Question

A source injects an electron of speed $\mathit{v}\mathit{=}\mathit{1}\mathit{.5}\mathit{\times }{\mathit{10}}^{\mathit{7}}$m/s into a uniform magnetic field of magnitude$\mathit{B}\mathbf{=}\mathbf{1}\mathbf{.0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}$T. The velocity of the electron makes an angle$\mathit{\theta }\mathbf{=}\mathbf{}\mathbf{10}\mathbf{°}$with the direction of the magnetic field. Find the distance dfrom the point of injection at which the electron next crosses the field line that passes through the injection point.

Short answer

The distance from the point of injection at which the electron next crosses the field line that passes through the injection pointis $0.53m$.

Step-by-step solution

Listing the given quantities

• Speed of the electron$v=1.5\times {10}^7\text{\hspace{0.17em}m/s}$
• Magnetic field$B=1.0\times {10}^{-3}T$
• The angle of velocity with the direction of the magnetic field is$\theta ={10}^0$

Understanding the concept of the period of motion in terms of magnetic field

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:

$$\begin{gathered} T=2\pi m_e/qB \\ d=vt \\ F=1/T \end{gathered}$$

Calculation ofthe distance from the point of injection at which the electron next crosses the field line that passes through the injection point 

The period of the electron is:

$\begin{array}{c}T=\frac{2\pi m_e}{qB}\\ =\frac{2\pi (9.1\times {10}^{-31}\text{\hspace{0.17em}kg})}{(1.6\times {10}^{-19}\text{\hspace{0.17em}C})(1.0\times {10}^{-3}\text{\hspace{0.17em}T})}\\ =3.57\times {10}^{-8}\text{\hspace{0.17em}s}\end{array}$

And the velocity of the electron is given by

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

Therefore,

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

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