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JJ thomson's setup for the cathode ray tube

Explaining the setup:

The experiment is described in the picture. Instead of the magnets in the picture imagine two circular coils on both the sides with current running through it, this creates a magnetic field perpendicular to the loop which can be predicted using Biot-Savart law. Run the current in the loop in such a way so that the magnetic field is going into your screen, away from you. Without the magnets the beam is deflected up towards the positive plate proving the charge on the beam is -ve as +ve and -ve attract.

$q(v \times B) = \textrm{Magnetic force}$

so the magnetic force is acting downwards. The q/m ratio can be determined as well by combining both the magnets and the charged plates and changing the electric and magnetic fields until there seems to be no deflection on the beam so net force acting on beam must be close to 0 and we solve taking all the forces, more details here.

I had a problem with how Thomson finds the q/m ratio which is why the gravitational force on the electron is not considered? Only the magnetic force and electric force is considered. I thought maybe gravitational force is negligible? I included gravitational force and got an expression which relates q/m to a constant value

Magnetic force + Gravitational force = Electric force

or

$$qvB +m \times \pu{9.8 m/s^2} = qE$$ $$v = E/B - (\pu{9.8 m/s^2} \times m)/(qB)$$

When we only have the magnetic force acting on the beam, it will always act perpendicular to the velocity due to the magnetic field going into the screen, so it will constantly change the direction of velocity but not its magnitude and the beam will move in a circle:

Centripetal force = Magnetic force

or

$$mv^2/r = qvB$$ $$mv/r = qB$$ $$\frac{m(E/B - (\pu{9.8 m/s^2} \times m)/(qB))}{r} = qB $$ $$\frac{mE}{Br} - \frac{\pu{9.8 m/s^2} \times m^2}{Bqr} -qB = 0$$

This can be treated as a quadratic equation where the variable is m. If we have an equation $ax^2 + bx + c = 0$ its roots are $\frac{-b +- \sqrt{b^2 - 4ac}}{2a}$, so that

$$m = \frac{ \left( -E/Br +- \sqrt{ \frac{E^2}{B^2 r^2} - \frac{4 \cdot \pu{9.8 m/s^2}}{r}}\right)Bqr }{2 \cdot \pu{9.8 m/s^2}}$$

$$\frac{m }{q}= \frac{ \left( -E/Br +- \sqrt{ \frac{E^2}{B^2 r^2} - \frac{4 \cdot \pu{9.8 m/s^2}}{r}}\right)Br }{2 \cdot \pu{9.8 m/s^2}}$$

But I don't know the magnetic and electric fields. I need the values of electric field of the charged plates and magnetic field of the coils at which there is no net deflection and the radius of the circle the beam goes through when only the magnetic field acts on it, provide the source as well. Maybe there is some other way to understand why gravitational force is not included if so could you explain that way.

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  • $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Chemistry Meta, or in Chemistry Chat. Comments continuing discussion may be removed. $\endgroup$
    – Buck Thorn
    Nov 27, 2023 at 10:04
  • $\begingroup$ The question should be clear, Why is effects of gravity not included in JJ thomson's finding of q/m ratio of electron where q is its charge and m its mass. What details are missing? $\endgroup$
    – Saif
    Nov 29, 2023 at 16:48
  • $\begingroup$ Pick some electron energy, and find the velocity and then the time spent traversing the experiment. From that time you can directly calculate the distance the electron 'falls' during the experiment. Say 1keV electron and 1 meter. I get a vertical displacement from gravity of ~10 femtometers, but may have made a mistake. $\endgroup$
    – Jon Custer
    Dec 7, 2023 at 17:37
  • $\begingroup$ @JonCuster Yes there is an answer on physicsstackexchange which i linked which has the same answer. That is the best answer I feel as we can also say by just looking at the beam that the deflection is negligible in absence of electric and magnetic fields. $\endgroup$
    – Saif
    Dec 7, 2023 at 17:45

1 Answer 1

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Various versions of Thomson's original article (Philosophical Magazine, 44, 293-316 (1897)) can be accessed freely online, for instance at Google books, or here (link liable to rot).

But you can work around not knowing the details of the experiment. Interestingly, you don't even need to know the magnitude of the magnetic field applied (therefore current).

Starting from your two equations we can compute $q/m$ (solving a quadratic equation in q instead of m):

$$\frac{q}{m} = \frac{E}{2B^2r}+\left[\frac{E^2}{4B^4r^2}-\frac{g}{B^2r} \right]^{1/2} \\ = \frac{E}{2B^2r}\left(1+\left[1-\frac{4gB^2r}{E^2} \right]^{1/2}\right) \tag{1}$$

We can use the fact that if we assume g is negligibly small then we know the solution is

$$\frac{q}{m} = \left(\frac{E}{B^2 r}\right)_{g=0}$$

Based on this, if we define

$$\left(\frac{q}{m}\right)_{\textrm{approx}} = \frac{E}{B^2 r}$$

we can use a Taylor expansion assuming the gravitational effect to be very small relative to the electromagnetic ones to recast Eq (1) as

$$\frac{q}{m} \approx \left(\frac{q}{m}\right)_{\textrm{approx}}\left[1-\frac{g}{E}\left(\frac{m}{q} \right)_{\textrm{approx}} \right] \\ \approx \left(\frac{q}{m}\right)_{\textrm{approx}} \left[1-\frac{mg}{qE} \right]$$

This result suggests using known constants to evaluate

$$E >> \frac{mg}{q} = \pu{5.57e-11 V/m}$$

as a condition for the value of the applied electric field above which g can be neglected. Using reasonable values for the distance between the electrodes we see that the applied potential required to satisfy the condition is bound to be quite small, much smaller than the range for instance reported in examples online, which are in the order of hundreds of volts.

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  • $\begingroup$ If you insist on wanting to know the current applied, work backward using the balance condition to solve for the applied magnetic field, knowing the range of electric fields that may have been used. $\endgroup$
    – Buck Thorn
    Nov 27, 2023 at 7:41
  • $\begingroup$ Can you explain how you got $\frac{q}{m} \approx \frac12\left(\frac{q}{m}\right)_{\textrm{approx}}\left[2+\frac{mg}{qE} \right]$ $\endgroup$
    – Saif
    Nov 28, 2023 at 5:35
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    $\begingroup$ I made a mistake in the centripetal force = magnetic force equation, I didn't include effects of gravity there Im trying to include it. If I can then there should be a change in the equations $\endgroup$
    – Saif
    Nov 28, 2023 at 5:39
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    $\begingroup$ I'm not familiar with Taylor expansion so I tried learning it. $$\frac{q}{m} = \frac{E}{2B^2r} +(\frac{E^2}{4B^4r^2}(1 - \frac{4B^2rg}{E^2}) )^{1/2} $$ $$ \frac{E}{2B^2r} + \frac{E^2}{2B^2r}(1 + x)^1/2 $$ $$ \frac{E}{2B^2r}+ \frac{E^2}{2B^2r}(1 + \frac{1}{2}x)$$ Here you are using the approximation around x = 0? so then I substituted for x I was typing this as you edited it, and got what you got in the new edit $\endgroup$
    – Saif
    Nov 30, 2023 at 6:29
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    $\begingroup$ I linked here from my answer to the HSM SE cross-posted version $\endgroup$
    – uhoh
    Dec 1, 2023 at 0:16

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