Question

(A) 2.5x10-10 N
(B) 7.6
A proton of mass m and charge te
should be the energy of a - particle (
radius
(A) 1 MeV
(B) 4 M
A deutron of kinetic energy 50 keV is
magnetic field B. The kinetic energy​

Answer 1

Explanation:

For a charged particle orbiting in a circular path in a magnetic field

rmv2=Bqv⇒v=mBqr

or, mv2=Bqvr

Also,

EK=21mv2=21Bqvr=Bq2r.mBqr=2mB2q2r2

For deuteron, E1=2×2mB2q2r2

For proton, E2=2mB2q2r2

E2E1=21⇒E250keV