The uniform $30mT$magnetic field in $FIGUREP29.65$points in the positive $Z$-direction. An electron enters the region of magnetic field with a speed of $5.0\times {10}^{6}m/s$and at an angle of ${30}^{\circ }$above the $xy$-plane. Find the radius $r$and the pitch $p$of the electron's spiral trajectory.

Since the electron moves in a circular manner, we call it cyclotron motion. The cyclotron motion is defined as a particle travelling in a uniform circular motion perpendicular to the magnetic field at a constant speed. The radius of the circular motion produced by the magnetic field is linked to it by equation in the form

$r_{\mathrm{cyc}}=\frac{m\overrightarrow{v}}{qB}$

Where localid="1648981926405" $q$denotes the particle's charge, localid="1648981702713" $v$denotes its speed, localid="1648981706252" $B$denotes the magnetic field, and mm denotes the particle's mass. Because the electron goes up at an angle oflocalid="1648981710448" $\theta ={30}^{\circ }$, equation will be true.

$r_{\mathrm{cyc}}=\frac{mv\cos \theta }{qB}$

To determine the radius, we input the values for localid="1648981718360" $m,v,q$and localid="1648981724670" $B$into equation.

$\begin{array}{l}{r}_{\mathrm{cyc}}=\frac{mv}{qB}=\frac{\left(9.1\times {10}^{-31}\mathrm{kg}\right)\left(5\times {10}^6\mathrm{m}/\mathrm{s}\right)\cos {30}^{\circ }}{\left(1.6\times {10}^{-19}\mathrm{C}\right)\left(30\times {10}^{-3}\mathrm{T}\right)}\\ r_{\mathrm{cyc}}=8.2\times {10}^{-4}\mathrm{m}.\end{array}$

In vertical velocity $v\sin \theta$, the pitch reflects the distance that is travelled in time $T$.

$p=(v\sin \theta )T$

For the vibration's period. The magnetic field causes the circular motion to create a circular frequency, which is described by an equation in the form

$f_{\mathrm{cyc}}=\frac{qB}{2\pi m}$

Now we plug in the proton's $B,m$and $q$values to get

$\begin{array}{l}{f}_{\mathrm{cyc}}=\frac{qB}{2\pi m}\\ f_{\mathrm{cyc}}=\frac{\left(1.6\times {10}^{-19}\mathrm{C}\right)\left(30\times {10}^{-3}\mathrm{T}\right)}{2\pi \left(9.1\times {10}^{-31}\mathrm{kg}\right)}\\ f_{\mathrm{cyc}}=8.4\times {10}^8{\mathrm{s}}^{-1}.\end{array}$

The reciprocal of frequency period is

$$\begin{gathered} T=\frac{1}{f_{\mathrm{cyc}}} \\ T=\frac{1}{8.4\times {10}^8{\mathrm{s}}^{-1}} \\ T=1.2\times {10}^{-9}\mathrm{s}. \end{gathered}$$

Substituting the values in equation to get pitch by

$\begin{array}{l}p=(v\sin \theta )T=\left(5\times {10}^6\mathrm{m}/\mathrm{s}\right)\sin {30}^{\circ }\left(1.2\times {10}^{-9}\mathrm{s}\right)\\ p=0.003\mathrm{m}\\ p=3\mathrm{mm}.\end{array}$