A proton enters a uniform magnetic field of 0.2 T with speed \( v \). It describes a semicircular path of radius 5 cm and comes out of the field. The final speed of the proton will be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( v \)
The magnetic force on a moving charge is always perpendicular to its velocity, so it does no work on it. By the work-energy theorem the kinetic energy does not change.
\( \displaystyle \Delta K = W = 0 \;\Rightarrow\; \tfrac{1}{2}mv_f^{2} = \tfrac{1}{2}mv^{2} \;\Rightarrow\; v_f = v \)
The field only turns the proton along the semicircle; the numbers \( 0.2\,\mathrm{T} \) and \( 5\,\mathrm{cm} \) are not needed.
NCERT (2026-27):
Motion in a Magnetic Field – “… In the case of motion of a charge in a magnetic field, the magnetic force is perpendicular to the velocity of the particle. So no work is done and no change in the magnitude of the velocity is produced (though the direction of momentum may be changed).”