Answer
Magnetic force, $ \vec{F}=q (\vec{V} \times \vec{B}) $
Magnitude of centripetal force $\frac{mV^2}{r}$
Magnitude of Linear momentum $p= mv$
Kinetic energy $K=\frac{1}{2}mV^2$
Work Step by Step
Magnetic force $ \vec{F}=q (\vec{V} \times \vec{B}) $
Magnitude = F=qVB
Now, centripetal force $\frac{mV^2}{r}= qVB$
Therefore $mV= qBr=P$ (This is magnitude of linear momentum, P)
From here we can get the radius as well $r=\frac{mV}{qB}$
Kinetic energy, $K=\frac{1}{2}mV^2=\frac{m^2V^2}{2m}=\frac{P^2}{2m}$
We get the answers by putting the known values(please see the documents attached)
