Question

An electron of kinetic energy 1.20 keV circles in a plane perpendicular to a uniform magnetic field. The orbit radius is 25.0 cm.

(a)Find the electron’s speed.

(b)Find the magnetic field magnitude.

(c)Find the circling frequency.

(d)Find the period of the motion.

Short answer

a) Speed of electron is.$2.05\times {10}^7\text{\hspace{0.17em}m/s}$

b) Magnitude of the magnetic field is .$4.67\times {10}^{-4}T$

c) Circling frequency is .$1.31\times {10}^{7}Hz$

d) Period of the motion is .$7.63\times {10}^{-8}s$$7.63\times {10}^{-8}s$

Step-by-step solution

Listing the given quantities

• Kinetic energyis$K=1.20\text{\hspace{0.17em}keV}=1.20\times {10}^3\times 1.6\times {10}^{-19}\text{\hspace{0.17em}J}$
• Orbit radius is$r=25.0\text{cm}=0.25\text{m}$

 Step 2: Understanding the concept of kinetic energy and magnetic field

We use the formula for kinetic energy to find speed. Then, we can find the frequency and period from speed and radius. To find the magnetic field, we have to use the formula for magnetic force and centrifugal force.

(a)Calculation of the speed of the electron 

The formula for kinetic energy is as follows:

$K=0.5\times m{v}^2$

Substituting the values in the above expression, and we get,

$\begin{array}{c}1.20\times {10}^3\times 1.6\times {10}^{-19}=0.5\times 9.11\times {10}^{-31}\times v^2\\ v=2.05\times {10}^7\text{\hspace{0.17em}m/s}\end{array}$

Speed of electron is $2.05\times {10}^7\text{\hspace{0.17em}m/s}$.

Step 4:(b)Calculation of the magnetic field

Now the magnetic force is balanced by centrifugal force so:

$\begin{array}{c}{F}_m=F_c\\ qvB=\frac{m{v}^2}{r}\\ B=\frac{mv}{qr}\end{array}$

Substituting the values in the above expression, and we get,

$\begin{array}{c}B=\frac{9.11\times {10}^{-31}\times 2.05\times {10}^7}{1.6\times {10}^{-19}\times 0.25}\\ =4.67\times {10}^{-4}T\end{array}$

The magnitude of the magnetic field is$4.67\times {10}^{-4}T$ .

(c) Calculation of the circling frequency

Now the frequency is as follows:

$\begin{array}{l}v=r\omega \\ v=r\times 2\pi \times f\\ f=\frac{v}{2\pi r}\end{array}$

Substituting the values in the above expression, and we get,

$\begin{array}{c}f=\frac{2.05\times {10}^7}{2\pi \times 0.25}\\ =1.31\times {10}^7\text{\hspace{0.17em}Hz}\end{array}$

Circling frequency is $1.31\times {10}^{7}Hz$.

(d) Calculation of the period of rotation

The period of rotation can be calculated as:

$T=\frac{1}{f}$

Substituting the values in the above expression, and we get,

$\begin{array}{c}T=\frac{1}{1.31\times {10}^7}\\ =7.63\times {10}^{-8}s\end{array}$

The period of the motion is $7.63\times {10}^{-8}s$.