Q186 / 240 · KCET 2026 · Physics
A proton, an electron and an $\alpha$-particle enter at right angles to a uniform magnetic field with the same velocity. If $R_p, R_e$ and $R_\alpha$ are the radii of circular paths of these particles, then
- A$R_\alpha = R_p = R_e$
- Correct answerB$R_\alpha > R_p > R_e$
- C$R_\alpha < R_p < R_e$
- D$R_\alpha > R_p = R_e$
Solution
The radius of the circular path of a charged particle in a uniform magnetic field is given by $R = \dfrac{mv}{qB}$.
Since the velocity $v$ and magnetic field $B$ are the same for all particles, $R \propto \dfrac{m}{q}$.
For a proton: $R_p \propto \dfrac{m_p}{e}$
For an electron: $R_e \propto \dfrac{m_e}{e}$
For an $\alpha$-particle: $R_\alpha \propto \dfrac{m_\alpha}{q_\alpha} = \dfrac{4m_p}{2e} = \dfrac{2m_p}{e}$
Since $m_e < m_p$, we have $\dfrac{m_e}{e} < \dfrac{m_p}{e} < \dfrac{2m_p}{e}$.
Therefore, $R_e < R_p < R_\alpha$, which can be written as $R_\alpha > R_p > R_e$.
Answer: $R_\alpha > R_p > R_e$
Since the velocity $v$ and magnetic field $B$ are the same for all particles, $R \propto \dfrac{m}{q}$.
For a proton: $R_p \propto \dfrac{m_p}{e}$
For an electron: $R_e \propto \dfrac{m_e}{e}$
For an $\alpha$-particle: $R_\alpha \propto \dfrac{m_\alpha}{q_\alpha} = \dfrac{4m_p}{2e} = \dfrac{2m_p}{e}$
Since $m_e < m_p$, we have $\dfrac{m_e}{e} < \dfrac{m_p}{e} < \dfrac{2m_p}{e}$.
Therefore, $R_e < R_p < R_\alpha$, which can be written as $R_\alpha > R_p > R_e$.
Answer: $R_\alpha > R_p > R_e$
Hence, the correct answer is option: 'B'.