.65 * 109* e' * Ec
Now since Ic = In
=>1.65 * 109* e' * Ec=1.74 * 107*e' * En
=>165 *Ec= 1.74 *En
Ec = 0.0105 En
Solving 1 & 2
Ec= 1.96 V/m
En= 187.46 V/m