KCET 2026 Physics Solved Paper

Concept:For a charged particle moving perpendicular to a uniform magnetic field, the magnetic force acts as the centripetal force, giving the radius of the circular path.Explanation:The magnetic force is given by $q v B$, and the centripetal force is $\frac{m v^2}{R}$.Equating them:$$q v B = \frac{m v^2}{R}$$This gives the radius:$$R = \frac{m v}{q B}$$Since all three particles enter with the same velocity $v$ in the same magnetic field $B$, the radius depends only on the ratio $\frac{m}{q}$.For a proton, $m = m_p$ and $q = e$, so:$$R_p = \frac{m_p v}{e B}$$For an electron, $m = m_e$ and $q = e$, so:$$R_e = \frac{m_e v}{e B}$$The electron has the same charge magnitude as the proton but a much smaller mass, hence:$$R_e < R_p$$For an $\alpha$-particle, mass is about $4 m_p$ and charge is $2e$, so:$$R_\alpha = \frac{4 m_p v}{2 e B} = 2 \frac{m_p v}{e B} = 2 R_p$$Therefore, comparing the three radii:$$R_\alpha > R_p > R_e$$Answer:The correct option is:$$\text{B. } R_{\alpha}>R_{p}>R_{e}$$