The Lorentz force acting on the proton is equal to:
[tex]F=qvB \sin \theta[/tex]
where
q is the proton charge
v is the proton speed
B is the intensity of the magnetic field
[tex]\theta[/tex] is the angle between the direction of v and B
since the proton travels perpendicularly to the magnetic field, in this case [tex]\theta=90^{\circ}[/tex], so the Lorentz force in this case is simply
[tex]F=qvB[/tex]
The magnetic force provides also the centripetal force that keeps the proton in circular motion:
[tex]m \frac{v^2}{r}=qvB [/tex]
where
m is the proton mas
r is the radius of the orbit
If we re-arrange this equation and we use the data of the problem, we can find the radius of the orbit:
[tex]r= \frac{mv}{qB}= \frac{(1.67 \cdot 10^{-27} kg)(2.0 \cdot 10^7 m/s)}{(1.6 \cdot 10^{-19} C)(0.1 T)} =2.09 m [/tex]
