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A proton carrying 1 MeV kinetic energy i...

A proton carrying `1 MeV` kinetic energy is moving in a circular path of radius `R` in unifrom magentic field. What should be the energy of an `alpha-` particle to describe a circle of the same radius in the same field?

A

2 MeV

B

1 MeV

C

0.5 MeV

D

4 MeV

Text Solution

Verified by Experts

The correct Answer is:
B
According to the principal of circular motion in a magnetic field ,
`F_c=F_mimplies(mv^2)/R=qVBimpliesR=(mv)/(qB) =P/(qB)=sqrt(2m.k)/(qB)`
`R_alpha=sqrt(2(4m)K.)/(2qB)" "R/R_alpha=sqrt(K/K)`
but `R = R_alpha ` (given)
Thus `K = K. = 1 MeV`
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