Moving Charges and Magnetism - Result Question 26

«««< HEAD:content/english/neet-pyq-chapterwise/physics/moving-charges-and-magnetism/moving-charges-and-magnetism-result-question-26.md

26. A deuteron of kinetic energy $50 keV$ is describing a circular orbit of radius 0.5 metre in a plane perpendicular to the magnetic field B. The kinetic energy of the proton that describes a circular orbit of radius 0.5 metre in the same plane with the same $B$ is

======= ####26. A deuteron of kinetic energy $50 keV$ is describing a circular orbit of radius 0.5 metre in a plane perpendicular to the magnetic field B. The kinetic energy of the proton that describes a circular orbit of radius 0.5 metre in the same plane with the same $B$ is

3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed:content/english/neet-pyq-chapterwise/physics/moving-charges-and-magnetism/moving-charges-and-magnetism—result-question-26.md (a) $25 keV$

(b) $50 keV$

(c) $200 keV$

(d) $100 keV$

[1991]

Show Answer

Answer:

Correct Answer: 26. (d)

Solution:

  1. (d) For a charged particle orbiting in a circular path in a magnetic field

$\frac{m v^{2}}{r}=B q v \Rightarrow v=\frac{B q r}{m}$

or, $m v^{2}=B q v r$

Also,

$E_K=\frac{1}{2} m v^{2}=\frac{1}{2} B q v r=B q \frac{r}{2} \cdot \frac{B q r}{m}=\frac{B^{2} q^{2} r^{2}}{2 m}$

For deuteron, $E_1=\frac{B^{2} q^{2} r^{2}}{2 \times 2 m}$

For proton, $E_2=\frac{B^{2} q^{2} r^{2}}{2 m}$

$\frac{E_1}{E_2}=\frac{1}{2} \Rightarrow \frac{50 keV}{E_2}=\frac{1}{2} \Rightarrow E_2=100 keV$