Answer:
I = 1.05x10⁻³ A
Explanation:
By definition, an electric current is the rate of charge flow at a given time:
[tex] I = \frac{q}{t} [/tex]
Where:
q: is the electrons charge = 1.602x10⁻¹⁹ C
t: is the time
In a circular motion, the time is given by:
[tex] t = T = \frac{2\pi}{\omega} = \frac{2\pi}{v/r} = \frac{2\pi r}{v} [/tex]
Where:
ω: is the angular speed = v/r
v: is the speed = 2.19x10⁶ m/s
r: is the radius = 5.29x10⁻¹¹ m
[tex] t = \frac{2\pi r}{v} = \frac{2\pi 5.29 \cdot 10^{-11} m}{2.19 \cdot 10^{6} m/s} = 1.52 \cdot 10^{-16} s [/tex]
Now, the effective current is:
[tex] I = \frac{q}{t} = \frac{1.602 \cdot 10^{-19} C}{ 1.52 \cdot 10^{-16} s} = 1.05 \cdot 10^{-3} A [/tex]
Therefore, the effective current associated with this orbiting electron is 1.05x10⁻³ A.
I hope it helps you!
