Question

A particle of mass \( m \) and charge \( q \) moving with velocity \( \vec{v} = v \hat{i} \) is subjected to a uniform electric field \( \vec{E} = E \hat{j} \). The particle will initially have a tendency to move in a circle of radius:

• A. \( \left( \dfrac{mv}{qE} \right) \) in x–y plane
• B. \( \left( \dfrac{mv^2}{qE^2} \right) \) in x–z plane
• C. \( \left( \dfrac{mv^2}{qE} \right) \) in x–y plane
• D. \( \left( \dfrac{mv}{qE^2} \right) \) in y–z plane

Hint:

When a charged particle moves perpendicular to an electric field, it undergoes circular motion. Equate centripetal force \( \frac{mv^2}{r} \) with electric force \( qE \) to find the radius.

Answer 1

The Correct Option is C

Given: - A charged particle with mass \( m \) and charge \( q \). - Initial velocity \( \vec{v} = v \hat{i} \) (along the x-direction). - Uniform electric field \( \vec{E} = E \hat{j} \) (along the y-direction). Since the electric field is perpendicular to the initial velocity, it applies a continuous force \( \vec{F} = q \vec{E} \) on the particle in the y-direction. This force results in acceleration in the y-direction, while the velocity component in the x-direction is maintained. Consequently, the particle is subjected to a centripetal force, causing its trajectory to curve in a manner analogous to uniform circular motion within the x–y plane. Applying the centripetal force equation: \[ F = \frac{mv^2}{r} \quad \text{and} \quad F = qE \] Equating these forces: \[ \frac{mv^2}{r} = qE \] Solving for the radius \( r \): \[ r = \frac{mv^2}{qE} \] Therefore, the particle follows a circular path with a radius: \[ r = \frac{mv^2}{qE} \] As both the velocity vector and the electric field vector lie within the x–y plane, the resulting circular motion also occurs within this plane.