Answer:
Speed of the proton is [tex]v=9.31\times 10^6\ m/s[/tex]
Explanation:
It is given that,
Magnetic field, B = 1.35 T
Radius of circular path, r = 0.072 m
The centripetal force is balanced by the magnetic force such that,
[tex]\dfrac{mv^2}{r}=qv[/tex]
[tex]v=\dfrac{qrB}{m}[/tex]
q and m is the charge and mass of a proton respectively
[tex]v=\dfrac{1.6\times 10^{-19}\times 0.072\times 1.35}{1.67\times 10^{-27}}[/tex]
[tex]v=9.31\times 10^6\ m/s[/tex]
So, the speed if the proton is [tex]9.31\times 10^6\ m/s[/tex]. Hence, this is the required solution.
