The Coulomb model is for Electrostatic force (key word: STATIC). Here the charge is DYNAMIC, so there is a "current" (relatively speaking). If there is a current, it will generate a magnetic field. The trajectory of the electron could be considered to be a coil. - What if you modeled the electron and nucleus as if they were electro-magnets? Nucleus charge is +e; electron charge is -e; mag F on "-e" charge due to "+e" = -e(v x B) mag F from "-e" over one loop =IL∑dB , L= length of loop =IL. ∑(k'. I.dLsin[θ]/R²)r^ , where k'= μ/(4π ) =(e.fL.k’ I) ∑ (dL/(R²+y²) )r^ , where f= electron orbital frequency. But y=0 , R=const => mag F=(e.fL.k’ I) (2πR/R²)r^ => mag F=(e.fL.k’ I) (2π /R)r^ mag F=μ(e.f) I (L/2R) r^ . (static) elec F= k(e/R)² r^ |elec F|/ |mag F|= k(e/R)² /(2πe.fL.k’ I /R) =(k/k’)(eT/I)/(2πR)² , using L=2πR. where: r^=unit vector from nucleus to electron, I= 'nucleus current' wrt static electron, f=I/e=electron orbital frequency. Lorentz F= Lor F=mag F+ elec F= (mag F)( 1+(k/k’)(eT/I)/(2πR)²)r^ k=1/4πε =c² . [μ]/10⁷ and k'= μ/2π - therefore (using μ/[μ]= μ' =4π/10⁷ ): k/k'=2πc² . (1/10⁷) /μ' =c²/2 and: Lor F= (mag F)( 1+(c²eT/2I)/(2πR)² )r^ Lor F= (mag F)( 1+((c/2πR)²eT/2I) )r^ So, find |v| : (mag F)/e = |B| |v| sin[θ] .r^ |v|=|mag F|./(e|B| |sin[θ] | ) |v|= [ |Lor F| /|(1+ ((c/2πR)²eT/2I) )r^| ] /(e|B| |sin[θ] | ) Drop the absolute value signs & put |Lor F|=F; |B| =B, and put c=N(L/T)=N(2πR)/T Then: v= [ F /( 1+(N/T)²eT/2I ) ] / [e|B| |sin[θ] | ] v= [(F/e) /( 1+N²/2) ] / |B| , assuming I=e/T and sin[θ] =1. This is v according to the Lorentz force model. Example: Presuming NL>L ,let N=3 & then: v=(F/e) / 5.5B. => F=5.5*Bev = 5.5 mag F F=5.5 * (elec F) /( (k/k’)(eT/I)/(2πR)² ) F=5.5 * I *(2πR)² *(elec F) /( (c²/2)(eT) ) F=5.5 * (I /eT) *8*(πR/c)² *(elec F) F=11 * (I /(e/T) ) *(elec F) F=11 *(elec F) So, the usual classical Coulomb model to get v could involve a MASSIVE underestimate of the total force and BINDING ENERGY involved !! Applied to your example, the actual speed would be about 3.3 times greater !!!
i was watching a previous video about conversation of energy and it was good and it was from 2015 keep up the good work and your videos are useful and helpful thanks for these videos :)
So now we need to know how the mass of the electron was discovered, calculated - it was likely not weighed - so we may have some very circular calculations and answers going on here. When at the end the speed of the electron in orbit about the proton nucleus was declared to be almost 5 million miles an hour - well that wasn't boring but it was also absolutely not believable - good night sweet prince Bohr, YOU BLEW IT.
Do you have any idea how difficult it is to find this calculation done online? Thank you! And how would we compensate for the assumption that the proton isn't going to stay perfectly stationary as the electron orbits as the force acts on it, as well?
You could use the derivation of binary orbits, but replace gravity with electrostatics. This is oversimplified, and neglecting quantum mechanics, and would only apply to atomic hydrogen. hyperphysics.phy-astr.gsu.edu/hbase/orbv.html#bo What would happen is that they both would co-orbit the center of mass of the two particles, which is a lot closer to the proton than to the electron.
@@PrintEngineering Use the formula for the energy of the electron in the Hydrogen atom, E = -Re/n^2, where Re is the Rydberg energy constant. Set this equal to the sum of EPE + KE, which will be negative for a stable orbit. From the same reasoning in this problem, you will have enough information to solve for the Bohr radius at all orbits. The limitations are that this is only valid for hydrogen, and unrealistic 1-electron ions of other elements. The multibody problem, plus lots of other quantum effects, become a problem, when solving it for other elements.
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"Not even remotely close to scale" haha, that got me
Glad you appreciated it. I thought it was subtle enough, but still an important thing to realize.
The Coulomb model is for Electrostatic force (key word: STATIC).
Here the charge is DYNAMIC, so there is a "current" (relatively speaking).
If there is a current, it will generate a magnetic field.
The trajectory of the electron could be considered to be a coil.
- What if you modeled the electron and nucleus as if they were electro-magnets?
Nucleus charge is +e;
electron charge is -e;
mag F on "-e" charge due to "+e"
= -e(v x B)
mag F from "-e" over one loop
=IL∑dB , L= length of loop
=IL. ∑(k'. I.dLsin[θ]/R²)r^ , where k'= μ/(4π )
=(e.fL.k’ I) ∑ (dL/(R²+y²) )r^ , where f= electron orbital frequency.
