Question

An electron of kinetic energy 1.59 keV circles in a plane perpendicular to a uniform magnetic field. The orbit radius is 35.4 cm. Find (a) the electron's speed, (b) the magnetic field magnitude, (c) the circling frequency, and (d) the period of the motion.

Answer 1

a)

b)

c)

d)

Step-by-step explanation:

a) We can find the electron's speed by knowing the kinetic energy:

Where:

K: is the kinetic energy = 1.59 keV

m: is the electron's mass = 9.11x10⁻³¹ kg

v: is the speed =?

b) The electron's speed can be found by using Lorentz's equation:

(1)

Where:

F: is the magnetic force

q: is the electron's charge = 1.6x10⁻¹⁹ C

θ: is the angle between the speed of the electron and the magnetic field = 90°

The magnetic force is also equal to:

(2)

By equating equation (2) with (1) and by solving for B, we have:

c) The circling frequency is:

Where:

T: is the period = 2π/ω

ω: is the angular speed = v/r

d) The period of the motion is:

I hope it helps you!