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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;
[tex]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[/tex]
Therefore, the number of electrons entering the thin wire every second is 1.75 x 10⁻³ mobile electrons / second
The number of electrons entering the thin wire every second is 1.75 x 10⁻³ mobile electrons / second
Calculation of the number of electrons:
Since
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
So here the magnitude should be
E = V/L
E = (1.5) / (0.06)
E = 25 N/C
Now the number of electrons should be
= 7 × 10⁻⁵ *25
= 1.75 x 10⁻³ mobile
hence, The number of electrons entering the thin wire every second is 1.75 x 10⁻³ mobile electrons / second
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