sir i really want to meet you but you are almost 90 and i am just 16 and i know that i don't have any right to do this but still may god give you a healthy 5000 years of life so that i can take blessing from you
Dear Dearest Professor I love you so much and can't thank you enough for your amazing closed-circuit videos from the 90s at MIT - they greatly helped me finish my research!
Thank you for the new problem, Professor! Will definitely give it a try:) Could you please make the next problem something based on your 8.01 lectures? That's more my syllabus and I'd love to rack my head with my current knowledge on one of your problems without any external help:)) Thank you again! Cheers and take care:)
Very interesting problem! I get x=2/B*sqrt(2V/q)*(sqrt(m2)-sqrt(m1)), m1=mass of U235 ion, m2=mass of U238. I get x=4.45cm, the radii of the circles are roughly 3.5m. I found images of these calutrons in the web, but they seemed to have a total size of maybe 4m instead of 2x3.5=7m. Or did they use different voltages/B-fields than indicated in your problem? I certainly will doublecheck my result.
Good morning, I always enjoy seeing your lectures. I got a laugh at your glasses, kudos to you for just being yourself. Please vary them whenever possible, they add to the flavour and experience of your lectures. I did gas chromatography design for real time engineering oil field applications, this lecture is a good addition to learn from. Cheers from Canada
@lecturesbywalterlewin.they9259 I appreciate your compliment, I only became a good engineer because of good teachers guiding me, my children are engineers as well and we all appreciate good teachers. Keep it up professor.
Professor Lewin, could you explain how time dilation near a black hole, as depicted in Interstellar, could practically affect human perception and survival? Is the depiction of time slowing down realistic ?
x = 2.22cm. The Lorentz force is equated to the centrifugal force in the rotating frame of the U ion. Hence, the Radius of curvature R = (sqrt(2eVm))/eB). The difference x in the radii of curvature of the U-238 and U-235 ions can thus be found.
Sir please can you tell me what is the god partical and please can you suggest me best books of physics related to quantum mechanics. Thank you so much sir 👍
Hey Dr. Lewin new viewer of yours from Arizona. I'm going to watch your 8.02 lecture on this topic and will attempt this problem. I just finished university physics: newtonian mechanics and passed with an A and now I'm starting electromagnetism. My Professor Casey Durandet has us watch all your lectures which is how I found out about you. I also wanted to ask, after I get a solution how do I send it to you?
Sir please tell me how to focus in study can you tell me the tips I always watching and playing games in phone and distract in my work I am class 10 ICSE
Hello Professor Lewin, wishing you a happy new year! Let q: electric charge of atom, V: potential difference, m: mass of atom, v: velocity of atom when entering magnetic field, B: magnetic induction of the B-field, R: radius. CLASSICAL Consider one singly ionized U atom. The electric potential energy created by the potential difference over which the atom is accelerated is converted to kinetic energy: qV = mv^2/2 -> v = sqrt(2qV / m). Next the atom enters the magnetic field and a Lorentz force acts upon the particle. Since the magnetic field is uniform with constant inductance and the Lorentz force is perpendicular to the particle's velocity, the Lorentz force acts as a centripetal force: qvB = mv^2/R -> R = mv / qB -> R = sqrt(2mV / qB^2). Since m(238) > m(235), R(238) > R(235) (intuitively, the U238 atoms have a higher inertia than U235 atoms hence also a larger radius): x = R(238) - R(235) -> x = sqrt(2V / qB^2) * (sqrt(m(238)) - sqrt(m(235))). Numerically, q = e = 1.6*10^-19 C (since only one electron was removed), V = 250 V, B = 0.01 T, m(238) = 92m(p) + 146m(n) + 91m(e) = 92*1.673*10^-27 + 146*1.675*10^-27 + 91*9.109*10^-31 kg = 3.9854889*10^-25 kg, m(235) = 92m(p) + 143m(n) + 91m(e) = 92*1.673*10^-27 + 143*1.675*10^-27 + 91*9.109*10^-31 kg = 3.9352389*10^-25 kg. Therefore, after rounding two two decimals, x = 2.23 cm. RELATIVISTIC Energy conservation states qV = (γ-1)mc^2 -> γ = qV/mc^2+1 with velocity v = c*sqrt(1-1/γ^2). Next, the Lorentz force acts as a centripetal force when entering the magnetic field: qvB = γmv^2/R -> R = γmv / qB -> R = m/qB * (qV/mc^2+1) * c*sqrt(1-1/( qV/mc^2+1)^2). Obviously, x = R(238) - R(235) -> x = m(238)/qB * (qV/m(238)c^2+1) * c*sqrt(1-1/( qV/m(238)c^2+1)^2) - m(235)/qB * (qV/m(235)c^2+1) * c*sqrt(1-1/( qV/m(235)c^2+1)^2).Note, however, that γ(238) and γ(235) are extremely close to one (approx. 1.000000001), so there is no need for relativistic corrections. By the way, the individual radii are R(238) = 3.529115018 m and R(235) = 3.506796481 m. Evidently, here we also obtain x = 2.23 cm.
