91影视

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

$r_{\mathrm{cyc}}=\frac{m\stackrel{鈫�}{v}}{qB}$

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" $胃={30}^{鈭�}$, equation will be true.

$r_{\mathrm{cyc}}=\frac{mv\cos 鈦�胃}{qB}$

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

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

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

$p=(v\sin 鈦�胃)T$

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

$f_{\mathrm{cyc}}=\frac{qB}{2蟺m}$

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

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

The reciprocal of frequency period is

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

Substituting the values in equation to get pitch by

$\begin{array}{l}p=(v\sin 鈦�胃)T=\left(5脳{10}^6\mathrm{m}/\mathrm{s}\right)\sin 鈦�{30}^{鈭�}\left(1.2脳{10}^{鈭�9}\mathrm{s}\right)\\ p=0.003\mathrm{m}\\ p=3\mathrm{mm}.\end{array}$