The radius of curvature of the path of a charged particle in a unifor…

The correct answer is Option (2) → momentum of the particle

A charged particle moving perpendicular to a uniform magnetic field moves in a circle, because the magnetic force provides the centripetal force:

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

So \( r \) is directly proportional to the momentum \( p \) of the particle.

Option (1) charge of the particle — Incorrect. \( r \propto 1/q \); a larger charge gives a smaller radius.

Option (2) momentum of the particle — Correct. \( r = p/(qB) \), so \( r \propto p \).

Option (3) kinetic energy of the particle — Incorrect. \( r = \sqrt{2mK}/(qB) \) varies as \( \sqrt{K} \), not directly as \( K \).

Option (4) flux density of the field — Incorrect. \( r \propto 1/B \); a stronger field gives a smaller radius.

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,

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

which gives

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

for the radius of the circle described by the charged particle. The larger the momentum, the larger is the radius and bigger the circle described.”