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 α-particle to describe a circle of same radius in the same field?

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 α-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

Tardigrade Admin · Accepted Answer

For a charged particle's motion in a magnetic field

F_{C} = F_{m} \Rightarrow \frac{m v ^{2}}{R} = q V B

\Rightarrow R_{p} = \frac{m _{p} v _{p}}{q _{p} B} = \frac{P _{p}}{q _{p} B} = \frac{m K _{p}}{qB}

R_{α} = \frac{2 ( 4 m ) K _{α}}{2 qB}

\frac{R _{p}}{R _{α}} = \frac{K _{p}}{K _{α}}

but

R_{p} = R_{α}

(given)
Thus

K_{p} = K_{α} = 1 M e V

Q. 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?

2 MeV

1 MeV

0.5 MeV

4 MeV

For a charged particle's motion in a magnetic field FC​=Fm​⇒Rmv2​=qVB ⇒Rp​=qp​Bmp​vp​​=qp​BPp​​=qBmKp​​​ Rα​=2qB2(4m)Kα​​​ Rα​Rp​​=Kα​Kp​​​ but Rp​=Rα​ (given) Thus Kp​=Kα​=1MeV

Q. A proton carrying $1 M e V$ 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?