Question

A proton initially moving west at speed of 1.00*10^6 m/s in a uniform magnetic field

of magnitude 0.300 T directed vertically upward. Describe in detail the proton's trajectory,

including its shape and orientation.

Answer 1

speed of the proton v = 1*10^6 m/s
The magnetic field B = 0.300 T
The magnetic force produce the centripetal force
so     F = mv^2/r
          qvB = mv^2/r
Therefore the radius of the path
            r = mv/qB
              = (1.67*10^-27)(1*10^6) / (1.6*10^-19 )(0.300)
              = 0.034 m

When the charged particle in the magnetic filed B directed out of the that plane. As a result a magnetic force FB = qv*B continuously deflects the particle and because v and B are perpendicular to each other, this deflection causes the particles to follow in circular path