Question

A beam of 30 MeV Alpha particles is to be obtained from a cyclotron of radius 50 cm then the strength of the magnetic field required to be applied will be

Answer 1

kinetic energy of alpha particle, K = 30MeV

we know, kinetic energy , K = 1/2 mv²

or, 2Km = (mv)²

or, mv = √{2Km}.........(1)

now, at equilibrium

centripetal force = magnetic force

or, mv²/r = Bqv

or, mv/r = Bq

or, B = mv/rq

from equation (1), we get B = √{2Km}/rq

here, K = 30MeV = 30 × 10^6 × 1.6 × 10^-19 J = 48 × 10^-13 J

mass of alpha particle , m = 6.645 × 10^-27 Kg

r = 50 cm = 0.5m

q = 2 × 1.6 × 10^-19 C = 3.2 × 10^-19 C

so, B = √{2 × 48 × 10^-13 × 6.645 × 10^-27}/(0.5 × 3.2 × 10^-19 )

= 1.5785 T

hence, magnetic field is 1.5785T

Answer 2

Answer:

the above-mentioned answer is correct