Chapter 30, Problem 35SP

An electron is accelerated from rest through a potential difference of 800 V. It then moves perpendicularly to a magnetic field of 30 G. Find the radius of its orbit and its orbital frequency.

To determine

The radius and orbital frequency of the electronthat is accelerated from rest through a potential difference of $800\text{V}$ and in perpendicular to the magnetic field of $30\text{G}$.

Answer to Problem 35SP

Solution:

$\text{3}\text{.2 cm}$, $\text{84 MHz}$

Explanation of Solution

Given data:

The electric potential difference is $\text{800 V}$.

The magnitude of the magnetic field acting on the electron is $30\text{G}$.

Formula used:

The radius of a circular path of a charged particle moving along in a magnetic field is given as

$r=\frac{mv}{qB}$

Here, $r$ is the radius of the circular path, $m$ is the mass of the charged particle, $v$ is the velocity of the charged particle, $q$ is the magnitude of charge, and $B$ is the magnitude of the magnetic field acting on the charge.

The velocity of a charged particle moving due an electric potential difference is given as

$v=\sqrt{\frac{2qV}{m}}$

Here, $V$ is the electric potential.

The expression for orbital frequency of an object is given as

$f=\frac{v}{2\pi r}$

Here, $f$ is the orbital frequency of the object.

Explanation:

Consider the expression for radius of the circular path of an electron:

$r=\frac{m_{e}v}{eB}$ …… (1)

Here, $v$ is the velocity of an electron, $e$ is the magnitude of an electron, $V$ is the electric potential, and $m_e$ is the mass of the electron.

Consider the expression for the velocity of an electronmoving due to an electric potential difference:

$v=\sqrt{\frac{2eV}{m_e}}$ …… (2)

From equation (1) and (2)

$\begin{array}{c}r=\frac{m_e}{eB}\left(\sqrt{\frac{2eV}{m_e}}\right)\\ =\frac{1}{B}\sqrt{\frac{2m_{e}V}{e}}\end{array}$

Keep in mind that $1\text{G}={10}^{-4}\text{ T}$, mass of an electron $m_e$ is $\text{9}\text{.1}\times {\text{10}}^{-\text{31}}\text{ kg}$, and charge on the electron $e$ is $\text{1}\text{.6}\times {\text{10}}^{-\text{19}}\text{ C}$.

Substitute $30\text{G}$ for $B$, $\text{9}\text{.1}\times {\text{10}}^{-\text{31}}\text{ kg}$ for $m_e$, $\text{800 V}$ for $V$, and $\text{1}\text{.6}\times {\text{10}}^{-\text{19}}\text{ C}$ for $e$

$\begin{array}{c}r=\frac{1}{30\text{G}}\sqrt{\frac{2\left(\text{9}\text{.1}\times {\text{10}}^{-\text{31}}\text{ kg}\right)\left(\text{800 V}\right)}{\left(\text{1}\text{.6}\times {\text{10}}^{-\text{19}}\text{ C}\right)}}\\ =\frac{1}{30\text{G}\left(\frac{{10}^{-4}\text{}\text{T}}{1\text{ G}}\right)}\sqrt{\frac{2\left(\text{9}\text{.1}\times {\text{10}}^{-\text{31}}\text{ kg}\right)\left(\text{800 V}\right)}{\left(\text{1}\text{.6}\times {\text{10}}^{-\text{19}}\text{ C}\right)}}\\ =\text{3}{\text{.18×10}}^{-\text{2}}\text{ m}\\ =\text{3}\text{.2 cm}\end{array}$

Consider the expression for the velocity of an electron moving due to an electric potential difference:

$v=\sqrt{\frac{2eV}{m_e}}$

Substitute $\text{9}\text{.1}\times {\text{10}}^{-\text{31}}\text{ kg}$ for $m_e$, $\text{800 V}$ for $V$, and $\text{1}\text{.6}\times {\text{10}}^{-\text{19}}\text{ C}$ for $e$

$\begin{array}{c}v=\sqrt{\frac{2\left(\text{1}\text{.6}\times {\text{10}}^{-\text{19}}\text{ C}\right)\left(\text{800 V}\right)}{\left(\text{9}\text{.1}\times {\text{10}}^{-\text{31}}\text{ kg}\right)}}\\ =\text{1}\text{.67}\times {\text{10}}^{\text{7}}\text{ m/s}\end{array}$

Consider the expression for the orbital frequency of the electron:

$f=\frac{v}{2\pi r}$

Here, $f$ is the orbital frequency of the electron, $v$ is the velocity of the electron, and $r$ is the radius of the circular path.

Substitute $\text{1}\text{.67}\times {\text{10}}^{\text{7}}\text{ m/s}$ for $v$ and $\text{3}{\text{.18×10}}^{-\text{2}}\text{ m}$ for $r$

$\begin{array}{c}f=\frac{\left(\text{1}\text{.67}\times {\text{10}}^{\text{7}}\text{ m/s}\right)}{2\pi \left(\text{3}{\text{.18×10}}^{-\text{2}}\text{ m}\right)}\\ =\text{83}\text{.62}\times {\text{10}}^{\text{6}}\text{ Hz}\\ =\text{84 MHz}\end{array}$

Conclusion:

Therefore, the radius and the orbital frequency of the electron orbit are $\text{3}\text{.2 cm}$ and $\text{84 MHz}$, respectively.