My dude. I sat a physics exam at university two years ago on electromagnetism. We had a poorly written paper and although the Biot Savart Law was only a small part of the syllabus, it came up on a quarter of the small questions and on two out of three of the long questions. I watched this video the night before. Thank you!
Some people teach you things without expecting anything in return. Thank you so much sir for such a simple experiment. In this lock down, these videos are really helpful against stupid online classes
Just wrote my PHY exam today (15 Nov 2018)... I am pretty confident that I did good and it is all thanks to you for taking your time to make these videos... And me taking the time to watch and listen to you. lol
Wish textbooks/professors could/would explain things in this manner, so many get way over bogged down in theory and formulas when almost everything can be explained in a way most people with common sense can understand. Currently studying EE, appreciate the video.
I am in a high school in India and this 11 year old video explained me law of Biot-Savart better than our syllabus DEDICATED youtube channels in 2021. Thankyou so much.
I saw in a book that you can solve the first integral with a little bit of trigonometrics ( no substitution) which leads to a intregal over the angle. Thanks for the good explanation!
Good sir, I give you all of my thanks. You're videos have helped me far more than I can express. You are an excellent teacher and I can only hope that other students like myself can utilize this wonderful resource you have provided for the general public.
Very helpful, but there is a way for this integral to be solved for those who want to know, you have to get a pen and write down this to fully understand : first you consider the ((perpendicular))* distance from a point P to the wire is a (as in the video) The distance from the origin to the the element of length equal to y and the distance r from the point P to the wire makes an angle theta (ø) with the wire ...all that makes a triangle , //you drew that!? :~D so we can get that: tan(ø) =a/y And so y=a/tan(ø) ,that is the same as y=a cot(ø)->* Then we can get dy=-a csc^2 (ø) dø->* Finally back to the integral (Without excluding ( a ) out ! ) You substitute the y and the dy The lower part of the fraction would be = a^3 csc^3 (ø) And the upper part: a (-a csc^2 (ø) dø) You can cancel terms till you get 1/csc (ø) And that integral to your surprise is just Į Sin(ø) from ø1 to ø2 :)) Which is solved easily. Cos(Ø2)-Cos (Ø1) and the minus canceled with the one that is earlier mentioned beside the a , in the case of this video since the wire is infinite ø1=pi and ø2=zero So we are left with “2 “ from the integration that is divided by the outside 4pi giving the 1/2pi in the solution written in the video This solution i knew from my professor at university so thanks to him :)) Hope that could help :)
Hi Prof, would love to make a donation, if u got patreon or anything? My lecturer in university fails to explain this as concisely as u did, despite having a greater luxury of time.
You cannot easily use Ampere's law for this because there is no Amperian loop that can be drawn where where the B's are parallel to the dl's and the B's magnitude is uniform at all points on the Amperian loop. Even if it were a full circle, the statement above would still apply.
Aside from exams, i wanted to know how this is applied technologically speaking... if a wire has a moving field coming off of it what can we do with it? Or about it, since this could be considered loss right?
For the first one the integral is complicated, there is an easier way. If you consider dy=a/cos²(Th)dTh you have then dB= µ0*I/(4pia)* cos(th)dth you integrate between -pi/2 pi/2 and you got the same result.
@lasseviren1 Would the formula we derived around 1:30 work at any point in space? Say, what if we picked a point that is that is above the wire and beyond it's length?
for some teachers/ whoever correct the exam...they want to know the direction of your Field. it's really important to establish the axis and directions just so you won't lose any marks/points. or if they really just want to know the magnitude. put an absolute value |B|.
Isn't the indefinite integral of 1/(y^2 +a^2) dy just 1/a arctan(y/a)? But then how would that simplify when you substitute in -infinity and infinity for y?
Hey, real quick, this is referring to the law in a 1d sense. I'm attempting to work with the law in 3D. In the 3D version of the law, you have to integrate the intergrals for the x, y, and z planes. Once you've done that, you have to combine them. I was wondering if when they are telling you to integrate along the different planes, are they saying to integrate 0 to 2π for the z-axis, and 0 to π for the other two? Keeping in mind that the shape of my electromagnet is a hemisphere.
