Question

Two particles of charges in the ratio \( 1:2 \) and masses in the ratio \( 2:3 \) moving along a straight line enter a uniform magnetic field at right angles to the direction of the field. If the radii of the circular paths of the particles in the magnetic field are in the ratio \( 3:4 \), then the ratio of the initial linear momenta of the two particles is:

• A. \( 1:1 \)
• B. \( 3:4 \)
• C. \( 3:8 \)
• D. \( 2:3 \)

Hint:

The linear momentum of a particle moving in a magnetic field depends on its charge, mass, and the radius of the circular path. The product of the charge-to-mass ratio and the radius determines the momentum.

Answer 1

The Correct Option is C

Step 1: The magnetic force on a charged particle moving in a circular path is given by: \[ F = \frac{mv^2}{r} \] where: - \( m \) is the mass of the particle,
- \( v \) is its velocity,
- \( r \) is the radius of the circular path.
The force experienced by the particle is also equal to the magnetic force: \[ F = qvB \] where:
- \( q \) is the charge,
- \( B \) is the magnetic field strength.
Equating the two expressions for force gives: \[ \frac{mv^2}{r} = qvB \] Step 2: Solving for the linear momentum \( p = mv \), we get: \[ p = \frac{qrB}{v} \] Since the velocity is proportional to the radius and the charge, the ratio of linear momenta of the two particles is proportional to the product of the charge and mass ratio. Step 3: Let the charge and mass ratio of the two particles be \( \frac{q_1}{q_2} = \frac{1}{2} \) and \( \frac{m_1}{m_2} = \frac{2}{3} \). The ratio of the linear momenta is given by: \[ \frac{p_1}{p_2} = \frac{q_1 m_1 r_1}{q_2 m_2 r_2} \] \[ \frac{p_1}{p_2} = \frac{1/2 \times 2/3 \times 3}{4} = \frac{3}{8} \]