First slide

Force on a charged particle moving in a magnetic field

Question

A deuteron of kinetic energy 50 keV is describing a circular orbit of radius 0.5 metre in a plane perpendicular to magnetic field $\vec{B}$ . The kinetic energy of the proton that describes a circular orbit of radius 0.5 metre in the same plane with the same $\vec{B}$ , is

Moderate

A

25 keV
B

50 keV
C

200 keV
D

100 keV

Solution

$\begin{array}{l} r = \frac{\sqrt{2 mK}}{qB} \Rightarrow K \propto \frac{q^{2}}{m} \\ \Rightarrow \frac{K_{p}}{K_{d}} = {(\frac{q_{p}}{q_{d}})}^{2} \times \frac{m_{d}}{m_{p}} = {(\frac{1}{1})}^{2} \times \frac{2}{1} = \frac{2}{1} \\ \Rightarrow K_{p} = 2 \times 50 = 100 keV \end{array}$

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