A charged particle moves along a circle under the action of constant…

The correct answer is Option (3) → \( E = 0,\ B \neq 0 \)

A charge can move in a circle at constant speed only if it feels a force that is always perpendicular to its velocity and constant in magnitude. The magnetic force \( \vec{F} = q(\vec{v}\times\vec{B}) \) is exactly such a force: it is perpendicular to \( \vec{v} \), does no work, and so it only turns the particle. When \( \vec{v} \) is perpendicular to \( \vec{B} \) it acts as the centripetal force.

    \( \displaystyle qvB = \frac{mv^{2}}{r} \Rightarrow r = \frac{mv}{qB} \)

So a circle needs a magnetic field alone, with no electric field.

Why the other options are incorrect

• Option (1) \( E = 0,\ B = 0 \) — Incorrect. With no field there is no force, so the particle moves in a straight line.

• Option (2) \( E \neq 0,\ B = 0 \) — Incorrect. A constant electric force accelerates the particle along a fixed direction, giving a straight line or a parabola, never a circle.

• Option (4) \( E \neq 0,\ B \neq 0 \) — Incorrect. The electric force is constant and does work, so the speed changes and the path becomes a helix or a cycloid-like curve, not a circle.

Note: The same question was also asked in CUET (UG) 2026 on 26-05-2026 (Slot 1).

NCERT (2026-27):
Motion in a Magnetic Field – “… If the velocity v is perpendicular to the magnetic field B, the magnetic force is perpendicular to both v and B and acts like a centripetal force. It has a magnitude q v B. Equating the two expressions for centripetal force, m v²/r = q v B, which gives

    \( \displaystyle r = \frac{mv}{qB} \qquad (4.5) \)

for the radius of the circle described by the charged particle.”