Question

A certain particle is sent into a uniform magnetic field, with the particle’s velocity vector perpendicular to the direction of the field. Figure 28-37 gives the period Tof the particle’s motion versus the inverseof the field magnitude B. The vertical axis scale is set by$\mathbf{}{\mathbf{T}}_s\mathbf{=}\mathbf{\text{40.0ns}}$, and the horizontal axis scale is set by ${\mathbf{\text{B}}}_{\mathbf{\text{s}}}^{\mathbf{\text{-1}}}\mathbf{=}{\mathbf{\text{5.0\hspace{0.17em}T}}}^{\mathbf{\text{-1}}}$what is the ratio m/qof the particle’s mass to the magnitude of its charge?

Short answer

The ratio m/qof the particle’s mass to the magnitude of its charge is $\frac{\text{m}}{\text{q}}=1.2\times {10}^{-9}\text{\hspace{0.17em}kg}/\text{C}$.

Step-by-step solution

Given

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

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

Figure 28-37 is the graph of $\text{T}\text{\hspace{0.17em}versus\hspace{0.17em}}{\text{B}}^{-1}$

Determining the concept

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:

$\text{T}=\frac{\text{2πm}}{\text{qB}}$

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

Determining the ratio m/q

Using the equation,

$\text{T}=\frac{\text{2πm}}{\text{qB}}$

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

$\frac{\text{m}}{\text{q}}=\frac{\text{TB}}{\text{2π}}$

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

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

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$\text{0.75\hspace{0.17em}ns}$, and for the x-axis, one unit is$1{\text{\hspace{0.17em}T}}^{-1}$

localid="1663950301260" $\begin{array}{c}\left(\frac{\text{T}}{{{\text{B}}_{\text{s}}}^{-1}}\right)=\frac{(0.75\times {10}^{-9}\text{\hspace{0.17em}s})}{1{\text{\hspace{0.17em}T}}^{\text{-1}}}\\ =7.5\times {\text{10}}^{\text{-9}}\text{\hspace{0.17em}T}\cdot \text{s}\end{array}$

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

$\begin{array}{c}\frac{\text{m}}{\text{q}}=\frac{1}{2\left(3.14\right)}(7.5\times {\text{10}}^{\text{-9}}\text{\hspace{0.17em}T}\cdot \text{s})\\ =1.2\times {10}^{-9}kg/C\end{array}$

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

$\frac{\text{m}}{\text{q}}=1.2\times {10}^{-9}\text{\hspace{0.17em}kg}/\text{C}$

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