For proton r=qB2m(KE) So q∝m(KE) Hence 2ee41KE=(4mp)(KE)(mp)(1MeV)=4KE1MeV=1MeV
A proton carrying 1MeV kinetic energy is moving in a circular path of radius R in uniform magnetic field. What should be the energy of an α-particle to describe a circle of same radius in the same field?
Held on 30 Apr 2012 · Verified 9 Jul 2026.
2MeV
1MeV
0.5MeV
4MeV
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