In JJ thomson's cathode ray experiment why is the effect of gravity on the electron not considered? [closed]

Question

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.

Answer 1

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.