An electron enters a region of space containing a uniform 0.0000213-T magnetic field. Its speed is 127 m/s and it enters perpendicularly to the field. As you know, the electron undergoes circular motion under these conditions. Find the radius of the electron/'s orbit, r, and the frequency of the motion, f r=? f=?

Answer:a) r = 3.37*10^-5m, b) v = 599782.72 Hz.

Explanation: when an electron enters a uniform magnetic field, the force exerted on it by the magnetic field is given as qvB. This same force is responsible for it centripetal force which is given mv²/r.

By relating both equation, we have that

qvB = mv²/r

m cancels out each other on both sides of the equation, hence we have that

qB = mv/r.

Where q = magnitude of electronic charge = 1.609*10^19 c, B = strength of magnetic field = 0.0000213 T, m = mass of electronic charge = 9.11*10 ^-31 kg, v = linear velocity = 127 m/s.

By substituting parameters we have that

1.609*10^-19 * 0.0000213 = 9.11*10^-31 * 127/r

By cross multiplication

1.609*10^-19 * 0.0000213 * r = 9.11*10^-31 * 127

r = (9.11*10^-31 * 127) / (1.609*10^-19 * 0.0000213)

r = 1.156*10^-28 / 3.427*10^-24

r = 3.37*10^-5m

b) v = ωr where ω = angular frequency.

127 = ω * 3.37*10^-5

ω = 127 / 3.37*10^-5

ω = 3768545.99 rad/s.

But ω = 2πv where v = frequency.

v = ω/2π

v = 376845.99/2π

v = 599782.72 Hz.