A beam of protons traveling at 1.20 km/s enters a uniform magnetic field, traveling perpendicular to the field. The beam exits the magnetic field, leaving the field in a direction perpendicular to its original direction. The beam travels a distance of 1.18 cm while in the field. what is the magnitude of the magnetic field?
Curved path length (0.0118m) = 1/4 circumference = 0.25 x 2πR
0.50πR = 0.0118 .. .. R = 0.0075m
Centripetal force (mv²/R) is provided by the magnetic force (Bqv)
mv²/R = Bqv .. .. (m=mass of proton, q=proton charge))
B = mv / qR .. .. (1.67^-27kg x 1.20^3m/s) / (1.60^-19C x 0.0075m) .. .. ►B = 1.67^-3 T