91Ó°ÊÓ

Y axis scale is${\text{T}}_{\text{s}}=\text{40ns}\left(\frac{10−9\text{ }s}{1\text{ }ns}\right)=4.0×{10}^{−8}\text{ }s$.

X axis scale is${\text{B}}_{\text{s}}^{\text{-1}}={\text{5.0T}}^{\text{-1}}$

Figure 28-37 is the graph of $\text{T}\text{ v±ð°ù²õ³Ü²õ }{\text{B}}^{−1}$

Use the concept of the period of the particle in a uniform magnetic field. Use the equation of period related to mass, charge, and magnetic field. Find the slope of the graph and plug it intotheequation. Rearranging the equation for $\frac{\text{m}}{\text{q}}$, find this ratio.

Formulae are as follows:

$\text{T}=\frac{\text{2Ï€³¾}}{\text{qB}}$

Where T is the time period, B is the magnetic field, m is mass, and q is the charge ofthe particle.

Using the equation,

$\text{T}=\frac{\text{2Ï€³¾}}{\text{qB}}$

Rearranging it for ,$\frac{\text{m}}{\text{q}}$

$\frac{\text{m}}{\text{q}}=\frac{\text{TB}}{\text{2Ï€}}$

We have the value of ${\text{B}}_{\text{s}}^{\text{-1}}$,

localid="1663950272832" $\begin{array}{c}\frac{m}{q}=\frac{T}{2\mathrm{π}{B}^{−1}}\\ =\frac{1}{2\mathrm{π}}\left(\frac{T}{B^{−1}}\right)\end{array}$

From the graph,find the slope localid="1663950285918" $\left(\frac{\text{T}}{{{\text{B}}_{\text{s}}}^{−\text{1}}}\right)\text{.}$

For the y-axis, one unit is$\text{0.75 n²õ}$, and for the x-axis, one unit is$1{\text{â€Í¿}}^{−1}$

localid="1663950301260" $\begin{array}{c}\left(\frac{\text{T}}{{{\text{B}}_{\text{s}}}^{−1}}\right)=\frac{(0.75×{10}^{−9}\text{ s})}{1{\text{â€Í¿}}^{\text{-1}}}\\ =7.5×{\text{10}}^{\text{-9}}\text{â€Í¿}â‹\dots \text{s}\end{array}$

Using this value in the above equation of $\frac{\text{m}}{\text{q}}$,

$\begin{array}{c}\frac{\text{m}}{\text{q}}=\frac{1}{2\left(3.14\right)}(7.5×{\text{10}}^{\text{-9}}\text{â€Í¿}â‹\dots \text{s})\\ =1.2×{10}^{−9}\text{ }kg/C\end{array}$

Hence, the ratio m/qof the particle’s mass to the magnitude of its charge is .

$\frac{\text{m}}{\text{q}}=1.2×{10}^{−9}\text{ k²µ}/\text{C}$

Therefore, by using the concept of the period of the particle in a uniform magnetic field and equations the ratio can be determined.