AP Physics B?

[hannahbannah]

help. I think what I'm screwing up is sin theta. when using F=qvbsin(theta)

In a nuclear research laboratory, a proton moves in a particle accelerator through a magnetic field of intensity 0.283 T at a speed of 3.41E7 m/s. If the proton is moving perpendicular to the field, what force acts on it? answer in N.
If the proton continues to move in a direction that is consistently perpendicular to the field, what is the radius of curvature of its path? (mass of proton = 1.67E-27 kg). Answer in M.

[Ben l]

since it is moving perpendicular to the field, sin(theta) = 1.
So F = qvB = (1.602E-19C)*(3.41E7m/s)*(0.283T) = 1.55E-12 N.

You know that a proton moving perpendicular to a magnetic field follows a circular path right? (Since the force is always perpendicular to the velocity and thus is a centripetal force). also, mv^2/R = F. (Circular Motion)

So R = mv^2/F = (1.67E-27kg)*(3.41E7m/s)^2/(1.55E-12N)

R = 1.25m.

Please redo the calculations yourself, it's hard computing such small/large digits on the windows calculator!