Physics
posted by Don .
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

Physics 
bobpursley
I thought I answered this.
you know velocity as 60m/s,but it is also a quarter circle in the given time
velocity=60=PI*radius/(2*timegiven)
solve that for radius.
a. centripetal force=mv^2/r and you know v, m, and r.
but magnetic force has to be the same as centripetal force, so you know magnetic force.
b. F=Bqv=centripetal force, solve for q.
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