91Ó°ÊÓ

• Speed of the electron$v=1.5×{10}^7\text{″¾/²õ}$
• Magnetic field$B=1.0×{10}^{−3}\text{ }T$
• The angle of velocity with the direction of the magnetic field is$\mathrm{θ}={10}^0$

We use the formula of the period of motion of charged particles into a magnetic field into the formula of velocity to find the distancefrom the point of injection at which the electron next crosses the field line that passes through the injection point.

Formula:

$$\begin{gathered} T=2\mathrm{Ï€}{m}_e/qB \\ d=vt \\ F=1/T \end{gathered}$$

The period of the electron is:

$\begin{array}{c}T=\frac{2\mathrm{Ï€}{m}_e}{qB}\\ =\frac{2\mathrm{Ï€}(9.1×{10}^{−31}\text{ k²µ})}{(1.6×{10}^{−19}\text{ C})(1.0×{10}^{−3}\text{â€Í¿})}\\ =3.57×{10}^{−8}\text{ s}\end{array}$

And the velocity of the electron is given by

$\begin{array}{c}{V}_{para}=v\cos \mathrm{θ}\\ =\left(1.5×{10}^7\text{ }\frac{\text{m}}{\text{s}}\right)\cos {10}^0\\ =1.48×{10}^7\text{ }\frac{\text{m}}{\text{s}}\end{array}$

Therefore,

$\begin{array}{c}d=V_{para}×T\\ =\left(1.48×{10}^7\text{ }\frac{m}{s}\right)(3.57×{10}^{−8}\text{ }s)\\ =0.53\text{ }m\end{array}$

Therefore, the distance from the point of injection at which the electron next crosses the field line that passes through the injection point is $0.53\text{ }m$.