Lewin prooblem 221 - mass spectrometer separation of singly ionized U235 and U238
U235 U238 mass/ kg 3.90E-25 3.95E-25 charge, C 1.60E-19 1.60E-19 B, Tesla 0.1 0.1 voltage 250 250 KE, joules = charge x voltage = m*v^2/2 4.00E-17 4.00E-17 v, m/s 14322.29748 14231.36134 Magnetic force F = qvB newtons 2.29E-16 2.28E-16 magnetic acceleration, a = F/mass, m/s^2, = (v^2)/R 5.88E+08 5.76E+08 R = v^2/a, meters 3.49E-01 3.51E-01 diameter, meters 6.98E-01 7.03E-01
x = difference in diameters, meters 4.46E-03 x = difference in diameters, cm 4.46E-01
sir i really want to meet you but you are almost 90 and i am just 16 and i know that i don't have any right to do this but still may god give you a healthy 5000 years of life so that i can take blessing from you
Dear Dearest Professor I love you so much and can't thank you enough for your amazing closed-circuit videos from the 90s at MIT - they greatly helped me finish my research!
You're very welcome
This is just incredible at your age. You deserve a nobel prize for the best physics professor in the world. Lots of love from Kenya 🇰🇪
Wow, thank you!
You are a legend, thanks for everything Mr. Lewin:)
You are very welcome
We got a profile picture change! Missed seeing the videos prof!
Thank you for noticing!
Your determination and your love for physics and teaching is just flawless at this age ... Hats off you sir... lots of love ❤
So nice of you
Thank you for the new problem, Professor! Will definitely give it a try:) Could you please make the next problem something based on your 8.01 lectures? That's more my syllabus and I'd love to rack my head with my current knowledge on one of your problems without any external help:))
Thank you again!
Cheers and take care:)
many of my problems are about Newtonian Mechanics which is covered in my 8.01 lectures
Sir, Thank you from the core of my heart, ❤❤❤❤❤❤❤❤
Welcome Prof!
Very interesting problem! I get x=2/B*sqrt(2V/q)*(sqrt(m2)-sqrt(m1)), m1=mass of U235 ion, m2=mass of U238. I get x=4.45cm, the radii of the circles are roughly 3.5m. I found images of these calutrons in the web, but they seemed to have a total size of maybe 4m instead of 2x3.5=7m. Or did they use different voltages/B-fields than indicated in your problem?
I certainly will doublecheck my result.
Good morning, I always enjoy seeing your lectures. I got a laugh at your glasses, kudos to you for just being yourself. Please vary them whenever possible, they add to the flavour and experience of your lectures.
I did gas chromatography design for real time engineering oil field applications, this lecture is a good addition to learn from. Cheers from Canada
You must be a very talented engineer.
@lecturesbywalterlewin.they9259 I appreciate your compliment, I only became a good engineer because of good teachers guiding me, my children are engineers as well and we all appreciate good teachers. Keep it up professor.
Professor Lewin, could you explain how time dilation near a black hole, as depicted in Interstellar, could practically affect human perception and survival? Is the depiction of time slowing down realistic ?
Thanks sir for making us love physics again
my pleasure
Wow. Great to see you.
Congrats professor 🎉😮😮
Thank you! 😃
x = 2.22cm. The Lorentz force is equated to the centrifugal force in the rotating frame of the U ion. Hence, the Radius of curvature R = (sqrt(2eVm))/eB). The difference x in the radii of curvature of the U-238 and U-235 ions can thus be found.
Hello professor, greetings to you, you are a fan of physics from Iran❤❤😊
Hey, thanks!
Sir please can you tell me what is the god partical and please can you suggest me best books of physics related to quantum mechanics. Thank you so much sir 👍
the god particle is the Higgs Boson. It was predicted already in the sixties and discovered with the Hadron Collider. Google *Higgs Boson*
Hey Dr. Lewin new viewer of yours from Arizona. I'm going to watch your 8.02 lecture on this topic and will attempt this problem. I just finished university physics: newtonian mechanics and passed with an A and now I'm starting electromagnetism. My Professor Casey Durandet has us watch all your lectures which is how I found out about you. I also wanted to ask, after I get a solution how do I send it to you?
