Moving Charges and Magnetism - Result Question 2
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:
(a) $0.5 MeV$
(b) $1.5 MeV$
(c) $1 MeV$
(d) $4 MeV$
[2015 RS]
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Answer:
Correct Answer: 2. (c)
Solution:
- (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
$=\frac{q^{2}}{m}=\frac{(2)^{2}}{4}=1 MeV$
