Chapter 30, Problem 26SP

To determine

The motion of the electron inside a magnetic field of field strength of $2.0\text{ mT}$ in +x direction if the initial velocity of electron is $5.0\times {10}^6\text{m/s}$ at an angle of $20°$ to the $+x-\text{axis}$.

Answer to Problem 26SP

Solution:

Electron will follow helical path of $\text{0}\text{.49 cm}$ radius and $\text{8}\text{.4 cm}$ pitch.

Explanation of Solution

Given data:

The speed of the electron is $\text{5}\text{.0}\times {\text{10}}^{\text{6}}\text{ m/s}$ at an angle of $20°$ to the $+x-\text{axis}$.

The magnitude of the magnetic field acting on the electron is $\text{2}\text{.0}\times {\text{10}}^{\text{-3}}\text{ T}$ in the $+x-\text{direction}$.

Formula used:

The radius of a charge moving along a circular path 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 charge, $v$ is the velocity of the charge, $q$ is the magnitude of charge, and $B$ is the magnitude of the magnetic field acting on the charge.

The period of each revolution of a charge travelling in a circular path is given as

$\text{Period}=\frac{2\pi r}{v_{\perp }}$

Here, $r$ is the radius of the circular path and $v_{\perp }$ is the perpendicular component of the velocity.

The horizontal (x) distance travelled by the charge during a given time (also known as pitch)is given as

$\text{Pitch}=v_{\left|\right|}\text{(Period)}$

Here, $v_{\left|\right|}$ is the parallel component of the velocity.

Explanation:

As the electron is moving at an angle with respect to the horizontal (+x) direction, its vertical and horizontal components shall be determined as follows:

The horizontal (parallel) component of velocity will be:

$v_{\left|\right|}=v\cos \theta$

Here, $v_{\left|\right|}$ is the parallel component of the velocity and $\theta$ is the angle with respect to the horizontal (+x) direction.

Similarly, the vertical (perpendicular) component of velocity will be:

$v_{\perp }=v\sin \theta$

Here, $v_{\perp }$ is the perpendicular component of the velocity and $\theta$ is the angle with respect to the horizontal (+x) direction.

The radius of the electron moving along a circular path in a magnetic field is given as

$\begin{array}{c}r=\frac{m{v}_{\perp }}{qB}\\ =\frac{mv\sin \theta }{qB}\end{array}$

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

Substitute $\text{9}\text{.1}\times {\text{10}}^{\text{-31}}\text{ kg}$ for $m$, $\text{5}\text{.0}\times {\text{10}}^{\text{6}}\text{ m/s}$ for $v$, $20°$ for $\theta$, $\text{1}\text{.6}\times {\text{10}}^{\text{-19}}\text{ C}$ for $q$, and $\text{2}\text{.0}\times {\text{10}}^{\text{-3 }}\text{T}$ for $B$

$\begin{array}{c}r=\frac{\left(\text{9}\text{.1}\times {\text{10}}^{-\text{31}}\text{ kg}\right)\left(\text{5}\text{.0}\times {\text{10}}^{\text{6}}\text{ m/s}\right)\left(\sin 20°\right)}{\left(\text{1}\text{.6}\times {\text{10}}^{-\text{19}}\text{ C}\right)\left(\text{2}\text{.0}\times {\text{10}}^{-\text{3 }}\text{ T}\right)}\\ \text{= }\left(\text{4}{\text{.86×10}}^{-\text{3 }}\text{ m}\right)\left(\frac{100\text{ cm}}{1\text{ m}}\right)\\ =\text{ 0}\text{.49 cm}\end{array}$

The period of each revolution of the electron travelling in a circular path is given as

$\begin{array}{l}\text{Period}=\frac{2\pi r}{v_{\perp }}\\ =\frac{2\pi r}{v\sin \theta }\end{array}$

Substitute $\text{4}\text{.86}\times {\text{10}}^{\text{-3}}\text{ m}$ for $r$, $\text{5}\text{.0}\times {\text{10}}^{\text{6}}\text{ m/s}$ for $v$, and $20°$ for $\theta$

$\begin{array}{c}\text{Period}=\frac{2\pi \left(\text{4}\text{.86}\times {\text{10}}^{\_\text{3}}\text{ m}\right)}{\left(\text{5}\text{.0}\times {\text{10}}^{\text{6}}\text{ m/s}\right)\sin \left(20°\right)}\\ =\text{ 17}{\text{.84×10}}^{\text{-9}}\text{ s}\end{array}$

The horizontal (x) distance travelled by the electron during the given time (also known as pitch) is given as follows:

$\begin{array}{c}\text{Pitch}=v_{\left|\right|}\left(\text{Period}\right)\\ =v\cos \theta \left(\text{Period}\right)\end{array}$

Here, $v_{\left|\right|}$ is the parallel component of the velocity.

Substitute $\text{17}\text{.84}\times {\text{10}}^{\text{-9}}\text{ s}$ for $\text{period}$, $\text{5}\text{.0}\times {\text{10}}^{\text{6}}\text{ m/s}$ for $v$, and $20°$ for $\theta$

$\begin{array}{c}\text{Pitch}=\left(\text{5}\text{.0}\times {\text{10}}^{\text{6}}\text{ m/s}\right)\left(\cos 20°\right)\left(\text{17}\text{.84}\times {\text{10}}^{\_\text{9}}\text{ s}\right)\\ =\left(\text{83}\text{.82}\times {\text{10}}^{-\text{3}}\text{ m}\right)\left(\frac{100\text{ cm}}{1\text{ m}}\right)\\ =\text{ 8}\text{.4 cm}\end{array}$

Conclusion:

The motion of the electron inside a magnetic fieldwill be a helix having a radius of $\text{0}\text{.49 cm}$ and a pitch of $\text{8}\text{.4 cm}$.