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
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.
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
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 :)
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!
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.
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.
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.
@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?
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.
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?
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.
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?
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|.
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.
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.
You don't really have to integrate because conceptually you're adding up the all the dl's. But if you want to integrate it, you can change the dl to a(d(theta)) and then your integral is the S a(d(theta) with limits from 0 to pi. The "S" is an integral sign and "a" is the radius of the half circle and is a constant.
Really wish you were able to evaluate that integral... I'm disappointed because this looked like it was going to be a thorough proof coming from part 1. It's almost all for nothing if a step gets skipped
look up integration by "trigonometric substitution." that particular integral is actually quite common. you basically allow the term raised to the three halves power to be written as the sine or cosine of an angle. Youll then solve the integral in terms of an angle (dθ)
The intergral, using a trig. substitution. Let y = x tan t, dy = x sec^2(t) dt solves to: (1/x^2) * y/sqrt(x^2 + y^2) + C, since tan t = y/x. Now how the hell he reaches the result is anybody's guess. He must have simply copied it from somewhere.
Superheros don't have to wear capes, kids.
You're right
You're right
yes, because they are busy to teach physics
Lots of textbooks fail at explaining this as concisely as you did. Well done and thank you very much!
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.
Looked all over the internet for this simply explained. This is great. Thank you!
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
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.
Sir, you have no idea how much you are saving my life
Best explaination if heard so far, thank you very much my captain
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
Your explanations are terrific!
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 :)))
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!
You are a Great Teacher!! Im gonna ace my exams thanks to this video!!
you're my new favorite online tutor! thank you
Thanks a million, I was studying for an exam and really needed the explanation! Why can't professors do this???
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.
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
What you do bro is deep...you have no idea...knowledge is indeed power
wow i wish i would have found this channel in the beginning of the semester! now i have my final in 2 days..
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'RE AN ACTUAL LIFE SAVER OMFG THANK YOU
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?
you are an amazing person. Thanks for sharing your knowledge
I love your calm voice. Thanks for your help!
Good god man, this is helping me pass my college electromagnetism class XD
@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?
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.
Thanks from India!
2010 to 2020, this is gold!
thank you sir,you helped me understand the biot-savard law within 17 min.
I've got a question. How do you use the law of Biot-Savart on a toroidal coil?
Excellent, but how did we come up or how did he concluded it ?bio savarts law?
So what is it that sin(theta) matters for an electric field but not for a magnetic field?
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 do you go from the integral to the final result?
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?
Superb sir.. Well done👌👍
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.
Does the distance “a” have to be perpendicular to I?
So happy to have found this! Thank you :)
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
0:12 instead of dB there should be B
what about straight wire? on both side of hemi circle
My professor and my textbook uses B for magnetic flux density. We use H for the magnetic field.
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|.
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.
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.
Sen yaşayan efsanesin = U re living legend
How'd he do that integral?
to solve integral, say y=a*tan(phi) and dy=a*dphi/cos^2(phi)
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
Excellent teaching technique
I have my college exam on this tmrw. This video = awesome.
Integration constant due to indefinite integration?
This made my life loads easier, good sir!
Wouldn't B = 0 for the full circle at the middle
Thank You!!! Helped me alot!
sir, you explain things so beautifully and easily. thank you so much.
thank you, you explain eveything clearly and briefly
very nice but can some one tell me that it is for 2 coditons??
is (mu /2pi) (I/a) correct as well for B ?
A best explanation ..thank you✨️
so is it half of a full circle quantitiy
I have the AP Physics C exam in like 2 hours. Thanks!
Thnk U very Much......U dont know how much this will help me....Thanks again
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 makes what was explained to me in a very unintuitive way, seem a lot more intuitive.
How you got a'2 (I am really asking)
hoca büllügünü yerim sagol ya hakketen anlatyorsun.
Literally I thinking why should I go to University? When guys like him over internet, explains everything so clearly....
i really love your videos. appreciate
That sir was excellent. Thank you
Thank you so much! Simple and effective!
woah, its a rly cool tutorial doode, so pretty helping for my exam 😂
how do you integrate this?
You don't really have to integrate because conceptually you're adding up the all the dl's. But if you want to integrate it, you can change the dl to a(d(theta)) and then your integral is the S a(d(theta) with limits from 0 to pi. The "S" is an integral sign and "a" is the radius of the half circle and is a constant.
beautifully explained thank You very much
Thanks! I finally understand this law!
Thank you, sir, for helping me work on my thesis
Thank u sir u deserve more followers
thank you so much all your videos are fantastic
Totally dig your style.
you ARE my physics teacher
thanks, you are saving my life!
why does sin theta equal to one?
No matter where your dl is in the semi circle the theta is always equal to 90 degrees.
And when u put sin 90 into a calculator u will get 1.
Sir Meowington thanks for the detailed explanation.
Sen ADAMSIN. Allah'ına kurban...
Really wish you were able to evaluate that integral... I'm disappointed because this looked like it was going to be a thorough proof coming from part 1. It's almost all for nothing if a step gets skipped
This video was awesome! Thank you so much.
One of the best, thank you!
muy buena explicación, me ayudo mucho!
good job dude thanks for making understand MUCH BETTER
can anyone tell me how to solve the intergral????
look up integration by "trigonometric substitution." that particular integral is actually quite common. you basically allow the term raised to the three halves power to be written as the sine or cosine of an angle. Youll then solve the integral in terms of an angle (dθ)
The intergral, using a trig. substitution.
Let y = x tan t, dy = x sec^2(t) dt
solves to:
(1/x^2) * y/sqrt(x^2 + y^2) + C, since tan t = y/x. Now how the hell he reaches the result is anybody's guess. He must have simply copied it from somewhere.
I missed my class but after i watch this i get it now
Awesome! This helped me a lot.
really thank u !! from KSA !
I think you probably want to use Ampere's law on a toroidal coil. It is much, much easier. I have a youtube on this.