Question

Two particles of masses \( m_1 = m \), \( m_2 = 2m \) and charges \( q_1 = q \), \( q_2 = 2q \) entered into uniform magnetic field. Find \( \frac{F_1}{F_2} \) (force ratio).

• A. \( \frac{1}{2} \)
• B. 1
• C. \( \frac{1}{3} \)
• D. 2

Hint:

In uniform magnetic fields, force on a charged particle is \( F = qvB \). Compare charges and velocities carefully to find ratios.

Answer 1

The Correct Option is A

Magnetic force: \( F = qvB \) Assume both particles enter perpendicular to the magnetic field and with equal kinetic energy (for fairness in ratio), then: \[ \text{Using } KE = \frac{1}{2}mv^2 \Rightarrow v \propto \frac{1}{\sqrt{m}} \Rightarrow v_1 = \frac{1}{\sqrt{m}}, \quad v_2 = \frac{1}{\sqrt{2m}} \] So, \[ F_1 = q_1 v_1 B = q \cdot \frac{1}{\sqrt{m}} \cdot B, \quad F_2 = q_2 v_2 B = 2q \cdot \frac{1}{\sqrt{2m}} \cdot B \] \[ \frac{F_1}{F_2} = \frac{q / \sqrt{m}}{2q / \sqrt{2m}} = \frac{1}{2} \cdot \sqrt{2} = \frac{1}{\sqrt{2}} \approx 0.707 \] But if we assume equal velocity, then: \[ F_1 = qvB, \quad F_2 = 2qvB \Rightarrow \frac{F_1}{F_2} = \frac{q}{2q} = \frac{1}{2} \] Thus, assuming same entry velocity, the force ratio is \( \frac{1}{2} \).