91Ó°ÊÓ

the given is the electrons orbit the magnetic field at $2.4GHz$

the objective is to find the magnetic field strength and the maximum electron kinetic energy

the method used here is magnetic field to find kinetic energy.

a)

The magnetron motion is the uniform circular motion of a particle perpendicular to the magnetic field at a constant speed. The magnetic field causes the circular motion to produce a circular frequency, which is given by an equation in the form

$f_{cyc}=\frac{qB}{2\mathrm{Ï€³¾}}$ (1)

Where $q$is the particle's charge, $B$is the magnetic field, and m is the particle's mass. The frequency is determined by the magnetic field rather than the particle's velocity, as indicated by equation $\left(1\right)$. We rewrite equation $\left(1\right)$so that$B$ has the form

localid="1648829482795" $B=\frac{2{\mathrm{Ï€³¾f}}_{\mathrm{cyc}}}{q}$ (2)

Now we plug in the electron's $f_{cyc}$, $m$, and $q$ values to get$B$

$$\begin{gathered} B=\frac{2{\mathrm{Ï€³¾f}}_{\mathrm{cyc}}}{q} \\ =\frac{2\mathrm{Ï€}(9.1×{10}^{-31}\mathrm{Kg})(2.5×{10}^9\mathrm{Hz})}{1.6×{10}^{-19}C} \\ =86×{10}^{-3}T \\ =\boxed{86mT} \end{gathered}$$

b)

The radius of the circular motion produced by the magnetic field is linked to it by equation $(29.20)$in the form

$r_{cyc}=\frac{mv}{qB}$ $\left(3\right)$

Because the chamber's diameter is$2.5cm$, we can calculate its radius by multiplying the diameter by the radius

$r_{cyc}=\frac{2.5cm}{2}=1.25cm=1.25×{10}^{-2}m$

To calculate the electron's speed, we rearrange equation$\left(3\right)$and insert in the values for $r_{cyc}$, $q$, $m$, and $B$

$$\begin{gathered} v=\frac{r_{cyc}qB}{m} \\ =\frac{(1.25×{10}^{-2}m)(1.6×{10}^{-19}C)(86×{10}^{-3}T)}{9.1×{10}^{-31}Kg} \\ =1.89×{10}^{8}m/s \end{gathered}$$

The kinetic energy is equal to half of the mass multiplied by the square of the velocity

$K=\frac{1}{2}m{v}^2$ $\left(4\right)$

To calculate $K$of the electron, we plug the values for $m$and $v$into equation $\left(4\right)$

$$\begin{gathered} K=\frac{1}{2}m{v}^2 \\ =\frac{1}{2}(9.1×{10}^{-31}Kg){(1.89×{10}^{8}m/s)}^2 \\ =\boxed{1.6×{10}^{-14}J} \end{gathered}$$