When you're taking the integral, dl, you are trying to sum up all the dls throughout. For a circle, dl, would be the circumference as you're adding up every little element through the entire circumference, giving you 2*pi*r. In this case, we are only adding up half a circle, giving pi*r, but in this case r=a so we get pi*a. You wouldn't integrate from 0 to pi, as that relates to the angle, and we're not looking at dtheta, we are looking at dl. That's a very dodgy explanation of why it's not 0 to pi, but I hope it somewhat helps (=
I can see the reasoning in this case, but how would you apply it generally? What if the path is not a circle. There is a step he skipped, but I can't figure out what it was.
Ali Etezadkhah integral of dl is l, which is the length of the path (in this case circle). You would need to set up the integral if the path length can't be computed directly like this
+Quientin Morrison oww, i see. so it is about following the direction of the current using the RHR. thanks. but i have a question. i encountered a problem... it says, there is 3 segments which is segment A, B and C. segment A forms a quarter circle in quadrant 4. segment B along the x-axis and segment C forms also a quarter circle but in quadrant 2.. the current starts at the segment A .. when i follow the RHR, the direction for A is away from the board, C is into the board and B is zero bec it is along the origin.. is it possible to get different directions?
+Denver Daño I guess it would depend on how he graph looked.. its hard to describe these things with words.... If both quarter circles are rotating counter clockwise with regards to the current than both should be into the board. Either orientation is possible. For the straight wire, remember to still do the RHR, one side will show the magnetic field as pointing into the board and the other out, you end up seeing the same relationship for a coil of wire or a solenoid, but the magnetic field outside the wire is usually not as important, as the magnitude tends to be less.
Lots of textbooks fail at explaining this as concisely as you did. Well done and thank you very much!
Superheros don't have to wear capes, kids.
You're right
You're right
yes, because they are busy to teach physics
My dude. I sat a physics exam at university two years ago on electromagnetism. We had a poorly written paper and although the Biot Savart Law was only a small part of the syllabus, it came up on a quarter of the small questions and on two out of three of the long questions. I watched this video the night before. Thank you!
Aerospace Engineering student and your explanation helped way more than my professor's. Kudos!
You are my real physics professor
You sir , deserve more subscribers.
Been tryin to understand this for a month till i stumble upon this video, this was a big help, major league!!! Keep doing what you do.Merci.
***** just realised. He should though, he's goood at this
Some people teach you things without expecting anything in return. Thank you so much sir for such a simple experiment. In this lock down, these videos are really helpful against stupid online classes
Looked all over the internet for this simply explained. This is great. Thank you!
Best explaination if heard so far, thank you very much my captain
Sir, you have no idea how much you are saving my life
Just wrote my PHY exam today (15 Nov 2018)... I am pretty confident that I did good and it is all thanks to you for taking your time to make these videos...
And me taking the time to watch and listen to you. lol
Wish textbooks/professors could/would explain things in this manner, so many get way over bogged down in theory and formulas when almost everything can be explained in a way most people with common sense can understand. Currently studying EE, appreciate the video.
You are a Great Teacher!! Im gonna ace my exams thanks to this video!!
Thanks a million, I was studying for an exam and really needed the explanation! Why can't professors do this???
I am in a high school in India and this 11 year old video explained me law of Biot-Savart better than our syllabus DEDICATED youtube channels in 2021. Thankyou so much.
Jee aspirant?
Thanks from Turkey!
Ben de ben de
Emre kayan ben şimdi çıktım sınavdan. bir soru vardı 20 puanlık, bu video sayesinde çözdüm :D
Bugun 5te de benim sinavim var hadi bakalim hayırlısı
I saw in a book that you can solve the first integral with a little bit of trigonometrics ( no substitution) which leads to a intregal over the angle. Thanks for the good explanation!
Your explanations are terrific!
you're my new favorite online tutor! thank you
wow i wish i would have found this channel in the beginning of the semester! now i have my final in 2 days..
Good sir, I give you all of my thanks. You're videos have helped me far more than I can express. You are an excellent teacher and I can only hope that other students like myself can utilize this wonderful resource you have provided for the general public.
