A proton of charge \( 1.6\times10^{-19} \) C is moving with a velocity of \( 5\times10^{3} \) m/s in a magnetic field of 4 T. The work done by the magnetic field on the proton is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → zero
The magnetic force on a moving charge, \( \vec{F} = q\,(\vec{v}\times\vec{B}) \), is always perpendicular to the velocity. The work done by a force is \( W = \vec{F}\cdot\vec{d} = Fd\cos90^\circ = 0 \), since the displacement is along \( \vec{v} \).
The kinetic energy of the proton therefore does not change; the field only changes the direction of its motion. The values of \( q \), \( v \) and \( B \) are not needed for the answer.
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).”