Question

A charged particle of mass m having kinetic energy K passes undeflected through a region with electric field \(\vec{E}\) and magnetic field \(\vec{B}\) acting perpendicular to each other. The mass m of the particle will be:

• A. \(\frac{KB^2}{2E^2}\)
• B. \(\frac{2KB^2}{E^2}\)
• C. \(\frac{2KE^2}{B^2}\)
• D. \(\frac{KE^2}{2B^2}\)

Hint:

This principle is used in a "Velocity Selector." Only particles with the specific speed \(v = E/B\) will pass through the fields without curving.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For a particle to pass undeflected in crossed fields, the electric force must be balanced by the magnetic force (\(qE = qvB\)). This determines the velocity of the particle. We then relate this velocity to kinetic energy and mass.

Step 2: Key Formula or Approach:
1. Velocity selector condition: \(v = E/B\). 2. Kinetic Energy: \(K = \frac{1}{2}mv^2\).

Step 3: Detailed Explanation:
1. From the undeflected condition: \(v = \frac{E}{B}\). 2. Substitute \(v\) into the kinetic energy formula: \[ K = \frac{1}{2} m \left( \frac{E}{B} \right)^2 \] \[ K = \frac{m E^2}{2 B^2} \] 3. Solve for mass \(m\): \[ m = \frac{2 K B^2}{E^2} \]

Step 4: Final Answer:
The mass \(m\) is \(\frac{2KB^2}{E^2}\).