Question

The emf of the battery is 1.5 V. In Nichrome there are 9 × 1028 mobile electrons per m3, and the mobility of mobile electrons is 7 × 10−5 (m/s)/(N/C). Each thick wire has length 29 cm = 0.29 m and cross-sectional area 9 × 10−8 m2. The thin wire has length 6 cm = 0.06 m and cross-sectional area 1.3 × 10−8 m2. (The total length of the three wires is 64 cm.) In the steady state, calculate the number of electrons entering the thin wire every second. Do not make any approximations, and do not use Ohm's law or series-resistance equations.

Answer 1

Answer:

The number of electrons entering the thin wire every second is 1.75 x 10⁻³ mobile electrons / second

Explanation:

Given;

emf of the battery, V = 1.5 V

electron density, = 9 × 10²⁸ mobile electrons per m³

mobility of electron, u = 7 × 10⁻⁵ (m/s)/(N/C)

length of thin wire, L = 6 cm = 0.06 m

cross sectional area of the thin wire, A = 1.3 x 10⁻⁸ m²

The magnitude of the electric field in the thin wire is given by;

E = V/L

E = (1.5) / (0.06)

E = 25 N/C

the number of electrons entering the thin wire every second is given by;

$e/s = mobility \ x \ Electric \ field\\\\number \ of \ electrons \ per \ second =\frac{7*10^{-5} (m/s)}{N/C} *25 (N/C)\\\\number \ of \ electrons \ per \ second = 1.75*10^{-3} \ m/s$

Therefore, the number of electrons entering the thin wire every second is 1.75 x 10⁻³ mobile electrons / second