A positively charged particle of mass 5.60 10-8 kg is traveling due east with a speed of 60 m/s and enters a 0.49-T uniform magnetic field. The particle moves through one-quarter of a circle in a time of 1.20 10-3 s, at which time it leaves the field heading due south. All during the motion the particle moves perpendicular to the magnetic field.

Question

A positively charged particle of mass 5.60  10-8 kg is traveling due east with a speed of 60 m/s and enters a 0.49-T uniform magnetic field. The particle moves through one-quarter of a circle in a time of 1.20  10-3 s, at which time it leaves the field heading due south. All during the motion the particle moves perpendicular to the magnetic field.

(a) What is the magnitude of the magnetic force acting on the particle? 
1  N

(b) Determine the magnitude of its charge. 
2  C

bobpursley · Answer

the path of the particle has length PI/2 * radius.

velocity of the particle: PI*radius/2timegiven=60m/s solve for r

Knowing the velocity as that, solve for r.

centripetal force=mv^2/r

then magnetic force, Bqv, solve for that as it equals centripetal force.
then solve for q.