A) 1 MeV
B) 2 MeV
C) 4 MeV
D) 0.5 MeV
Correct Answer: A
Solution :
When a charged particle moves in a circular path in a magnetic field, then magnetic force provides the centripetal force to the particle. i.e., \[Bqv=\frac{m{{v}^{2}}}{r}\] or \[mv=Bqr\] or \[p=Bqr\] Kinetic energy of proton \[=\frac{{{p}^{2}}}{2m}\] or \[KE=\frac{{{B}^{2}}{{q}^{2}}{{r}^{2}}}{2\,m}\] or \[KE\propto \frac{1}{m}\] (for same circular path). As mass of deuteron is twice that of proton, hence it should have half the energy of proton, i.e., 1 MeV.
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