Moving Charges and Magnetism - Result Question 26
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
(a) $25 keV$
(b) $50 keV$
(c) $200 keV$
(d) $100 keV$
[1991]
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Answer:
Correct Answer: 26. (d)
Solution:
- (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, $E_2=\frac{B^{2} q^{2} r^{2}}{2 m}$
$\frac{E_1}{E_2}=\frac{1}{2} \Rightarrow \frac{50 keV}{E_2}=\frac{1}{2} \Rightarrow E_2=100 keV$
