The magnetic force acting on the proton is
[tex]F=qvB \sin \theta[/tex]
where
q is the proton charge
v is its 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 is moving perpendicular to the magnetic field, [tex]\theta=90^{\circ}[/tex] and [tex]\sin \theta=1[/tex], so the force becomes
[tex]F=qvB[/tex]
this force provides the centripetal force that keeps the proton in circular motion:
[tex]m \frac{v^2}{r} = q v B [/tex]
where the term on the left is the centripetal force, with
m being the mass of the proton
r the radius of its orbit
Re-arranging the previous equation, we can find the radius of the proton's orbit:
[tex]r= \frac{mv}{qB}= \frac{(1.67 \cdot 10^{-27} kg)(6.5 m/s)}{(1.6 \cdot 10^{-19} C)(1.8 T)}=3.77 \cdot 10^{-8}m [/tex]
And now we can calculate the centripetal acceleration of the proton, which is given by
[tex]a_c = \frac{v^2}{r}= \frac{(6.5 m/s)^2}{3.77\cdot 10^{-8}m}=1.12 \cdot 10^9 m/s^2 [/tex]
