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:

zero $8 \times 10^{-16}$ J $8 \times 10^{16}$ J $3.2 \times 10^{-16}$ J

zero

The correct answer is Option (1) → zero