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I recently read that J.J. Thompson used a velocity selector to find the charge to mass ratio of an electron.

Having learnt the principle behind the velocity selector, I see no relation between these two things at all.

How did a velocity selector help Thompson do this?

My guess is that the velocity of an electron was known beforehand and the selector was used to separate electrons from other particles.

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    $\begingroup$ This link (nyu.edu/classes/tuckerman/adv.chem/lectures/lecture_3/…) provides a good explanation. The velocity selector initially finds the velocity of the electron, then the charge to mass ratio is solved for with the $B$-field switched off and the Electric field on. Its explained well in the link. $\endgroup$
    – tmwilson26
    Feb 12, 2016 at 14:16

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If you balance the forces due to the electric and the magnetic fields on the charged particle in such a way that there is no resultant force on the charged particle, then that is called a velocity selector. It means that the Lorentz force on the particle is 0.

$$ F =Q(E + v \times B) = 0$$

This allows you to measure the velocity of the charged particles emitted (incoming cathode rays into the setup in Thomson's case), which let us assume is in the $y$-direction, with the potential difference in the $x$-direction and and magnetic field in the $z$ direction. After that Thomson, switched off the magnetic field and measured the deflection as the cathode rays came out of the setup. The deflection was given by

$$\tan\theta = \dfrac{qVa}{mhv^2}$$

where $V$ is the applied potential difference, $h$ is the separation of the plates of exerted potential difference, where $a$ is the distance it travels in the y direction through the electric field in the setup and $v$ is already found out using the velocity selector. So Thomson could find out $q/m$ as he had knowledge of all other quantities.

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  • $\begingroup$ I just calculated deflection and I found that it is: qV/2mv^2, all the quantities are the same. How did you get your formula? Your formula is also dimensionally wrong. Thanks, though. I got the point. $\endgroup$ Feb 13, 2016 at 1:19
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    $\begingroup$ @user93868 edited, as I had missed a factor...i used $v = u + at $ for finding the y component of velocity...please recheck.... $\endgroup$
    – Bruce Lee
    Feb 13, 2016 at 1:37
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It meant that all the electrons that he deflected with electric and magnetic field were all travelling at the same velocity. This meant that they were all deflected by the same amount and travelled along the same trajectory to hot the screen at the same place ie produced a well defined spot of light.

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  • $\begingroup$ I'm sorry, but I how is this related to charge/mass $\endgroup$ Feb 12, 2016 at 14:09

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