Applications of Ampere's Law (part II)

Part 2 of a series on applications of Ampere's law. This is at the AP Physics level.

[@puffpuffshare]

Phew! Now I can sleep better at night knowing the magnetic field outside my cable wire is 0. Thanks alot ;)

[@Mrsteisag2]

I know for a fact this guy has a mustache

[@NarcoIXTV]

How ?

[@jellyboiii6269]

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[@comTeste]

Sir you are the best physics professor I've ever had. It's a shame your videos don't have as many views and likes as they should. I posted a comment about you on a PatrickJMT video, he is the guy I go to for math and he is really famous on youtube. I really hope more people can benefit from your videos in the future. Your website is very organized and your videos are great. You should include the website in the description for people who don't know about it.
Thanks a lot for doing this

[@somedude5990]

This channel saved my grade. Thank you so much, teacher. I got a B, when I could have gotten a D or even failed.

[@omkararunteja2354]

u r simply ..awesome..sir
keep on going....make us knowledge ful

[@airsoft78910]

You have opened my third eye. Seriously thank you, you have saved my life for my emag exam

[@bossthai8805]

Thank you very much for every good videos, teacher.

[@lanlansk8]

Thanks Sr !! I´ll continuos watching your videos! thanks a lot

[@n3t912]

Learned so much from this! Thank you!

[@imenbenmahmoud1359]

Please can someone tell me why in this case he doesn't replace I by J*PI*r^2 ?

[@robsumner6872]

You don't really need to use current density (J) in this example because you know a current (I) is moving through the wire. You could use I = J*Pi*r^2 but its extra complication as I is known but J is not (so you would find J from I then substitute J in for I).

[@raikoshumi]

Don't really get why you would replace I by J.A because you're adding more variables and only removing one constant (pi). But I guess it all depends on what you use the formula for.

[@benjaminchan4078]

Consider this hypothetical scenario: suppose I have a wire with a cross section that is mapped out using the function y=f(x) with a uniform current density through it, how do I find B at an arbitrary distance away from the cross section?

[@billofrights5064]

Why is B, the magnetic flux density, used as though it were H, the magnetic field intensity?

[@selimyener2103]

Hi guys! at the time 8:46, I can not understand what is the lower r(r). Is it whole cable's radius or small loop's radius? Is there anybody who can explain it?

[@lasseviren1]

It is the radius of the small loop (the Amperian Loop). The right side of Ampere's equation, integral of B dot dl, is always referring to the Amperian Loop. The dl's always make up the Amperian Loop.

[@selimyener2103]

@@lasseviren1 Thanks Teacher :D. Hi from Turkey :D

[@towe02225]

Thanks for the video. It helped me a lot with understanding the concept.

[@nicko1110]

cool video; it'd be even cooler if you'd solve in terms of "I"

[@eyadsaleh12]

thanks very very much. my exam is tomorrow and those videos helped me a lot!

[@danamuise4117]

you are the BEST, thank you for making these videos!

[@awhu5696]

What’s J

[@dqueezy923]

I love you

[@arptics4264]

Thank you for your efforts and awesome videos and the way you explain the involved math so much easier to comprehend.

[@JTSeiber]

thank you. my teacher can't explain it like this at all

[@jesper7953]

can't pi.R^2.J be replaced by I since the inclosed current is equal to all the current through the wire?

[@adamcarcella6034]

+Jesper Fennema yes

[@rinap7428]

I really like your videos, thank you

[@mangoatree]

thank you! super cool :D

[@AhmedWaeln1]

greeeeeeeat effort > Thanks a lot
.

[@marammat1128]

I love you. Thanks

[@QuickLearnScience]

what does J stand for?

[@lasseviren1]

J is current density which is the current per cross-sectional area of the wire. J=I/Area and its units are A/m^2.

[@anthony-pi9ps]

Why is it that current goes down the sides? I didn't quite understand that.

[@jesset-p9225]

Think of this as a coaxial cable, which is made up of two conductors separated by some dielectric. The reason that the current, i, are in opposite directions is partly arbitrary but is based in application. He hinted at it with the sum of currents at a junctions being equal to zero ( just think of it as one being a propagating path and one being a return path).

[@naslamb94]

does the same apply for a copper conductor with an off center cylindrical hole?

[@jesset-p9225]

with an off center cylindrical hole, the B field will not be same on the amperian loop (since the conductor will be closer to one side) thus the B will be not be constant and you will have a different answer.

[@MrDanupa]

Thanks... useful stuff

[@Kevin-Pdog]

agree !!!

[@srinaath9845]

Thank you sir

[@Neeraj-is1jt]

can u plz explain me why the current going in outer cylinder is opposite to that of inside ????

[@omkararunteja2354]

since its a junction....current entering junction=current leaving junction

[@jesset-p9225]

Think of this as a coaxial cable, which is made up of two conductors
separated by some dielectric. The reason that the current, i, are in
opposite directions is partly arbitrary but is based in application. He
hinted at it with the sum of currents at a junctions being equal to zero
( just think of it as one being a propagating path and one being a
return path).

[@misssweethearted]

soooooo good!!!

[@TheArt832]

You are awesome.