AP Physics C - Biot Savart Law

A brief introduction to the Biot-Savart Law to determine magnetic fields due to current-carrying wires for students in calculus-based physics courses such as AP Physics C. For more information, please visit APlusPhysics.com.

[@ptyptypty3]

WOW !!.. thank you DAN !!!.. this derivation of the Biot Savart Equation was PERFECT from beginning to end !!.. I must have looked at 10 different Videos on this Subject and YOUR video was by far the BEST and Most Complete!!.. You left nothing to the imagination . You always come through on your topics!!!.. so appreciative.... Thank you!!

[@DanFullerton]

Thanks so much Philip -- great to hear. I think this is, by far, the hardest topic in E&M, and putting this video together was a long, long, long project for what amounts to a short little video. Thrilled to hear someone found benefit from it other than myself! :-D

[@wanjapaulsen797]

This video helped me so much, none of my textbooks explained sample 2 like you did. Thank you so much! :)

[@DanFullerton]

+Wanja Paulsen Glad it helped. I think this is by far the hardest part of AP Physics C (and I go and review my notes repeatedly every year before giving this lecture!)

[@nunhlimapeecee]

+Wanja Paulsen love u so much

[@Ollie2220]

This is really good, really explained all the bits that I didn't understand! Thank you for going through it slowly!

[@DanFullerton]

My pleasure!

[@harryli7557]

you explained it way better than my prof did!

[@MurasakiBubble]

Great explanation! I really like the usage of color to show where the variables came from! Being a visual person it helped so much! Thank you!

[@DanFullerton]

My mind works the same way. You're very welcome!

[@JeRRY-xw6xj]

I freaking love you. All your videos have helped me out so much.

[@DanFullerton]

So glad to hear it Gerardo! Make it a great day!

[@parveengoyal4566]

sir you are a great teacher these videos of yours help me a lot and improve my concepts thanks a lot

[@DanFullerton]

+parveen goyal Thrilled to hear you enjoyed them!

[@2C00L2LIVE]

do you have a video of applying ampere's law to toroids?

[@jackysub1941]

@12:36, shouldn't the angle be from theta equals pi/2 to 180 degrees? I assumed that since l goes from 0 to infinity, dl should move from the center of the wire to positive x-direction. Then shouldn't the angle be from pi/2 to 180 degrees?

[@DanFullerton]

We're not taking l as the length of the entire wire, as the wire is infinite. Instead, l is the distance along the wire from our reference point. In that way, we can integrate across all the little dl's to get the entire l, which gives us the length of the whole wire when we run from l=- infinity to l=+infinity.

[@anujkankani9860]

will dl * r hat always be equal to dl * sin theta? And if not why is it in this case? Also at 12:37 why is 0 on the top and not the bottom, since it's smaller than π/2?

[@DanFullerton]

In this case no... they look similar, however. Ampere's Law tells you the magnetic field due to the current flowing in a wire, but you have to use the current in a wire penetrating a virtual closed loop. In this problem, you actually have a loop of wire, and are looking at the magnetic field in the middle. The answers look similar, but it's a significantly different setup and application.

[@sarahmann6716]

We went over this in class, and I did not really understand it. The color coating really helped me on the second example :)

[@DanFullerton]

+Sarah Mann Glad to hear the video helped you out! I think this is the toughest part of the course...

[@berkcalabakan211]

This Video really helped alot , but i still got one question : dl x r equals dl * r * sin() , but why did the r disappear ? as seen at 09:29 f.ex.

[@ayberkkum9242]

yeah he probably made a mistake

[@ayberkkum9242]

sen dogrusunu bil kardeş:p

[@jadenngo1941]

when you take the cross product youre multiplying the magnitudes together times sintheta. the magnitude of r is 1 so it can be left out

[@eduardomora3565]

the r in the numerator in biot-savaart law is a unit vector. Its magnitude equals 1 and it points in the direction of r. That's why its not included or "disappears" in our integration here.