What you do bro is deep...you have no idea...knowledge is indeed power
Very helpful, but there is a way for this integral to be solved for those who want to know,
you have to get a pen and write down this to fully understand : first you consider the ((perpendicular))* distance from a point P to the wire is a (as in the video)
The distance from the origin to the the element of length equal to y and the distance r from the point P to the wire makes an angle theta (ø) with the wire ...all that makes a triangle , //you drew that!? :~D so we can get that:
tan(ø) =a/y
And so y=a/tan(ø) ,that is the same as
y=a cot(ø)->*
Then we can get dy=-a csc^2 (ø) dø->*
Finally back to the integral
(Without excluding ( a ) out ! )
You substitute the y and the dy
The lower part of the fraction would be = a^3 csc^3 (ø)
And the upper part:
a (-a csc^2 (ø) dø)
You can cancel terms till you get
1/csc (ø)
And that integral to your surprise is just
Į Sin(ø) from ø1 to ø2 :))
Which is solved easily.
Cos(Ø2)-Cos (Ø1) and the minus canceled with the one that is earlier mentioned beside the a , in the case of this video since the wire is infinite ø1=pi and ø2=zero
So we are left with “2 “ from the integration that is divided by the outside 4pi giving the 1/2pi in the solution written in the video
This solution i knew from my professor at university so thanks to him :))
Hope that could help :)
@William Han glad it helped :)))
YOU'RE AN ACTUAL LIFE SAVER OMFG THANK YOU
Good god man, this is helping me pass my college electromagnetism class XD
I love your calm voice. Thanks for your help!
2010 to 2020, this is gold!
Thank you so much. You help me a lot for my final exam. İ'll look at another videos that are made by you for physics.
thank you sir,you helped me understand the biot-savard law within 17 min.
you are an amazing person. Thanks for sharing your knowledge
Thanks from India!
Hi Prof, would love to make a donation, if u got patreon or anything? My lecturer in university fails to explain this as concisely as u did, despite having a greater luxury of time.
So happy to have found this! Thank you :)
You cannot easily use Ampere's law for this because there is no Amperian loop that can be drawn where where the B's are parallel to the dl's and the B's magnitude is uniform at all points on the Amperian loop. Even if it were a full circle, the statement above would still apply.
Thnk U very Much......U dont know how much this will help me....Thanks again
sir, you explain things so beautifully and easily. thank you so much.
thank you, you explain eveything clearly and briefly
Literally I thinking why should I go to University? When guys like him over internet, explains everything so clearly....
This made my life loads easier, good sir!
Thank you, this makes what was explained to me in a very unintuitive way, seem a lot more intuitive.
I have my college exam on this tmrw. This video = awesome.
oh man, thanksfor this, it helped me a lot. I was always pretty unsure because of that vector product, this sin version makes it easier.
woah, its a rly cool tutorial doode, so pretty helping for my exam 😂
Sen yaşayan efsanesin = U re living legend
Superb sir.. Well done👌👍
A best explanation ..thank you✨️
Aside from exams, i wanted to know how this is applied technologically speaking... if a wire has a moving field coming off of it what can we do with it? Or about it, since this could be considered loss right?
Thank you so much! Simple and effective!
Thank You!!! Helped me alot!
For the first one the integral is complicated, there is an easier way. If you consider dy=a/cos²(Th)dTh you have then dB= µ0*I/(4pia)* cos(th)dth you integrate between -pi/2 pi/2 and you got the same result.
Thanks! I finally understand this law!
Excellent teaching technique
Does the distance “a” have to be perpendicular to I?
So what is it that sin(theta) matters for an electric field but not for a magnetic field?
@lasseviren1 Would the formula we derived around 1:30 work at any point in space? Say, what if we picked a point that is that is above the wire and beyond it's length?
hoca büllügünü yerim sagol ya hakketen anlatyorsun.
I have the AP Physics C exam in like 2 hours. Thanks!
thanks, you are saving my life!
How do you go from the integral to the final result?
That sir was excellent. Thank you
i really love your videos. appreciate
One of the best, thank you!
