A beam of electrons whose kinetic energy is Kemerges from a thin-foil “window” at the end of an accelerator tube. A metal plate at distance dfrom this window is perpendicular to the direction of the emerging beam (Fig. 28-53). (a) Show that we can prevent the beam from hitting the plate if we apply a uniform magnetic field such that

$\mathbf{B}\mathbf{\ge }\sqrt{\frac{\mathbf{2}{\mathbf{m}}_{\mathbf{e}}\mathbf{K}}{\mathbf{\left(}{\mathbf{e}}^{\mathbf{2}}{\mathbf{d}}^{\mathbf{2}}\mathbf{\right)}}}$

in which mand eare the electron mass and charge. (b) How should $\overrightarrow{\mathbf{B}}$be oriented?

The kinetic energy of electron beam is K.

The metal plate distance from the window isd.

The window is perpendicular to the direction of emerging beam.

Mass of the electron is m.

Charge of the electron is e.

By using the equations for magnetic force F_B and magnitude of acceleration of the electron a and by applying Newton’s second law, we can find that the equation for the radius r of the circular path is in agreement with equation 28-16. By using equation for kinetic energy, we can find the speed v of the electron. Using this equation of v in the equation for the radius r, we can prove that $B\ge \sqrt{\frac{2m_{e}K}{\left(e^2{d}^2\right)}}$. Finally, from the direction of magnetic force, we can find the orientation of $\overrightarrow{B}$.

Formula:

• The magnetic field on an electron has magnitude F_B=evB
• The acceleration of the electron has magnitude a=v²/r
• The kinetic energy of the electron is K=1/2m_ev².

For the magnetic field to have an effect on the moving electrons, we need a non-negligible component of $\overrightarrow{B}$to be perpendicular to$\overrightarrow{v}$. It is most efficient, therefore, to orient the magnetic field so that it is perpendicular to the plane of the page. The magnetic field on an electron has magnitude

$F_B=evB$

And the acceleration of the electron has magnitude

localid="1662896876449" $a=v^2/r$

Therefore, Newton’s second law gives

$evB=\frac{m_e{v}^2}{r}$

So, the radius of the circle is given by

$r=\frac{\left(m_{e}v\right)}{eB}$

This is in agreement with equation 28-16.

The kinetic energy of the electron is

$K=\frac{1}{2}{m}_e{v}^2$

Therefore, the speed of the electron is

$v=\sqrt{\frac{2K}{m_e}}$

Thus,

localid="1662897468629" $r=\sqrt{\frac{2m_{e}K}{e^2{B}^2}}$

This must be less than d, so

$\sqrt{\frac{2m_{e}K}{e^2{B}^2}\le d}$

Or

localid="1662897375140" $B\ge \sqrt{\frac{2m_{e}K}{e^2{d}^2}}$

Hence, if we apply this magnetic field, we can prevent the beam from hitting the plate.

If the electrons are to travel as shown in Fig.28-53, the magnetic field must be out of the page. Then the magnetic force is towards the center of the circular path, as it must be in order to make the circular motion possible.