In JJ Thomson's cathode ray experiment I need values for the electric field and magnetic field when net force on the cathode beam = 0

Question

I asked here as well https://chemistry.stackexchange.com/questions/177889/in-jj-thomsons-cathode-ray-experiment-why-is-effects-of-gravity-on-electron-not https://physics.stackexchange.com/questions/789225/in-jj-thomsons-cathode-ray-experiment-why-is-effects-due-to-gravity-on-electron?noredirect=1#comment1773186_789225

$${}$$ 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 x B) = Magnetic force, so the magnetic force is acting downwards. 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 https://scienceready.com.au/pages/thomsons-discovery-of-the-electron

I had a problem with how he 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. So i thought maybe gravitational force is negligible? I included gravitational force and got an expression which relates q/m to a constant value $$Magnetic force + Gravitational force = Electric force$$ $$qvB +m9.8 = qE$$ $$v = E/B - (9.8m)/(qB)$$ 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. $$centripetal force = magnetic force$$ $$mv^2/r = qvB$$ $$mv/r = qB$$ $$\frac{m(E/B - (9.8m)/(qB))}{r} = qB $$ $$\frac{mE}{Br} - \frac{9.8m^2}{Bqr} -qB = 0$$ this can be treated as a quadratic equation where the variable is m
so if we have an equation ax^2 + bx + c = 0 its roots are = $\frac{-b +- \sqrt{b^2 - 4ac}}{2a}$ $$m = \frac{ ( -E/Br +- \sqrt{ \frac{E^2}{B^2 r^2} - \frac{4 \cdot 9.8}{r}})Bqr }{2 \cdot 9.8}$$ $$\frac{m}{q} = \frac{ ( -E/Br + \sqrt{ \frac{E^2}{B^2 r^2} - \frac{4 \cdot 9.8}{r}})Br }{2 \cdot 9.8}$$

but I dont know at what magnetic and electric fields so 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

Oh I see, normally an ExB filter is called a "velocity filter" - see Wikipedia's Wien filter where there be math.

In an ExB filter charged particles go in a straight line when $v = E/B$. So any combination with the same ratio will work.

But by adding the force of gravity which depends on the particle's mass explicitly, you've "broken" the Wien filter. At normal lab voltages and magnetic fields and accelerated electron velocities, the gravity term is so very tiny that it would be buried below all of the other experimental uncertainties in this device.

So here in HSM SE I think that the answer to

In JJ Thomson's cathode ray experiment I need values for the electric field and magnetic field when net force on the cathode beam = 0

is No you don't. For historical purposes of JJ Thomson's experiment, gravity was a way way negligible effect, and the problem reduces to the simple E/B ratio. The tube is an early Wien filter.

However, in Physics SE some math will be needed to see if there's an analytical solution to your problem, or if you have to use a numerical technique to find the roots (zeros) of the expression for total force.

So rewrite your equation in the form $F_{total} = ...$, set it to zero, and massage it mathemtatically to see if an analytical solution exist (probably does!) or if you have to find the zeros. There will of course be (at least) two because the gravitational force can be in the same direction of either the magnetic, or the electric force. You chose only one of the two possibilities in your example.

update: In your cross-post in Chemistry SE @BuckThorn used the first term in the Taylor expansion to get an approximate solution.