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

To solve this problem, we need to determine the mass of a charged particle that passes undeflected through regions of both electric and magnetic fields acting perpendicular to each other. Given that the kinetic energy of the particle is \( K \), we must use the relationship between the velocity of the particle, the electric field \( \vec{E} \), and the magnetic field \( \vec{B} \). We know:

• For a particle to pass undeflected, the electric force must equal the magnetic force:

\(qE = qvB\)

Where \( q \) is the charge of the particle, \( v \) is the velocity of the particle. This implies:

\(v = \frac{E}{B}\)

• The kinetic energy \( K \) is given by:

\(K = \frac{1}{2}mv^2\)

Substitute the expression for velocity:

\(K = \frac{1}{2}m\left(\frac{E}{B}\right)^2\)

Simplify to find the mass \( m \):

• Multiply through by 2 to eliminate the fraction:

\(2K = m\frac{E^2}{B^2}\)

• Rearrange to solve for \( m \):

\(m = \frac{2KB^2}{E^2}\)

Thus, the correct option is:

• \(\frac{2KB^2}{E^2}\)

This verifies the chosen option as outlined in the question.