Answer:
2.28 cm
Explanation:
Radius of the proton if v and B is perpendicular will be,
[tex]r=\frac{mv}{qB}[/tex]
Now,
mass of proton, [tex]m=1.67\times10^{-27} kg[/tex].
Radius of the orbit, [tex]r=1.7\times10^{6} m/s[/tex].
Magnetic field, [tex]B=0.8 T[/tex].
Charge on the proton, [tex]q=1.6\times10^{-19} C[/tex].
Now substitute all the variables in r.
[tex]r=\frac{1.67\times10^{-27} kg(1.7\times10^{6} m/s) }{0.8 T(1.6\times10^{-19} C) }[/tex].
[tex]r=2.28 \times10^{-2} m\\r=2.28 cm[/tex].
Therefore the radius of the orbit of moving proton is 2.28 cm.
