Physics
A positively charged particle of mass 5.60 108 kg is traveling due east with a speed of 60 m/s and enters a 0.49T uniform magnetic field. The particle moves through onequarter of a circle in a time of 1.20 103 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

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.posted by bobpursley
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