Answer:
[tex]2.4\times 10^{12}[/tex]
Explanation:
We are given that
Speed of light,v=[tex]2.998\times 10^8 m/s[/tex]
Diameter of ring,d=92 m
Radius,r=[tex]\frac{d}{2}=\frac{92}{2}=46 m[/tex]
Current, I=0.40 A
We have to find the number of electrons in the beam.
We know that
Current,I=[tex]\frac{q}{t}[/tex]
Where q= ne
[tex]e=1.6\times 10^{-19} C[/tex]
Using the formula
[tex]0.40=\frac{1.6\times 10^{-19}n}{\frac{2\pi r}{v}}[/tex]
[tex]0.40=\frac{1.6\times 10^{-19}n\times v}{2\pi r}[/tex]
[tex]0.40=\frac{1.6\times 10^{-19}n\times 2.998\times 10^8}{2\pi\times 46}[/tex]
[tex]n=\frac{0.40\times 2\pi\times 46}{1.6\times 10^{-19}\times 2.998\times 10^8}=2.4\times 10^{12}[/tex]