[@kushalg.4231]

10:55, In the tan theta we took, the adjacent side as l. But it looks like, we initially, took l as the length of the entire wire. So, how can the adjacent side be possibly taken as l in the tan theta ratio?

[@LaLakrs]

Does any of this change if you have a rectangle?

[@yusram.9949]

Can you explain the difference between the three r notations at 3:00?

[@zort9367]

In the second problem i put dlr^ in the integral as rd(x)sinx and got the same answer. I am wondering if that was a shortcut or i was lucky to find the same answer.

[@danielawesome12]

At 6:45 when you state the answer to the first example, you should specify it's in the Z (or upward) direction. This is because B is a vector. Technically when you evaluated the cross product of dl and r, you should have a direction and not just a magnitude.

[@DanFullerton]

You'll note that I solved for the magnitude of B, not the B vector itself, having already determined its direction, but regardless, I think the video and explanation conveys what I was hoping to communicate to those interested in learning more about application of the Biot-Savart Law.

[@jackysub1941]

Hello, what is the difference between the formula for B you talked about in the last video (one that involves 2pir in the denominator) and the formula for B in the Biot-Savart Law? Does integrating the formula for Biot-Savart law result in the same formula as the formula in the last video?

[@jackysub1941]

oh wait, I get it now you explained it in the last part. So do we use the equation that involves 2pi only in the case of an infinitely long wire?

[@DanFullerton]

Yes, you get the same answer.

[@jackysub1941]

@10:50, why is tangent theta R over l? Didn't you set up l as the total length of the wire?

[@DanFullerton]

It's a trig identity. Tangent of the angle is equal to the opposite side over the adjacent side. The opposite side of our triangle is R, the adjacent side is l.

[@jackysub1941]

I mean doesnt l denote the total length not the length of the adjacent side?

[@DanFullerton]

@@jackysub1941 No, the length of the wire is infinite.

[@jackysub1941]

aha got it

[@mechwarreir2]

6:39 -> mu I / 2 r doesn't that defy ampere's law? From ampere's law you get B = (mu I) divided by (2 pi r). HELP!!

[@RonaldinhoOwnz]

i was wondering, why do we integrate from pi/2 to 0? Like how do we get that?

[@DanFullerton]

As we change the variable of integration from dl to theta at 12:35, you look at the problem and see that if you were to go from l=0 to l=infinity, the angle theta would change from Pi/2 to 0. So as you switch variables of integration, you switch the corresponding limits.

[@gurlgurl05]

Dan Fullerton sorry, i still dont get it.. where is I=0 and how do you measure the angle?

[@DanFullerton]

Yen l l=0 would be one end of the infinite wire, l=infinity would be the other end. If you start at the very left an infinite distance away, and draw a line to point P, your angle would be 0. As you move closer and closer to P along the wire, that angle gets larger and larger until you are right under P, at which point the angle would be Pi/2 (all in radians, of course).

[@gurlgurl05]

Dan Fullerton it is much clearer to me now, thank you so much!!!! :) and thanks for the fast reply!!

[@shloakpatil9217]

hey Mr fullerton, how did you come up with the equation dB = (mew)(current)(dL x r)/4(pie)(r^2)

[@DanFullerton]

+shloak patio I looked it up. That is the Biot-Savart Law.

[@JacobSmithodc]

What was the reasoning behind replacing 0 and infinity with 0 and π/2 when all the substitutions took place?

[@DanFullerton]

+Jacob Smith We ran a change of variables (explained at 12:30), if we were going to integrate from l = 0 to infinity, when we change to our variable theta, 0 to infinity across l corresponds to theta going from 0 to Pi/2 (180 degrees).

[@ericzhu9264]

Is sample 2 basically a harder version of Ampere's law? Or is it just by coincidence that the two give the same answer?

[@DanFullerton]

Sample 2 is the "brute force" method of obtaining the magnetic field due to a current-carrying wire using the Biot-Savart Law. In the Ampere's Law lecture, you find that by using Ampere's Law you can find this same result much more simply.

