Please help with 234 A charged particle of mass m 46 times

Please help with #2,3,4 :)

A charged particle of mass m - 4.6 times 10^-8 kg, moving with constant velocity in the y-direction enters a region containing a constant magnetic field B = 2.3T aligned with the positive z-axis as shown. The particle enters the region at (x, y) - (0.79 m, 0) and leaves the region at (x, y) - 0. 0.79 m a time t - 412 Mus after it entered the region. With what speed v did the particle enter the region containing the magnetic field? What is F_x. the x-component of the force on the particle at a time t_1 = 137.3 Mus after it entered the region containing the magnetic field. What is F_y. the y-component of the force on the particle at a time t_1 = 137.3 Mus after it entered the region containing the magnetic field. What is q. the charge of the particle? Be sure to include the correct sign.

Solution

given that, m = 4.6*10^-8 kg,

                  B = 2.3 T,

                   r = 0.79 m,

                   t = 412 µs

1)

in this case, period is, t = (2*pi*r/4) / v
so the velocity is,

       v =2*pi*r/(4t) = 3010.4 m/s
2)

The angle turned is ,

    @ = (pi/2)*t1/t = 0.52347 rad

thus, the x-component of force F is,

         Fx = -(mv^2/r)cos@ = -0.457 N
3)

the y-component of force F is,

      Fy = -(mv^2/r)sin@ = -0.26378 N = -0.264 N
4)

           q = mv/(Br) =7.62*10^-5 C
by the right hand rule q is negative, so q = -7.62*10^-5 C