A proton moves in a circle of radius 2 cm in a plane perpendicular to a uniform magnetic field and takes a time \( T \) to complete one revolution. If the kinetic energy of the proton is made four times its earlier value, the radius of its circle and its time of revolution in the same field become, respectively,
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 4 cm and \( T \)
In a uniform magnetic field the magnetic force \( qvB \) provides the centripetal force. So the radius of the circle is \( r=mv/qB \). The radius grows with the momentum of the proton. The period of revolution \( T=2\pi m/qB \) does not depend on its speed or energy.
The kinetic energy is \( K=p^{2}/2m \), so:
\( \begin{aligned} r &= \frac{p}{qB} \\ &= \frac{\sqrt{2mK}}{qB} \end{aligned} \)
With \( m \), \( q \) and \( B \) unchanged, \( r\propto\sqrt{K} \). For four times the kinetic energy:
\( \begin{aligned} r' &= \sqrt{4}\times2\,\mathrm{cm} \\ &= 4\,\mathrm{cm} \end{aligned} \)
The faster proton goes round a larger circle in the same time, and its period stays \( T \).
NCERT (2026-27):
Motion in a Magnetic Field – “… Equating the two expressions for centripetal force, m v2/r = q v B, which gives
\( r=mv/qB \qquad (4.5) \)
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. If ω is the angular frequency, then v = ω r. So,
\( \omega=2\pi\nu=qB/m \qquad [4.6(a)] \)
which is independent of the velocity or energy.”