A proton enters a constant magnetic field of magnitude 0.050 t and traverses a semicircle of radius 1.0 mm before leaving the field. what is the proton's speed?

Velocity of proton = 4782 m/s

Explanation:

For proton, the centripetal force required for circular motion is provided by the magnetic force,

so Fm = Fc

q v B = m v²/r

m = mass of proton

v = velocity

B = magnetic field=0.05 T

q = charge = 1.6 x 10⁻¹⁹

r = radius = 1 mm = 0.001 m

v = q B r/m

V = (1.6 x 10⁻¹⁹) (0.05) (0.001) / (1.673 x 10⁻²⁷)

V = 4782 m/s