A proton is projected horizontally from east to west in a uniform magnetic field of strength $3.0 \times 10^{-3} T$, with a speed of $2 \times 10^6 ms^{-1}$. The magnetic field exists in vertically upward direction, then the force on proton due to the magnetic field will be:

$9.6 \times 10^{-16} N$ towards north

The correct answer is Option (1) → $9.6 \times 10^{-16} N$ towards north

According to Lorentz force,

$F=q(\vec V×\vec B)$

$=qVB\sin θ$

$θ=90°,q=1.6×10^{-19}C,B=3×10^{-3}T$

$F=(1.6×10^{-19})(2×10^6)(3×10^{-3})$

$=9.6 \times 10^{-16} N$ towards north [given by $\vec V×\vec B$]