PLEASE PLEASE help me with physics..?

Question

Find the radius of a proton's orbit when it moves perpendicular to a magnetic field of 0.33 T with a speed of 6.16 105 m/s.

__cm

Answer 1

Remember the Lorentz force equation:

F=q (vXB)

in this case, the cross product is just,

F=q v B,

q=1.602 x 10^-19C

B=0.33T

v=6.16 x 10^5 m/s (I guess this is your velocity, it's hard to tell)

F=(1.602 x 10^-19C)*(0.33T)*(6.16 x 10^5 m/s)

F=3.26 × 10^-14 N

Hope this helps =)