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=?

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.

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be

that would be because of both gravity and friction. so     c