A deuteron of kinetic energy 50 keV is describing a circular orbit of radius 0.5 m in a plane perpendicular to magnetic field B the kinetic energy of the proton that describes a circular orbit of radius 0.5m and in the same plane with the same magnetic field B is?
Answers
Solution:-
When charged particles move on circular path. The force F on the charged particle due to magnetic field provides the required centripetal force (=mv^2/r ) necessary for motion along the circular path.
So,
Where,
- m = mass of particle
- v = velocity of particle
- q = charge on particle
- B = external magnetic field
- r = radius of circular path
▶Kinetic Energy:-
= Ek = 1/2 mv^2 = 1/2Bqvr
= Bq × r/2 (Bqr/m) B^2q^2r^2/ 2m
▶For deutron :-
E1 = (B^2 q^2 r^2)/ 2×2m [mass = 2m]
▶For proton :-
E2 = (B^2 q^2 r^2)/ 2m [mass = m]
⟹ E1/E2 = 1/2
⟹ 50 keV/E2 = 1/2
⟹ E2 = 100 keV
∴ The kinetic Energy of the proton is 100 keV !
Answer:
Solution:-
When charged particles move on circular path. The force F on the charged particle due to magnetic field provides the required centripetal force (=mv^2/r ) necessary for motion along the circular path.
So,
{\huge{\boxed{\mathcal{{\frac{m {v}^{2}}{r} = qvB}}}}}
r
mv
2
=qvB
Where,
m = mass of particle
v = velocity of particle
q = charge on particle
B = external magnetic field
r = radius of circular path
▶Kinetic Energy:-
= Ek = 1/2 mv^2 = 1/2Bqvr
= Bq × r/2 (Bqr/m) B^2q^2r^2/ 2m
▶For deutron :-
E1 = (B^2 q^2 r^2)/ 2×2m [mass = 2m]
▶For proton :-
E2 = (B^2 q^2 r^2)/ 2m [mass = m]
⟹ E1/E2 = 1/2
⟹ 50 keV/E2 = 1/2
⟹ E2 = 100 keV
∴ The kinetic Energy of the proton is 100 keV !