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thomsons experiment (1 Viewer)
- Thread starter mat
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He got a CRT and balanced magnetic field with a electric field in such a way that Electrons were able to travel striaght.
He then equated F = eq with F = qvbsin @ and since the angle was known, the strength of the and b field were known, he calculated the velocity. Removing the electric field, he measured the radius that the magnetic field created (r) which then led him to equate F = mv squared with F = qvbsin @. With all the variables except m and q known, he figured out the ratio of charge to mass.
He then equated F = eq with F = qvbsin @ and since the angle was known, the strength of the and b field were known, he calculated the velocity. Removing the electric field, he measured the radius that the magnetic field created (r) which then led him to equate F = mv squared with F = qvbsin @. With all the variables except m and q known, he figured out the ratio of charge to mass.
![+:: $i[Q]u3 ::+](/data/avatars/m/2/2831.jpg?1559318848){kind=avatar}
+:: $i[Q]u3 ::+
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chill.. chill....~~~~ it's over..
it's pretty much textbook material... he stuck capacitor plates and coils (electromagnets) around a cathode ray tube, so the forces they produced on the rays directly opposed each other. he varied the fields (the electirc and magnetic) til the rays remained undeflected, then he equated the forces. and got a q/m ratio
it's pretty much textbook material... he stuck capacitor plates and coils (electromagnets) around a cathode ray tube, so the forces they produced on the rays directly opposed each other. he varied the fields (the electirc and magnetic) til the rays remained undeflected, then he equated the forces. and got a q/m ratio
Constip8edSkunk
Joga Bonito
lol A is just wrong man
special cathode ray tube blah blah blah helmholtz coil+ || E plates, blah blah
equate E and B forces
qE=qvB
v=E/B
turn off E field
calculate r of deflection by measuring displacement and arc
equate B and centripetal force
mv^2/r=qvB
m(E/B)/r=qB
q/m=E/B^2r
edit: lol shit im a slow typer, maybe typing and talking on the phone isnt a good idea lol
special cathode ray tube blah blah blah helmholtz coil+ || E plates, blah blah
equate E and B forces
qE=qvB
v=E/B
turn off E field
calculate r of deflection by measuring displacement and arc
equate B and centripetal force
mv^2/r=qvB
m(E/B)/r=qB
q/m=E/B^2r
edit: lol shit im a slow typer, maybe typing and talking on the phone isnt a good idea lol
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those people who say that he just equated magnetic and electric field forces, its actually really important to mention that he also turned off the electric field, because otherwise there is no way mathematically he could determine the ratio
SoCal
Hollywood
I wrote most of what people here have said but I didn't use any equations.
.
sugaryblue
Living on deficit
This was the question i love actually, as much as I disliked that test. coz it's one of the few that I knew the answers.
smeyo
Member
yeah i did the same as merethround. i wrote the E and B equating then further experiments with magnetic deflection yo derive at q/m
Viper
Member
The answer to 14 can't be A!!!!
Hertz didn't call them RADIO WAVES, therefore he couldn't have referred to them as radio waves!!! I spoke to my teacher about this question, and we agree that the correct answer is D
Cheers
Hertz didn't call them RADIO WAVES, therefore he couldn't have referred to them as radio waves!!! I spoke to my teacher about this question, and we agree that the correct answer is D
Cheers
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smeyo
Member
i ahvent talked to my teacher yet, but ill ask my tutor 2morow about it but as far as i know so far it is D an observation
+:: $i[Q]u3 ::+
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what was it.. a four mark question? they won't be that picky... well i doubt it anyway. i can hardly see someone going "oh u got the order wrong.. only 3 for u~"
Under-stated
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did you actualy have to give the charge to mass ratio of electron?
that is q/m=1.75x10^11??
that is q/m=1.75x10^11??
freaking_out
Saddam's new life
yeah, same here...i just said that he equated the magnetic force on the electron wif the magnetic force...Originally posted by Merethrond
I wrote most of what people here have said but I didn't use any equations
dnt stop runnin
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from wat ive read.......he equalled them first.....adjusting the electric field so that it can go in straight.......then he took out the electric field...........then again, i dnt think it matters.......as long as he found wat he had to
SmokedSalmon
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Actually he applied the electric field FIRST, then magnetic field... I know so cause my father is a highly commended physics professor from UTS but as long as you pointed out what the point of either one was then you'd be fine .
.Yeah. putting the magnetic field first to measure the radius would be stupid.. as you have to adjust the magnetic field to equal the electric field anyway (i believe he left the electric field constant strength the whole time .. and thus was able to equate mv2/r at the end)