F=qvBsinα=
=1.6•10⁻¹⁹•4•10⁶•0.5•sin60=
=2.77•10⁻¹³ N
a=F/m=2.77•10⁻¹³/1.67•10⁻²⁷=1.66•10¹⁴ m/s²
1.A proton travels at 4 x 10^6 m/s in a direction that is at an angle of 60 degrees to a magnetic field of 0.5 T. What is the magnitude of the force on the proton in Newtons?
2.What is the acceleration of the proton?
1 answer