In the circuit shown in the diagram below, two thick wires connect a 1.3 volt ba
ID: 3894648 • Letter: I
Question
In the circuit shown in the diagram below, two thick wires connect a 1.3 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 6 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?
V/m
What is the magnitude of the electric field in the thin Nichrome wire?
V/m
Explanation / Answer
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* e28 (where e is exponential)
5. e' = charge on one electron
Now
Ec * 2 *(14/1000) + En * (8/1000) = 1.6 [ Since Voltage = Ed]
=> 28 Ec + 8En = 1600 => 14Ec + 4 En = 800 --------------------- 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 ----------- eqn 2
Solving 1 & 2
Ec = 2.03 V/m
En = 192.88 V/m