CUET Preparation Today
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 Since, the force is perpendicular to the direction of motion, the work done by the magnetic force is zero. |