You'll have to figure that out on your own
@@lecturesbywalterlewin.they9259 just sent you my solution
Sir please tell me how to focus in study can you tell me the tips I always watching and playing games in phone and distract in my work
I am class 10 ICSE
eat yogurt every day but *never on Fridays* that also worked for Einstein and for me.
Sir, I read somewhere that whatever we think is happening somewhere in this universe. Do you believe in this? 🙃
question unclear
Hello Professor Lewin, wishing you a happy new year!
Let q: electric charge of atom, V: potential difference, m: mass of atom, v: velocity of atom when entering magnetic field, B: magnetic induction of the B-field, R: radius.
CLASSICAL
Consider one singly ionized U atom. The electric potential energy created by the potential difference over which the atom is accelerated is converted to kinetic energy: qV = mv^2/2 -> v = sqrt(2qV / m). Next the atom enters the magnetic field and a Lorentz force acts upon the particle. Since the magnetic field is uniform with constant inductance and the Lorentz force is perpendicular to the particle's velocity, the Lorentz force acts as a centripetal force: qvB = mv^2/R -> R = mv / qB -> R = sqrt(2mV / qB^2). Since m(238) > m(235), R(238) > R(235) (intuitively, the U238 atoms have a higher inertia than U235 atoms hence also a larger radius): x = R(238) - R(235) -> x = sqrt(2V / qB^2) * (sqrt(m(238)) - sqrt(m(235))).
Numerically, q = e = 1.6*10^-19 C (since only one electron was removed), V = 250 V, B = 0.01 T, m(238) = 92m(p) + 146m(n) + 91m(e) = 92*1.673*10^-27 + 146*1.675*10^-27 + 91*9.109*10^-31 kg = 3.9854889*10^-25 kg, m(235) = 92m(p) + 143m(n) + 91m(e) = 92*1.673*10^-27 + 143*1.675*10^-27 + 91*9.109*10^-31 kg = 3.9352389*10^-25 kg. Therefore, after rounding two two decimals, x = 2.23 cm.
RELATIVISTIC
Energy conservation states qV = (γ-1)mc^2 -> γ = qV/mc^2+1 with velocity v = c*sqrt(1-1/γ^2). Next, the Lorentz force acts as a centripetal force when entering the magnetic field: qvB = γmv^2/R -> R = γmv / qB -> R = m/qB * (qV/mc^2+1) * c*sqrt(1-1/( qV/mc^2+1)^2). Obviously, x = R(238) - R(235) -> x = m(238)/qB * (qV/m(238)c^2+1) * c*sqrt(1-1/( qV/m(238)c^2+1)^2) - m(235)/qB * (qV/m(235)c^2+1) * c*sqrt(1-1/( qV/m(235)c^2+1)^2).Note, however, that γ(238) and γ(235) are extremely close to one (approx. 1.000000001), so there is no need for relativistic corrections. By the way, the individual radii are R(238) = 3.529115018 m and R(235) = 3.506796481 m. Evidently, here we also obtain x = 2.23 cm.
Sir i am a jee aspirant currenrly in 10th grade i love watching your videos.on physics where should i start from to build a strong base
12 seconds ago is crazy
Hello sir today I again come how are you all is fine😂😂🇮🇳🇮🇳
Respect your efforts sir❤❤❤
Thank you for the kind words!
Lewin prooblem 221 - mass spectrometer separation of singly ionized U235 and U238
U235 U238
mass/ kg 3.90E-25 3.95E-25
charge, C 1.60E-19 1.60E-19
B, Tesla 0.1 0.1
voltage 250 250
KE, joules = charge x voltage = m*v^2/2 4.00E-17 4.00E-17
v, m/s 14322.29748 14231.36134
Magnetic force F = qvB newtons 2.29E-16 2.28E-16
magnetic acceleration, a = F/mass, m/s^2, = (v^2)/R 5.88E+08 5.76E+08
R = v^2/a, meters 3.49E-01 3.51E-01
diameter, meters 6.98E-01 7.03E-01
x = difference in diameters, meters 4.46E-03
x = difference in diameters, cm 4.46E-01
Hello sir I'm a big fan from India ❤ love yr lectures
Thanks Sir 👍
Approximately 2cm.
2.22 to be more precise.
x = 2.1 cm
Sir i admire you are lot and always love your lectures ❤❤❤❤❤ from 🇮🇳 🇮🇳 🇮🇳 india
keep it up
I struggle with physics what should i do love from india ❤
eat yogurt every day but *never on Fridays* that also worked for Einstein and for me
@@lecturesbywalterlewin.they9259what is so special about Friday???