Answer:
The diameter is [tex]d = 0.00123 \ m[/tex]
Explanation:
From the question we are told that
The number of electrons is [tex]N = 1.40 *10^{16} \ electrons[/tex]
The time taken is [tex]t = 290 \mu s = 290 *10^{-6} \ s[/tex]
The drift speed is [tex]v = 7.*10^{-4} \ m/s[/tex]
Generally the number of electrons that flows through the cross-sectional area at the given time t is mathematically represented as
[tex]N= n * A * v_d * t[/tex]
Here n is the number of conduction electrons present in and the value is
[tex]n = 5.8*10^{28} / m^3[/tex]
A is the cross-sectional area of the wire
So
[tex]1.4*10^{16} = 5.80 *10^{28} * A * 7*10^{-4} * 290 *10^{-6}[/tex]
=> [tex]A = 1.189 *10^{-6} \ m^2[/tex]
Gnerally the area is mathematically represented as
[tex]A = \pi* \frac{d^2}{4}[/tex]
=> [tex]1.189 *10^{-6} = \pi* \frac{d^2}{4}[/tex]
=> [tex]1.189 *10^{-6} = \pi* \frac{d^2}{4}[/tex]
=> [tex]d = \sqrt{1.51369 *10^{-6}}[/tex]
=> [tex]d = 0.00123 \ m[/tex]
The diameter of the silver wire in which the given number of electrons flows through is 1.22 × 10⁻³m
Given the data in the question;
From the expression of drift velocity, the number of electrons passing in through an area can be expressed:
[tex]N_e = n Avt[/tex]
Where [tex]N_e[/tex] is the Number of electron, n is the conduction electron density in silver( [tex]5.8*10^{28} electrons/m^3[/tex] ), A is the cross-sectional area, v is drift velocity or speed and t is the time.
We know that, Area; [tex]A = \pi r^2[/tex]
So
[tex]N_e = n (\pi r^2) vt[/tex]
We make "r", the subject of the formula
[tex]N_e = n (\pi r^2) vt\\\\r = \sqrt{\frac{N_e}{\pi nvt} }[/tex]
We substitute our values into the equation
[tex]r = \sqrt{\frac{1.4*10^{16}}{\pi *(5.8*10^{28}m^{-3})*(7.10*10^{-4}m/s)*(290*10^{-6}s)} } \\\\r = \sqrt{3.73*10^{-7}m^2}\\\\r = 6.1* 10^{-4}m[/tex]
We know that, Diameter;[tex]d = 2*r[/tex]
So,
[tex]d = 2 * (6.1*10^{-4}m)\\\\d = 1.22 * 10^{-3}m[/tex]
Therefore, the diameter of the silver wire in which the given number of electrons flows through is 1.22 × 10⁻³m.
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