Excellent, but how did we come up or how did he concluded it ?bio savarts law?
My professor and my textbook uses B for magnetic flux density. We use H for the magnetic field.
Thank you, sir, for helping me work on my thesis
Thank u sir u deserve more followers
what about straight wire? on both side of hemi circle
good job dude thanks for making understand MUCH BETTER
thank you so much all your videos are fantastic
Totally dig your style.
This video was awesome! Thank you so much.
I missed my class but after i watch this i get it now
for some teachers/ whoever correct the exam...they want to know the direction of your Field.
it's really important to establish the axis and directions just so you won't lose any marks/points.
or if they really just want to know the magnitude. put an absolute value |B|.
Isn't the indefinite integral of 1/(y^2 +a^2) dy just 1/a arctan(y/a)? But then how would that simplify when you substitute in -infinity and infinity for y?
Use limitations to find the boundry
Sen ADAMSIN. Allah'ına kurban...
Hey, real quick, this is referring to the law in a 1d sense. I'm attempting to work with the law in 3D. In the 3D version of the law, you have to integrate the intergrals for the x, y, and z planes. Once you've done that, you have to combine them. I was wondering if when they are telling you to integrate along the different planes, are they saying to integrate 0 to 2π for the z-axis, and 0 to π for the other two? Keeping in mind that the shape of my electromagnet is a hemisphere.
How did the integral of dl become pi*a? If you integrate from 0 to pi, the answer should be pi.
When you're taking the integral, dl, you are trying to sum up all the dls throughout. For a circle, dl, would be the circumference as you're adding up every little element through the entire circumference, giving you 2*pi*r. In this case, we are only adding up half a circle, giving pi*r, but in this case r=a so we get pi*a. You wouldn't integrate from 0 to pi, as that relates to the angle, and we're not looking at dtheta, we are looking at dl. That's a very dodgy explanation of why it's not 0 to pi, but I hope it somewhat helps (=
I can see the reasoning in this case, but how would you apply it generally? What if the path is not a circle. There is a step he skipped, but I can't figure out what it was.
Ali Etezadkhah integral of dl is l, which is the length of the path (in this case circle). You would need to set up the integral if the path length can't be computed directly like this
I've got a question. How do you use the law of Biot-Savart on a toroidal coil?
Thank you so much, that made a ton of sense!
very nice but can some one tell me that it is for 2 coditons??
you ARE my physics teacher
is (mu /2pi) (I/a) correct as well for B ?
Great! Tnx! this gonna much help me in my physics Long quiz today
Thank u so much. I got everything. Hope I will be able to cope with it in the exam tomorrow)
Awesome! This helped me a lot.
beautifully explained thank You very much
Wouldn't B = 0 for the full circle at the middle
really thank u !! from KSA !
Integration constant due to indefinite integration?
How you got a'2 (I am really asking)
How'd he do that integral?
So, what is the direction for the magnetic field?
+Denver Daño
Into the board by the RHR (right hand rule).
+Quientin Morrison oww, i see. so it is about following the direction of the current using the RHR. thanks. but i have a question. i encountered a problem... it says, there is 3 segments which is segment A, B and C. segment A forms a quarter circle in quadrant 4. segment B along the x-axis and segment C forms also a quarter circle but in quadrant 2.. the current starts at the segment A .. when i follow the RHR, the direction for A is away from the board, C is into the board and B is zero bec it is along the origin.. is it possible to get different directions?
+Denver Daño I guess it would depend on how he graph looked.. its hard to describe these things with words.... If both quarter circles are rotating counter clockwise with regards to the current than both should be into the board. Either orientation is possible. For the straight wire, remember to still do the RHR, one side will show the magnetic field as pointing into the board and the other out, you end up seeing the same relationship for a coil of wire or a solenoid, but the magnetic field outside the wire is usually not as important, as the magnitude tends to be less.
Thank you!! This video helped me alot!
so is it half of a full circle quantitiy
that was awesome. thank you digital prof :)
Thank you so much! i understood very clear!
You are a hero
Thank you! very informative tutorial :)
to solve integral, say y=a*tan(phi) and dy=a*dphi/cos^2(phi)