Question

a proton moves at a speed of 2.0 x 10^7 m/s at right angles to a magnetic field that is directed into the page with a magnitude of 0.1T, Calculate the radius of the orbit

Answer 1

The Lorentz force acting on the proton is equal to:
$F=qvB \sin \theta$
where
q is the proton charge
v is the proton speed
B is the intensity of the magnetic field
$\theta$ is the angle between the direction of v and B

since the proton travels perpendicularly to the magnetic field, in this case $\theta=90^{\circ}$, so the Lorentz force in this case is simply
$F=qvB$

The magnetic force provides also the centripetal force that keeps the proton in circular motion:
$m \frac{v^2}{r}=qvB$
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:
$r= \frac{mv}{qB}= \frac{(1.67 \cdot 10^{-27} kg)(2.0 \cdot 10^7 m/s)}{(1.6 \cdot 10^{-19} C)(0.1 T)} =2.09 m$