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

Answer:

a) $v = 2.36 \cdot 10^{7} m/s$

b) $B = 3.80 \cdot 10^{-4} T$

c) $f = 1.06 \cdot 10^{7} Hz$

d) $T = 9.43 \cdot 10^{-8} s$

Explanation:

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

$K = \frac{1}{2}mv^{2}$

Where:    

K: is the kinetic energy = 1.59 keV

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

v: is the speed =?

$v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2*1.59 \cdot 10^{3} eV*\frac{1.602 \cdot 10^{-19} J}{1 eV}}{9.11 \cdot 10^{-31} kg}} = 2.36 \cdot 10^{7} m/s$

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

$F = q(v\times B) = qvBsin(\theta)$   (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:

$F = ma_{c} = m\frac{v^{2}}{r}$   (2)

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

$B = \frac{mv}{rq} = \frac{9.11 \cdot 10^{-31} kg*2.36 \cdot 10^{7} m/s}{0.354 m*1.6 \cdot 10^{-19} C} = 3.80 \cdot 10^{-4} T$

c) The circling frequency is:

$f = \frac{1}{T} = \frac{\omega}{2\pi} = \frac{v}{2\pi r}$

Where:

T: is the period = 2π/ω

ω: is the angular speed = v/r

$f = \frac{v}{2\pi r} = \frac{2.36 \cdot 10^{7} m/s}{2\pi*0.354 m} = 1.06 \cdot 10^{7} Hz$

d) The period of the motion is:

$T = \frac{1}{f} = \frac{1}{1.06 \cdot 10^{7} Hz} = 9.43 \cdot 10^{-8} s$

I hope it helps you!