But y=0 , R=const
=> mag F=(e.fL.k’ I) (2πR/R²)r^
=> mag F=(e.fL.k’ I) (2π /R)r^
mag F=μ(e.f) I (L/2R) r^ .
(static) elec F= k(e/R)² r^
|elec F|/ |mag F|= k(e/R)² /(2πe.fL.k’ I /R) =(k/k’)(eT/I)/(2πR)² , using L=2πR.
where:
r^=unit vector from nucleus to electron,
I= 'nucleus current' wrt static electron,
f=I/e=electron orbital frequency.
Lorentz F= Lor F=mag F+ elec F= (mag F)( 1+(k/k’)(eT/I)/(2πR)²)r^
k=1/4πε =c² . [μ]/10⁷ and k'= μ/2π
- therefore (using μ/[μ]= μ' =4π/10⁷ ):
k/k'=2πc² . (1/10⁷) /μ' =c²/2
and: Lor F= (mag F)( 1+(c²eT/2I)/(2πR)² )r^
Lor F= (mag F)( 1+((c/2πR)²eT/2I) )r^
So, find |v| :
(mag F)/e = |B| |v| sin[θ] .r^
|v|=|mag F|./(e|B| |sin[θ] | )
|v|= [ |Lor F| /|(1+ ((c/2πR)²eT/2I) )r^| ] /(e|B| |sin[θ] | )
Drop the absolute value signs & put |Lor F|=F; |B| =B,
and put c=N(L/T)=N(2πR)/T
Then: v= [ F /( 1+(N/T)²eT/2I ) ] / [e|B| |sin[θ] | ]
v= [(F/e) /( 1+N²/2) ] / |B| , assuming I=e/T and sin[θ] =1.
This is v according to the Lorentz force model.
Example: Presuming NL>L ,let N=3 & then: v=(F/e) / 5.5B.
=> F=5.5*Bev = 5.5 mag F
F=5.5 * (elec F) /( (k/k’)(eT/I)/(2πR)² )
F=5.5 * I *(2πR)² *(elec F) /( (c²/2)(eT) )
F=5.5 * (I /eT) *8*(πR/c)² *(elec F)
F=11 * (I /(e/T) ) *(elec F)
F=11 *(elec F)
So, the usual classical Coulomb model to get v could involve a
MASSIVE underestimate of the total force and BINDING ENERGY involved !!
Applied to your example, the actual speed would be about 3.3 times greater !!!
Damn that a lot of work which I do not understand
Thank you for making this video. It helped me grasp the concept quickly and saved me a lot of time.
i was watching a previous video about conversation of energy and it was good and it was from 2015 keep up the good work and your videos are useful and helpful thanks for these videos :)
Just so you know, this physics has not changed a whole lot in 5 years. Actually, I do not think it has changed at all.
Sir, are there any videos about electric fields?
Great video Mr. P! It was a very enjoyable video and a really cool problem!
Thanks!
You just helped me solve my homework problem thank you!!! It seemed so difficult but you made it seem so simple, thank you again!!!
You are welcome!
So now we need to know how the mass of the electron was discovered, calculated - it was likely not weighed - so we may have some very circular calculations and answers going on here.
When at the end the speed of the electron in orbit about the proton nucleus was declared to be almost 5 million miles an hour - well that wasn't boring but it was also absolutely not believable - good night sweet prince Bohr, YOU BLEW IT.
Do you have any idea how difficult it is to find this calculation done online? Thank you!
And how would we compensate for the assumption that the proton isn't going to stay perfectly stationary as the electron orbits as the force acts on it, as well?
You could use the derivation of binary orbits, but replace gravity with electrostatics. This is oversimplified, and neglecting quantum mechanics, and would only apply to atomic hydrogen.
hyperphysics.phy-astr.gsu.edu/hbase/orbv.html#bo
What would happen is that they both would co-orbit the center of mass of the two particles, which is a lot closer to the proton than to the electron.
Thank u for saving my physics grade
You saved your physics grade. I just provided tools to help you.
Congrats!
How can we calculate the second orbit? Is it the same process?
I'm also very interested in this topic for *all* orbits
@@PrintEngineering Use the formula for the energy of the electron in the Hydrogen atom, E = -Re/n^2, where Re is the Rydberg energy constant. Set this equal to the sum of EPE + KE, which will be negative for a stable orbit. From the same reasoning in this problem, you will have enough information to solve for the Bohr radius at all orbits.
The limitations are that this is only valid for hydrogen, and unrealistic 1-electron ions of other elements. The multibody problem, plus lots of other quantum effects, become a problem, when solving it for other elements.
Would you still arrive at the same answer if you plugged in the formula for centripetal force or mv^2/r instead of v^2/r?
yes sure because you actually get mv^2/r anyways.
You're a awesome teacher
Congratulations Mr P for 50k subscribers
Thank you. It's been a long 7 years. But totally worth it.
Thank you so much for this
Greattttttt videoooo!!!
You are a savior!!!! 😂💝
amazing
The nerd one is my favorite
Looking for oil, struck gold
Wonderful!
Nice hair
You look like the transgender version of the dude from The Big Bang Theory.