Answer:
The magnetic field strength of the proton is 0.026 Tesla.
Explanation:
It is given that,
Speed of the proton, [tex]v=0.25\times 10^7\ m/s[/tex]
The radius of circular path, r = 0.975 m
It is moving perpendicular to a magnetic field such that the magnetic force is balancing the centripetal force.
[tex]qvB\ sin90=\dfrac{mv^2}{r}[/tex]
[tex]B=\dfrac{mv}{qr}[/tex]
q = charge on proton
[tex]B=\dfrac{1.67\times 10^{-27}\ kg\times 0.25\times 10^7\ m/s}{1.6\times 10^{-19}\ C\times 0.975\ m}[/tex]
B = 0.026 Tesla
So, the magnetic field strength of the proton is 0.026 Tesla.
