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The Tolman-Stewart experiment in 1916 demonstrated that
Chapter 25, Problem 25.85(choose chapter or problem)
The Tolman-Stewart experiment in 1916 demonstrated that the free charges in a metal have negative charges and provided a quantitative measurement of their charge-to-mass ratio, \(|q| / m\). The experiment consisted of abruptly stopping a rapidly rotating spool of wire and measuring the potential difference that this produced between the ends of the wire. In a simplified model of this experiment, consider a metal rod of length \(L\) that is given a uniform acceleration \(\vec{a}\) to the right. Initially the free charges in the metal lag behind the rod's motion, thus setting up an electric field \(\overrightarrow{\boldsymbol{E}}\) in the rod. In the steady state this field exerts a force on the free charges that makes them accelerate along with the rod.
(a) Apply \(\sum \vec{F}=m \vec{a}\) to the free charges to obtain an expression for \(|q| / m\) in terms of the magnitudes of the induced electric field \(\overrightarrow{\boldsymbol{E}}\) and the acceleration \(\vec{a}\).
(b) If all the free charges in the metal rod have the same acceleration, the electric field \(\overrightarrow{\boldsymbol{E}}\) is the same at all points in the rod. Use this fact to rewrite the expression for \(|q| / m\) in terms of the potential \(V_{b c}\) between the ends of the rod (Fig. P25.85).
(c) If the free charges have negative charge, which end of the rod, b or c, is at higher potential?
(d) If the rod is 0.50 m long and the free charges are electrons (charge \(q=-1.60 \times 10^{-19}\) C, mass \(9.11 \times 10^{-31}\) kg), what magnitude of acceleration is required to produce a potential difference of 1.0 mV between the ends of the rod?
(e) Discuss why the actual experiment used a rotating spool of thin wire rather than a moving bar as in our simplified analysis.
Questions & Answers
QUESTION:
The Tolman-Stewart experiment in 1916 demonstrated that the free charges in a metal have negative charges and provided a quantitative measurement of their charge-to-mass ratio, \(|q| / m\). The experiment consisted of abruptly stopping a rapidly rotating spool of wire and measuring the potential difference that this produced between the ends of the wire. In a simplified model of this experiment, consider a metal rod of length \(L\) that is given a uniform acceleration \(\vec{a}\) to the right. Initially the free charges in the metal lag behind the rod's motion, thus setting up an electric field \(\overrightarrow{\boldsymbol{E}}\) in the rod. In the steady state this field exerts a force on the free charges that makes them accelerate along with the rod.
(a) Apply \(\sum \vec{F}=m \vec{a}\) to the free charges to obtain an expression for \(|q| / m\) in terms of the magnitudes of the induced electric field \(\overrightarrow{\boldsymbol{E}}\) and the acceleration \(\vec{a}\).
(b) If all the free charges in the metal rod have the same acceleration, the electric field \(\overrightarrow{\boldsymbol{E}}\) is the same at all points in the rod. Use this fact to rewrite the expression for \(|q| / m\) in terms of the potential \(V_{b c}\) between the ends of the rod (Fig. P25.85).
(c) If the free charges have negative charge, which end of the rod, b or c, is at higher potential?
(d) If the rod is 0.50 m long and the free charges are electrons (charge \(q=-1.60 \times 10^{-19}\) C, mass \(9.11 \times 10^{-31}\) kg), what magnitude of acceleration is required to produce a potential difference of 1.0 mV between the ends of the rod?
(e) Discuss why the actual experiment used a rotating spool of thin wire rather than a moving bar as in our simplified analysis.
ANSWER:Step 1 of 5
(a) The free electrons only move under the electric force, which is given by
\(\overrightarrow{\mathbf{F}}_{e}=q \overrightarrow{\mathbf{E}}\)
So, by applying Newton's second law, we get
\(\begin{aligned} \sum F & =F_{e}=m a \\ |q| E & =m a \\ \therefore \frac{|q|}{m} & =\frac{a}{E} \end{aligned}\)