I haven't slept coz I have exam tommorow and I know everything but Biot-saviort and Ampere's law but this vid made me understand everything thank u so much sir
@lasseviren1 I am amazed of how simple you are able to explain things. What is your background and how do you manage to break down explanations so elegantly?
I appreciate these videos, but I wonder why you don't define the differential lengths (dL), as differential arc lengths(dS) and then have dS=r*d(theta). Then plug that into the integral with the bounds set from 0 to 2pi. Since r is constant on this loop (circle), this evaluates to 2*pi*r. This way the viewer has a more rigorous understanding of Ampere's Law.
13 years ago.. a little late, but you wouldn't do that because ampere's law is supposed to be simple. Using biot-sarvart's, on the other hand, you might be required to use that approach. There is no need to add unnecessary steps to something that's utilizes symmetry for efficiency.
Thank you for the great video. However, is it possible to rename the video to a more convenient name? I had a bit of difficulty finding part 2. Once again, thank you for the video! It helped a lot.
Its all good but I want to point out that the graph at the end should be linear, not exponential because all the variables in the formula are raised to the power of one.
Words cannot describe how helpful you are.
This might be the best channel for physics on TH-cam
You are amazing! I wish i found your channel sooner
Thank you for your comment. I agree with it and, with your permission, would like to add it to the video in the form of a caption.
Some of the best I've seen on TH-cam (and I've seen my fair share). Thank you so much!
I haven't slept coz I have exam tommorow and I know everything but Biot-saviort and Ampere's law but this vid made me understand everything thank u so much sir
@lasseviren1 I am amazed of how simple you are able to explain things. What is your background and how do you manage to break down explanations so elegantly?
isnt he not making videos anymore??His videos were very simple and clear!
I appreciate these videos, but I wonder why you don't define the differential lengths (dL), as differential arc lengths(dS) and then have dS=r*d(theta). Then plug that into the integral with the bounds set from 0 to 2pi. Since r is constant on this loop (circle), this evaluates to 2*pi*r. This way the viewer has a more rigorous understanding of Ampere's Law.
13 years ago.. a little late, but you wouldn't do that because ampere's law is supposed to be simple. Using biot-sarvart's, on the other hand, you might be required to use that approach. There is no need to add unnecessary steps to something that's utilizes symmetry for efficiency.
@@yeahhh-pn8wi 😂😂 this will help his son probably
so if the point you're finding is inside the wire, just multiply r^3/R^3 to the current to get only the portion of the current going through the wire
this video really helped me to pass my physics exam, thank you very much, greetings from argentina!
if you watched the last video, skip ahead to 3:42
You are great. Thank you for all your great work.
Watching this in 2018. Thanks for your hardwork!
2019
@@avi-ii5qs 2020
@@yr-mrribani9869 2021😂
@@avi-ii5qs 22dec 2023
If I pass Phys102, it will be thanks to you
+Alper. Uçar dostum geçtin mi ve koçta mıydın
+Hsgngr Hsgngr Gectim, be Bilkentteyim
Amazing video, thanks for the help in understanding
Thank you for the great video. However, is it possible to rename the video to a more convenient name? I had a bit of difficulty finding part 2. Once again, thank you for the video! It helped a lot.
What if amperian loop is of irregular shape?
Understood a lot...thnx
wow thank you sir. you did it.
how do u learn this stuff yourself?
how is this different from part 1
Great Explanation...ppl need this..=)
Thank you man..It would have been good if you have take more than one wire
A great way .... Finally its done. I suppose. Thanks
what is the value of 'mu not' ?
'mu not' is equal to 4*pi*10^-7
Very good
thank you actually i got a lot of advantages thank you very much.
Thanks a lot for your videos!!!!!!!!
THANK U SM SIR
You are amazing! :D
Thank you Sir
This video is just a repeat of last video haha. But thank you!
Ur amazing!
@SideShockGame
ah ok thanks.
Thanks!
really thank u !
Its all good but I want to point out that the graph at the end should be linear, not exponential because all the variables in the formula are raised to the power of one.
by any chance are u the guy from kahn academy cause boi do u guys sound the same HAHHAHAH
you should be lazy and write b(dot)dl => b*dl*cos(theta). and since angle between the vectors is 0, then cos0=1.
I agree. It is negligible, but are the viewers aware of it?
More precisely, mu naught = 1/(C^2 Epsilon naught) where C is the speed of light and epsilon naught is the permittivity of free space
apphysicslectures . com /E_M_Videos.html all of the kinds of the videos
This is exactly the same as the fist part. Doesn't add anything new
geçersem aboneyim
Thank you Sir
Thanks!