Applications of Ampere's Law (part II)

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  • เผยแพร่เมื่อ 23 ธ.ค. 2024

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  • @comTeste
    @comTeste 11 ปีที่แล้ว +18

    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

  • @puffpuffshare
    @puffpuffshare 12 ปีที่แล้ว +24

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

  • @somedude5990
    @somedude5990 4 ปีที่แล้ว +2

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

  • @airsoft78910
    @airsoft78910 5 ปีที่แล้ว +2

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  • @omkararunteja2354
    @omkararunteja2354 8 ปีที่แล้ว +9

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

  • @bossthai8805
    @bossthai8805 9 ปีที่แล้ว +2

    Thank you very much for every good videos, teacher.

  • @arptics4264
    @arptics4264 11 ปีที่แล้ว

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

  • @n3t912
    @n3t912 10 ปีที่แล้ว +1

    Learned so much from this! Thank you!

  • @Mrsteisag2
    @Mrsteisag2 7 ปีที่แล้ว +55

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  • @lanlansk8
    @lanlansk8 11 ปีที่แล้ว +1

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

  • @benjaminchan4078
    @benjaminchan4078 7 ปีที่แล้ว

    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?

  • @selimyener2103
    @selimyener2103 4 ปีที่แล้ว

    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
      @lasseviren1  4 ปีที่แล้ว +1

      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
      @selimyener2103 4 ปีที่แล้ว

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

  • @eyadsaleh12
    @eyadsaleh12 12 ปีที่แล้ว

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

  • @danamuise4117
    @danamuise4117 11 ปีที่แล้ว

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

  • @billofrights5064
    @billofrights5064 4 ปีที่แล้ว

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

  • @towe02225
    @towe02225 6 ปีที่แล้ว

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

  • @JTSeiber
    @JTSeiber 11 ปีที่แล้ว

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

  • @imenbenmahmoud1359
    @imenbenmahmoud1359 7 ปีที่แล้ว +1

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

    • @robsumner6872
      @robsumner6872 7 ปีที่แล้ว

      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).

  • @jesper7953
    @jesper7953 9 ปีที่แล้ว

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

  • @rinap7428
    @rinap7428 7 ปีที่แล้ว

    I really like your videos, thank you

  • @Neeraj-is1jt
    @Neeraj-is1jt 8 ปีที่แล้ว

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

    • @omkararunteja2354
      @omkararunteja2354 8 ปีที่แล้ว

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

    • @jesset-p9225
      @jesset-p9225 7 ปีที่แล้ว

      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
    @naslamb94 11 ปีที่แล้ว

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

    • @jesset-p9225
      @jesset-p9225 7 ปีที่แล้ว

      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.

  • @raikoshumi
    @raikoshumi 7 ปีที่แล้ว

    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.

  • @anthony-pi9ps
    @anthony-pi9ps 7 ปีที่แล้ว

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

    • @jesset-p9225
      @jesset-p9225 7 ปีที่แล้ว

      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).

  • @QuickLearnScience
    @QuickLearnScience 3 ปีที่แล้ว

    what does J stand for?

    • @lasseviren1
      @lasseviren1  3 ปีที่แล้ว

      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.

  • @MrDanupa
    @MrDanupa 9 ปีที่แล้ว

    Thanks... useful stuff

  • @misssweethearted
    @misssweethearted 11 ปีที่แล้ว

    soooooo good!!!

  • @AhmedWaeln1
    @AhmedWaeln1 11 ปีที่แล้ว

    greeeeeeeat effort > Thanks a lot
    .

  • @TheArt832
    @TheArt832 10 ปีที่แล้ว

    You are awesome.

  • @mangoatree
    @mangoatree 6 ปีที่แล้ว

    thank you! super cool :D

  • @awhu5696
    @awhu5696 5 ปีที่แล้ว

    What’s J

  • @marammat1128
    @marammat1128 7 ปีที่แล้ว

    I love you. Thanks

  • @nicko1110
    @nicko1110 4 ปีที่แล้ว

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

  • @srinaath9845
    @srinaath9845 6 ปีที่แล้ว

    Thank you sir

  • @Kevin-Pdog
    @Kevin-Pdog 12 ปีที่แล้ว

    agree !!!

  • @dqueezy923
    @dqueezy923 4 ปีที่แล้ว

    I love you