What is the magnetic force acting on a proton (charge $+1.6 × 10^{-19}C$) moving at an angle of 30° across a magnetic field of $5.3 × 10^{-3} T$ at a speed of $3.0 × 10^5 m s^{-1}$?

$1.3 × 10^{-16} N$

The correct answer is Option (2) → $1.3 × 10^{-16} N$

According to Lorentz force,

$F=qvB\sin θ$

where,

q = Charge of particle = $+1.6×10^{-19}C$

v = Velocity of proton = $3×10^5m/s$

B = Magnetic field = $5.3×10^{-3}T$

θ = Angle = 30°

$F=(1.6×10^{-19})×(3×10^5)×(5.3×10^{-3})×\sin 30°$

$=1.27× 10^{-16} N$