A proton is fired with a speed of 2×106 m/s at an angle of 60o to the X- axis. If a uniform magnetic field of 0.1T is applied along the y-axis, then what is the force acting on the proton ?

1.2×10−14 N 2.7×10−14 N 3.3×10−14 N 4.3×10−14 N

2.7×10−14 N

Magnetic field is along Y-axis. So, θ=60o = angle between velocity & magnetic field F = q (v X B) = qvB Sinθ