A proton of charge $1.6 \times 10^{-19}$ C is moving with a velocity of $5 \times 10^3$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
Since, the force is perpendicular to the direction of motion, the work done by the magnetic force is zero.