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 of mass \( m \) and charge \( q \) - Initial velocity \( \vec{v} = v \hat{i} \) (i.e., along x-direction) - Uniform electric field \( \vec{E} = E \hat{j} \) (i.e., along y-direction) As the electric field is perpendicular to the initial velocity, it exerts a continuous force \( \vec{F} = q \vec{E} \) on the particle in the y-direction. This force causes acceleration in the y-direction while the velocity remains in the x-direction. Thus, the particle experiences a centripetal force that curves its path, similar to uniform circular motion in the x–y plane. Using centripetal force: \[ F = \frac{mv^2}{r} \quad \text{and} \quad F = qE \Rightarrow \frac{mv^2}{r} = qE \Rightarrow r = \frac{mv^2}{qE} \] Hence, the particle moves in a circular path of radius: \[ r = \frac{mv^2}{qE} \] And since both the velocity and electric field lie in the x–y plane, the circular motion also occurs in the x–y plane.