Electron in magnetic field, quick fix

[pleasehelpme6]

An electron with speed of 3.7e5 m/s enters a uniform magnetic field of magnitude 0.042 T at an angle 39 degrees to magnetic field lines. The electron will follow a helical path.

Determine the radius of the helical path.

F = (mv^2)/r
F = vq * Bsin(39)
set F's equal and solve for r


r = (mv)/(qBsin(39))


I solved this using
m = 9.109e-31 kg
q = 1.602e-19 C

the answer i got was 7.906e-5 m
the correct answer is 3.1e-5 m

can someone show me what I am doing wrong?

 

[ehild]

The electron moves along a helical path. The projection of this path in the plane which is normal to B is a circle. The speed along this circle is not the same as the total speed of the electron.

ehild