First slide

Force on a charged particle moving in a magnetic field

Question

A proton is moving in a circular path with kinetic energy 8 MeV in a uniform magnetic field. What will be the kinetic energy of an $α -$ particle if it moves in a circular path of same radius in the same magnetic field?

Moderate

A

16 MeV
B

32 MeV
C

8 MeV
D

2 MeV

Solution

$KE = \frac{1}{2} {mv}^{2} = \frac{1}{2} {mω}^{2} r^{2} = \frac{1}{2} m {(\frac{QB}{m})}^{2} r^{2} = \frac{Q^{2} B^{2} r^{2}}{2 m}$
$∴ \frac{{(KE)}_{α}}{{(KE)}_{p}} = \frac{{(2 e)}^{2}}{{(e)}^{2}} \times (\frac{m_{p}}{4 m_{p}}) = 1 \Rightarrow {(KE)}_{α} = 8 MeV$

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