JEE Mains · Physics · STD 12 - Moving charges and magnetism
- JEE Mains 2018
- Medium
- MCQ
An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits of radii \(r_e,r_p\) and \({r_\alpha }\) respectively in a uniform magnetic field \(B\). The relation between \(r_e,r_p\) and \(\;{r_\alpha }\) is
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(r_e < r_p\) \(=\) \(\;{r_\alpha }\)
Step-by-step Solution
Detailed explanation
As we know, radius of circular path in magnetic field \(\mathrm{r}=\frac{\sqrt{2 \mathrm{Km}}}{\mathrm{qB}}\)
For electron, \(\mathrm{r}_{\mathrm{e}}=\frac{\sqrt{2 \mathrm{Km}_{\mathrm{e}}}}{\mathrm{eB}}.........(i)\)
For proton, \(\mathrm{r}_{\mathrm{p}}=\frac{\sqrt{2 \mathrm{Km}_{\mathrm{p}}}}{\mathrm{eB}}.........(ii)\)
For \(\alpha\) particle,
\(\mathrm{r}_{\alpha}=\frac{\sqrt{2 \mathrm{Km}_{\mathrm{a}}}}{\mathrm{q}_{\alpha} \mathrm{B}}=\frac{\sqrt{2 \mathrm{K} 4 \mathrm{m}_{\mathrm{p}}}}{2 \mathrm{eB}}=\frac{\sqrt{2 \mathrm{Km}_{\mathrm{p}}}}{\mathrm{eB}} .........(iii)\)
\(\mathrm{r}_{\mathrm{e}}<\mathrm{r}_{\mathrm{p}}=\mathrm{r}_{\alpha} \quad\left(\because \mathrm{m}_{\mathrm{e}}<\mathrm{m}_{\mathrm{p}}\right)\)
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