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A particle of charge $q$ and mass $m$ has momentum (magnitude) $p = mv$ and kinetic energy $K = \frac{1}{2}m v^2 = \frac{p^2}{2m}$. If the particle moves in a circular orbit of radius $r$ perpendicular to a uniform magnetic field of magnitude $B$, show that:

• $p = Bqr$
• $K = \frac{B^2q^2r^2}{2m}$
• The angular momentum magnitude $L = Bqr^2$.