Moving Charges and Magnetism - Result Question 2

«««< HEAD:content/english/neet-pyq-chapterwise/physics/moving-charges-and-magnetism/moving-charges-and-magnetism-result-question-2.md

2. A proton and an alpha particle both enter a region of uniform magnetic field $B$, moving at right angles to field $B$. If the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton is $1 MeV$ the energy acquired by the alpha particle will be:

======= ####2. A proton and an alpha particle both enter a region of uniform magnetic field $B$, moving at right angles to field $B$. If the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton is $1 MeV$ the energy acquired by the alpha particle will be:

3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed:content/english/neet-pyq-chapterwise/physics/moving-charges-and-magnetism/moving-charges-and-magnetism—result-question-2.md (a) $0.5 MeV$

(b) $1.5 MeV$

(c) $1 MeV$

(d) $4 MeV$

[2015 RS]

Show Answer

Answer:

Correct Answer: 2. (c)

Solution:

  1. (c) As we know, $F=q v B=\frac{m v^{2}}{R}$

$\therefore R=\frac{m v}{q B}=\frac{\sqrt{2 m(k E)}}{q B}$

Since $R$ is same so, $KE \propto \frac{q^{2}}{m}$

Therefore KE of $\alpha$ particle

$=\frac{q^{2}}{m}=\frac{(2)^{2}}{4}=1 MeV$