Magnetic Field Physics in Circular motion (i'm over complicating it)?

Question

[b]1. The problem statement, all variables and given/known data[/b]

A positively charged particle of mass 7.40 10-8 kg is traveling due east with a speed of 60 m/s and enters a 0.38 T uniform magnetic field. The particle moves through one-quarter of a circle in a time of 4.40 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?
N

(b) What is the magnitude of the charge of the particle?
C

[b]2. Relevant equations[/b]
v = 2* pi * r / T
F_c_ = mv^2 / r

Circumference = 2 * pi * r
B = F / (q*v*sin theta)

[b]3. The attempt at a solution[/b]

First I found r by using the first equation to be 58 m, but since it only went 1/4 of the way i took the the circumference (2 * 58 * 3.14) = 364.24 * .25 (1/4 of the circle was travelled) and then got lost

but then also tried this way which i really thought was right:

Answer 1

Figure out the acceleration. (the particle is in uniform circular motion). Then use Newton's second law to figure out the force. Finally, use F = qv x B to find the charge.

EDIT:
Your radius is incorrect. Sorry, I didn't check it before my first reply. You used the right formula but didn't solve for r correctly.
F_c = mv^2 / r : this formula is actually F = ma in disguise. The magnitude of the acceleration of an object traveling in a circle at constant speed is v^2/r. You can get the force directly from F_c = mv^2/r.

Answer 2

So, would it be ((7.4 e- 8 kg ) (60 m/s)) / 4.4e-3s (F=ma)

which gives 1.01e-3 N
then take 1.01e-3N = q * 60 m/s * .38T
therefore q = 4.43e-5?

If not then can you be more detailed in the guidance? Thanks!