First slide

Force on a charged particle moving in a magnetic field

Question

A proton of mass m and charge +e is moving in a circular orbit in a magnetic field with energy 1 Me V. What should be the energy of $α - particle$ (mass = 4m and charge = + 2e), so that it can revolve in the path of same radius?

Moderate

A

1 MeV
B

4 MeV
C

2 MeV
D

0.5 MeV

Solution

$\begin{array}{l} r = \frac{\sqrt{2 mK}}{qB} \Rightarrow K \propto \frac{q^{2}}{m} \Rightarrow \frac{K_{p}}{K_{α}} = {(\frac{q_{p}}{q_{α}})}^{2} \times \frac{m_{α}}{m_{p}} \\ \Rightarrow \frac{1}{K_{α}} = {(\frac{q_{p}}{2 q_{p}})}^{2} \times \frac{4 m_{p}}{m_{p}} = 1 \Rightarrow K_{α} = 1 MeV \end{array}$

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