Answer:
n = 3.1x10¹²
Explanation:
To find the number of electrons we need to find first the charge (q):
[tex] I = \frac{q}{\Delta t} \rightarrow q = I*\Delta t [/tex] (1)
Where:
I: is the electric current = 0.59 A
t: is the time
The time t is equal to:
[tex]v = \frac{\Delta x}{\Delta t} \rightarrow \Delta t = \frac{\Delta x}{v}[/tex] (2)
Where:
x: is the displacement
v: is the average speed = 2.998x10⁸ m/s
The displacement is equal to the perimeter of the circumference:
[tex] \Delta x = 2\pi*r = \pi*d [/tex] (3)
Where d is the diameter = 80.0 m
By entering equations (2) and (3) into (1) we have:
[tex]q = I*\Delta t = I*\frac{\Delta x}{v} = \frac{I\pi d}{v} = \frac{0.59 A*\pi*80.0 m}{2.99 \cdot 10^{8} m/s} = 4.96 \cdot 10^{-7} C[/tex]
Now, the number of electrons (n) is given by:
[tex] n = \frac{q}{e} [/tex]
Where e is the electron's charge = 1.6x10⁻¹⁹ C
[tex] n = \frac{q}{e} = \frac{4.96 \cdot 10^{-7} C}{1.6 \cdot 10^{-19} C} = 3.1 \cdot 10^{12} [/tex]
Therefore, the number of electrons in the beam is 3.1x10¹².
I hope it helps you!