You are a research scientist working on ahigh-energy particle accelerator. Using a modern version of the Thomson e/m apparatus, you want to measure the mass of a muon(a fundamental particle that has the same charge as an electronbut greater mass). The magnetic field between the two charged plates is 0.340T . You measure the electric field for zeroparticle deflection as a function of the accelerating potential V. This potential is largeenough that you can assumethe initial speed of the muonsto be zero. Figure is an ${\mathbf{E}}^{\mathbf{2}}$versus graph of your data. (a) Explain why the data points fall close to a straight line. (b) Use the graph in Figure to calculate the mass m of a muon. (c) If the twocharged plates are separated by, what must be the voltage between the plates in order for the electric field between the plates to be $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{V}\mathbf{/}\mathbf{m}$? Assume that the dimensions ofthe plates are much larger than the separation between them.(d) When V = 400 V, what is the speed of the muons as theyenter the region between the plates?

The given data can be listed below as

• The magnetic field is, B = 0.340 T .
• The two charged plates are separated by d = 6.00 m
• The electric field is, E = $2.00\times {10}^{5}V/m$.

The term magnetic field may be defined as the area around the magnet behave like a magnet.

The Thomson experiment says$\frac{e}{m}=\frac{E^2}{2V{B}^2}$

And the relation between kinetic energy and potential energy is$V=\sqrt{\frac{2qV}{m}}$

So

$E^2=2\left(\frac{e}{m}\right)B^{2}V$

Here the slope is 2 (e/m) $B^2$.

Hence,the equation of straight line is${\mathrm{E}}^2=2\left(\frac{\mathrm{e}}{\mathrm{m}}\right){\mathrm{B}}^2\mathrm{V}$

Here the slope is $2\left(\frac{e}{m}\right)B^2$.

The mass of the muon is calculated (from figure)

$\begin{array}{r}\text{Slope}=\frac{600\times {10}^8{\mathrm{V}}^2/{\mathrm{m}}^2-200\times {10}^8{\mathrm{V}}^2/{\mathrm{m}}^2}{300\mathrm{V}-100\mathrm{V}}\\ =2\times {10}^8\mathrm{V}/{\mathrm{m}}^2\end{array}$

And

$\text{Slope}=2\left(\frac{e}{m}\right)B^2$

So,

$\frac{e}{m}=\frac{\text{slope}}{2B^2}$

Substitute all the value in the above equation.

$\begin{array}{r}\frac{\mathrm{e}}{\mathrm{m}}=\frac{\text{slope}}{2{\mathrm{B}}^2}\\ \frac{\mathrm{e}}{\mathrm{m}}=\frac{2.00\times {10}^8\mathrm{V}/{\mathrm{m}}^2}{2(0.340\mathrm{T}{)}^2}\\ \frac{\mathrm{e}}{\mathrm{m}}=8.65\times {10}^8\mathrm{C}/\mathrm{kg}\\ \mathrm{m}=\frac{16\times {10}^{-20}\mathrm{C}}{8.65\times {10}^8\mathrm{C}/\mathrm{kg}}\\ \mathrm{m}=1.85\times {10}^{-28}\mathrm{kg}\end{array}$

Hence, the mass of muon is$\mathrm{m}=1.85\times 10-28\mathrm{kg}$ .

The charge between two plates can be calculated as

V = Ed

Substitute all the value in the above equation.

$\begin{array}{r}\mathrm{V}=\left(2.00\times {10}^5\mathrm{V}/\mathrm{m}\right)(0.00600\mathrm{m})\\ \mathrm{V}=1.20\mathrm{kV}\end{array}$

Hence, the voltage between the plates is V = 1.20 kV.

The velocity of muon can be calculated as

$V=\sqrt{\frac{2eV}{m}}$

Substitute all the value in the above equation.

$\begin{array}{r}\mathrm{v}=\sqrt{2\times \frac{\mathrm{e}}{\mathrm{m}}\times \mathrm{V}}\\ \mathrm{v}=\sqrt{2\left(8.65\times {10}^8\mathrm{C}/\mathrm{kg}\right)\left(400\mathrm{V}\right)}\\ \mathrm{v}=8.32\times {10}^5\mathrm{m}/\mathrm{s}\end{array}$

Hence, the speed of the muons is$V=8.32\times {10}^5\mathrm{m}/\mathrm{s}$