Physics, asked by nidashah29, 27 days ago

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

Answered by Csilla
59

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}}}}}

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 !

Answered by jitendra999
2

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 !

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