Question

a certain particle is sent into a uniform magnetic field, with the particles velocity vector perpendicular to the direction of the field. the figure gives the period t of the particles motion versus the inverse of the field magnitude b. the vertical axis scale is set by $t_s = 43.1$ ns, and the horizontal axis scale is set by $b_s^{-1}=6.8$ t$^{-1}$. what is the ratio m/q of the particles mass to the magnitude of its charge?

Explanation:

Step1: Recall the formula for the period of a charged - particle in a magnetic field

The period of a charged particle moving in a uniform magnetic field with its velocity perpendicular to the field is given by $T=\frac{2\pi m}{qB}$, which can be rewritten as $T = 2\pi\frac{m}{q}B^{-1}$. This is a linear equation of the form $y = mx + c$, where $y = T$, $x = B^{-1}$, and the slope $m = 2\pi\frac{m}{q}$.

Step2: Find the slope of the line from the graph

The slope of the line $m_{slope}$ in the $T - B^{-1}$ graph can be calculated using two - point formula. Let's assume two points on the line: $(B_1^{-1},T_1)$ and $(B_2^{-1},T_2)$. The slope $m_{slope}=\frac{T_2 - T_1}{B_2^{-1}-B_1^{-1}}$. From the graph, we can also note that the slope of the line $T$ versus $B^{-1}$ is related to $\frac{m}{q}$ by $m_{slope}=2\pi\frac{m}{q}$.
We know that the vertical - axis scale is set by $T_s = 43.1\ ns$ and the horizontal - axis scale is set by $B_s^{-1}=6.8\ T^{-1}$.
If we consider the full range of the graph (assuming the line passes through the origin and the end - point of the visible line segment), when $B^{-1}=B_s^{-1}=6.8\ T^{-1}$, $T = T_s = 43.1\ ns$.
Since $T = 2\pi\frac{m}{q}B^{-1}$, then $\frac{m}{q}=\frac{T}{2\pi B^{-1}}$.
Substituting $T = 43.1\times10^{-9}\ s$ and $B^{-1}=6.8\ T^{-1}$ into the formula:
\[

$$\begin{align*} \frac{m}{q}&=\frac{43.1\times 10^{-9}}{2\pi\times6.8}\\ &=\frac{43.1\times 10^{-9}}{42.73}\\ &\approx1.01\times 10^{-9}\ kg/C \end{align*}$$

\]

Answer:

$1.01\times 10^{-9}\ kg/C$