Question

A proton carrying $1 \,MeV$ kinetic energy is moving in a circular path of radius $R $ in uniform magnetic field. What should be the energy of an $\alpha$-particle to describe a circle of same radius in the same field?

• A. 2 MeV
• B. 1 MeV
• C. 0.5 MeV
• D. 4 MeV

Answer 1

Answer: (B)

For a charged particle's motion in a magnetic field
$F _{ C }= F _{ m } \Rightarrow \frac{ mv ^{2}}{ R }= qVB$
$\Rightarrow R _{ p }=\frac{ m _{ p } v _{ p }}{ q _{ p } B }=\frac{ P _{ p }}{ q _{ p } B }=\frac{\sqrt{ mK _{ p }}}{ qB }$
$R _{\alpha}=\frac{\sqrt{2(4 m ) K _{\alpha}}}{2 qB }$
$\frac{ R _{ p }}{ R _{\alpha}}=\sqrt{\frac{ K _{ p }}{ K _{\alpha}}}$
but $R _{ p }= R _{\alpha}$ (given)
Thus $K _{ p }= K _{\alpha}=1 MeV$