[@ericzhu9264]

Ah thank you. I was always confused what the difference between the Biot-Savart Law and Ampere's Law was.

[@anniehodges2531]

where does the sin of theta come from in the second sample problem?

[@DanFullerton]

It comes from the definition of the cross product.

[@anniehodges2531]

..wow now I feel silly for thinking that over so much.. That's what I get for over-analyzing haha. Thanks for responding so quickly!

[@DanFullerton]

Annie Greer Not at all, and you're very welcome.

[@yepyep266]

I don't understand the reasoning of calculating half of B. I can see that if we tried to calculate it all at first going from -pi/2 to pi/2 the integral would give 0, but I can't find an intuitive or physical way to see it.

[@DanFullerton]

Hi Yep Yep. If you do that final integral of the sin of threat from Pi/2 to -Pi/2, and carry all your negative signs carefully, you end up with the same thing... I'm betting you're losing a negative sign somewhere in your integration.

[@MP-vy2mi]

Thanks for the help Dan!

[@DanFullerton]

+Maxime Perusse My pleasure!

[@sikolu5580]

for the last sample why pi/2 as one of the boundaries

[@chriswinnard7739]

I might be a bit late helping out here, but that's for when l = 0 and r is just a vertical line.

[@splashcity2k10]

Hey Dan Im on my 4th video of the day haha. I was wondering when the equation b=Moi/2pieR is used vs the equation b=Moi/4pieR

[@DanFullerton]

Ben Eckardt Hi Ben. This video is actually about deriving magnetic fields using Biot-Savart Law, so instead of just memorizing equations for specific uses, instead you use the Biot-Savart Law to derive the magnetic field equation for those instances. (Though I gotta tell you, Biot-Savart is probably the hardest part of AP Physics C)...

[@pradeeppaudel763]

In fig(sample 1), you named radius as R but in the derivation you put dB=(u°I)/4 pie r^2 - - - - why small r?

[@DanFullerton]

Big R is the constant radius of the loop, little r is my variable of integration, the vector from the loop to the center point.

[@pradeeppaudel763]

Its a circular loop then isn't the radius always constant?

[@DanFullerton]

yes, but we're using little r as the vector from the hoop to the center.

[@zheeweylee]

great explanation, thank you very much!

[@DanFullerton]

You're welcome!

[@DanFullerton]

Steve Michaels Thanks Steve!

[@DanFullerton]

The law still works, the application of the law may be more complex.

[@Razib10000]

what if in sample 2 the wIre was not infinitely long? What would the equation be?

[@DanFullerton]

You could still use the Biot-Savart Law, but you'd probably want a computer to do calculate the answer, as it would be extremely tedious to do by hand.

[@anamericanarmysoldierhooah7782]

YOU ARE A VERY GOOD TEACHER. THANKFUL GOD BLESS YOU FRIEND:)

[@DanFullerton]

My pleasure, and god bless you as well!

[@DanFullerton]

Great question!

[@GonzaloBelascuen]

Thank you! great video!

[@DanFullerton]

You're very welcome!

[@ShugeLuo]

I LOVE YOUR VIDEOS!!!!!!!!!!!!!!!!!!1

[@DanFullerton]

Thanks so much!!!

[@Issuemasi]

I want application of boot savart law Magnetic field axis of current carrying circular loop

[@Issuemasi]

Biot *

[@davisjohn-d6h]

@@Issuemasi yes

[@mdtanzidhossain6476]

nice explanation

[@royalfunfilm7794]

thank you sir.

[@linusbao8650]

Dan the MVP

[@DanFullerton]

You are too kind...

[@mankedek6588]

they said that electromagnetics are one of the most difficult subject in electrical engineering curriculum. they were right :'(

[@Uzairtak]

thanks...it helped

[@titirmajumdar3730]

helpful !

[@marebearzzz027]

thanks!!!!

[@nilo034]

dan fullerton is the goat

[@ICOD73]

That was harsh