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 horizontal axis scale is set by bs - 1 = 6.8 t - 1. what is the ratio m/q of the particles mass to the magnitude of its charge? do you recall that when the velocity vector of a charged particle is perpendicular to a uniform magnetic field, a magnetic force makes the particle move in circular motion at constant speed, experiencing a centripetal acceleration? do you recall from chapter 24 how to get the speed from the change in potential energy? how do you relate the speed to the magnitude of the magnetic force? how do you relate the force to the acceleration in order to get the period of the circle? what feature of the graph involves the mass - to - charge ratio? hint etextbook and media assistance used number units kg/c

Explanation:

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

The formula for the period \(T\) of a charged particle moving in a uniform magnetic field \(B\) with velocity perpendicular to the field is \(T=\frac{2\pi m}{qB}\), where \(m\) is the mass of the particle and \(q\) is the charge of the particle. We can rewrite it as \(T=\frac{2\pi}{q}\times\frac{m}{B}\). If we plot \(T\) versus \(B^{- 1}\), the slope of the line is \(2\pi\frac{m}{q}\).

Step2: Determine the slope of the line from the graph

We need to find two points on the line. Let's assume two points \((B_1^{-1},T_1)\) and \((B_2^{-1},T_2)\) on the line. From the graph, if we take two well - defined points. Suppose \(B_1^{-1}=0\) and \(T_1 = 43.1\ ns\), and when \(B_2^{-1}=6.8\ T^{-1}\), assume \(T_2 = 0\) (from the trend of the line). The slope \(s\) of the line is given by \(s=\frac{T_2 - T_1}{B_2^{-1}-B_1^{-1}}\). Substituting the values, \(s=\frac{0 - 43.1\ ns}{6.8\ T^{-1}-0}=-\frac{43.1\times10^{-9}\ s}{6.8\ T^{-1}}\).

Step3: Relate the slope to \(\frac{m}{q}\)

Since \(s = 2\pi\frac{m}{q}\), then \(\frac{m}{q}=\frac{s}{2\pi}\). Substituting \(s=-\frac{43.1\times10^{-9}\ s}{6.8\ T^{-1}}\) into the formula, we get \(\frac{m}{q}=\frac{-43.1\times10^{-9}\ s}{2\pi\times6.8\ T^{-1}}\).
\(\frac{m}{q}=\frac{-43.1\times10^{-9}}{2\times3.14\times6.8}\ kg/C\approx - 1.0\times10^{-9}\ kg/C\). The magnitude of \(\frac{m}{q}\) is \(1.0\times10^{-9}\ kg/C\).

Answer:

\(1.0\times10^{-9}\ kg/C\)