A particle with a charge of 0.08 C is moving at right angles to a uniform magnetic field with a strength of 0.8 T. The velocity of the charge is 800 m/s.
What is the magnitude of the magnetic force exerted on the particle?
Answer is in N
F = Q ( v cross B)
cross is the cross product in general v B sin theta
where theta is the angle between v and B
since here it is 90 degrees, sin theta = 1
and it is just
F =Q v B = .08 * 800 * 0.8 Newtons