Question

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?

Answer 1

Velocity of proton= 4782 m/s

Step-by-step 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