Kerala KEAM 5 June 2024 Solved Paper

The radius of the circular path of a charged particle moving in a uniform magnetic field is given by the formula: $$r = \frac{mv}{qB}$$where:- $r$ is the radius of the circular path,- $m$ is the mass of the particle,- $v$ is the velocity of the particle,- $q$ is the charge of the particle,- $B$ is the magnetic field strength.For a proton, the kinetic energy $K_e$ is related to its velocity by: $$K_e = \frac{1}{2} m v^2$$For a deuteron, the mass $m_d$ is twice the mass of the proton $m_p$, i.e., $m_d = 2 m_p$, and the charge $q_d$ is twice the charge of the proton, i.e., $q_d = 2 q_p$.Since both particles are moving in the same circular path, we can equate their radii: $$r_p = r_d$$Thus: $$\frac{m_p v_p}{q_p B} = \frac{m_d v_d}{q_d B}$$Simplifying, we get: $$\frac{m_p v_p}{q_p} = \frac{2 m_p v_d}{2 q_p}$$ $$v_p = v_d$$Therefore, the velocities of the proton and deuteron are the same. Now, using the relation between kinetic energy and velocity for both particles: $$K_e = \frac{1}{2} m v^2$$For the proton, we know the kinetic energy is 2 MeV: $$2 \, MeV = \frac{1}{2} m_p v_p^2$$For the deuteron, the kinetic energy $K_d$ is: $$K_d = \frac{1}{2} m_d v_d^2 = \frac{1}{2} (2 m_p) v_p^2 = 2 \times \frac{1}{2} m_p v_p^2 = 2 \times 2 \, MeV = 1 \, MeV$$Thus, the correct answer is option (B), 1 MeV. Quick Tip: For particles of different masses but the same radius in a magnetic field, their kinetic energies are proportional to their masses. The velocity is the same for both particles in the same magnetic field and radius.