Question

A proton with kinetic energy of 2 MeV is describing a circular path of radius \( R \) in a uniform magnetic field. The kinetic energy of the deuteron to describe the same circular path in the same field is:

• A. 0.5 MeV
• B. 1 MeV
• C. 2 MeV
• D. 4 MeV
• E. 0.25 MeV

Hint:

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.

Answer 1

The Correct Option is B

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.