Ampere's Law: Magnetic Field in a Coaxial Cable

Physics Ninja applies Ampere's law to calculate the magnetic field in a coaxial cable. The inner and outer conductors carry current in opposite directions.
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[@johnstanford4610]

Beautifully explained - better than my book! Thanks

[@PhysicsNinja]

john stanford thanks!

[@ztac_dex]

*Cries in Jackson*

[@brianrook6207]

An outstanding presentation. It has cleared up all the problems I had understanding the physics behind coaxial cables.
Thank you
Brian Rook

[@gideonkyeremeh868]

Thanks Ninja... Just got it at the right time for my presentation the next morning

[@ronnyalfons7096]

Without watching about 2 minutes of the video, i can already tell that there's good teaching in this video

[@libertyprime656]

Excellent video, thank you for creating this. Studying for my E&M final and this has been extremely helpful

[@lagado1lag]

Amazing! I really understood the exercise. Thank you!

[@lNDlANSPlCE]

Why doesn't this have more views... I'm gonna share it with all my emag friends. Thank you, Ninja! When I graduate and have an engineer's salary I will be able to contribute and quantify my gratitude for your videos!

[@PhysicsNinja]

Thanks for the kind words!

[@dromero8127]

Hope you made it bud

[@dareadegun2460]

Thanks for the clear and smooth explanation

[@brayschway6746]

This was beautifully done. Thank you.

[@karthikkrishnan7558]

simple explanation...helped me a lot.. Thank you Sir

[@shaunwu3609]

Thanks for going into the details! You have a nice voice as well!

[@juliazhang3600]

simple explanation! I love it!

[@JH-ux1re]

Perfect! I was always confused by these things. Thank you very much! Saved me!

[@user-xo7bq4cz6b]

It is the best video for a college student studying electromagnetism. It was very helpful. Thank you.

[@sealol5209]

Thanks man! Was having trouble understanding this question!

[@rakibuddin385]

Thank you so much .you explained so easily

[@hasranman]

This really helped me. Thanks!

[@masouddayaghi5923]

Thanks a lot man, it was wonderful.

[@jennykeeling747]

This was explained so well thank you so much!

[@lilac.2858]

So well explained!

[@sor3na369]

Thanks for great explanation.

[@maxdougherty3429]

Great explanation! I wish I had you teaching my physics class!

[@taladiv3415]

Thank you for this clear solution.

[@etanetan101]

I think you just saved me before my midterm tomorrow. 😂 Thanks so much!!!

[@user-ny1vt3ek4i]

Thank you Physics Ninja!!!!

[@igorsantos-qj3zd]

Thx from Brazil!

[@kabandajamir9844]

The world's best teacher thanks sir

[@asgoodasbread5559]

Great Explanation !! TY

[@boysenbeary]

Great explanation!

[@pierretobgui1792]

Really helpful

Hey dude I guess you made a mistake. It will be good for you if you check the b

[@jorgemillan4041]

these are great please make more

[@rifusaki]

You've just saved an anonymous student on the internet! This explanation is awesome!

[@matthew_bang]

Fantastic work

[@EnzoChai]

Tq very much man...save my day 🙌🙏

[@linjisoo9975]

Very helpful video!

[@peytonkliewer1997]

Thank you so much! This was very helpful!

[@JD867]

You're awesome. This helped me a lot.

[@yigitarsoy1944]

This is awesome!

[@MayankLC93]

GREAT HELP SIR !!!
💞INDIA

[@betulguncaldi4676]

Beautiful explanation! You're great! :)

[@saffyt348]

This was so helpful! Thank you!

[@PhysicsNinja]

Glad the video was helpful for you.

[@martinsibanda8026]

thank you well explained

[@kateystevenson7912]

Great video, thank you!

[@negative258]

Thank you so much

[@bobross5716]

Thank you so much! Not even Griffith’s could properly go over these types of problems

[@estebansingh9411]

In my experience TH-cam tutorials work better than books. At least better than griffiths, purcell and thermodynamics books

[@HakaiKhaled]

Literally saved my life
God bless your soul

[@3washoka]

Thank you 🙏🏼

[@kholoudhakim7887]

thanks so much

[@nabeelljassim6652]

Great good teacher

[@user-gs9oq7hi6x]

You’re better than my professor, thank you

[@asmartguycodename4721]

Thanks for this great help. 😃

[@leandrogarcia6396]

This is the best video on the god damn internet

[@aysenuratagun6534]

Thanks a lot, this video helps me a lot anytime i need :)

[@laisguizelli25]

Congratulations from Brazil, your video help me a lot!!

[@PhysicsNinja]

Thanks for watching and the support. It's great to hear from students.

[@THEWouter1212]

Thanks a lot!

[@ferhat1962]

man you are the best, I tried to find the answer to the same question

[@ozanyldz445]

thanks a lot

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Thanks brother!

[@Zakirhossain-wx8qc]

Thank you ♥️♥️

[@samogx86]

Thank you.

[@HD-dw7zk]

thank you

[@AndreaGarcia-rl8cs]

THANK SO MUCH!

[@xhelixshotx]

Very clear explanation, thank you! On a side note though, in the final screen you accidentally put another I in the equation for b < r < c

[@PhysicsNinja]

Marc thanks, looks like an extra I got in there from my cut and paste.

[@ezefferth]

Brasil love's you!! grateful!!

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Thank you :3

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Nice one

[@killirina]

dude, thank you!

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God Bless You

[@shimabarati7963]

tnx a lot! very well !

[@MaDcAmStErGaming]

Thanks physics ninja, you might help me not fail uni xxx

[@akibmostabe98]

Well explained,but when u do the closed integration it should be integration over rho product with dfi according to cylindrical co-ordinate system(Here rho is a product with dfi because rho is the perimeter across the arc dfi).Above all its good.
Another thing is that u may use H(Magnetic field intensity) instead of B (Magnetic flux density) because when we r considering free space only then they aren't equal that is B=meu(o).H.But in practical cases medium may vary according to products choice

[@cosmolearner5562]

@21:05 Which current is taken positive and which one negative is also included in ampere's law -
First arbitrarily consider the direction of path that you have choosen (generally taken same as magnetic field)
The apply right hand thumb rule to the path choosen the direction of thumb is taken as positive that mean on applying RHT rule on path all the enclosed current in the direction of thumb are taken as positive and all other currents opposite to direction of thumb are taken as negative..

[@aryanviolett4606]

Nice! Thanks alot :)

[@birahnugmedhin1934]

You are special❤❤❤❤❤

[@alpertemizel9892]

Thank you :))

[@sanskararora6425]

as the inner cylinder is a conductor so all charges would be at the surface of the cylinder and hence the current should flow through the surface (as current is the flow of charges) so I would be zero for r

[@johngordon8513]

ur a legend

[@ShrabantiDeb]

better explanation than books..

[@tanishkhandal4759]

Thank u

[@uchennaubeh9498]

Thanks so much, this clarified my Delima about the magnetic field everywhere in space.

[@lukedimitriades9835]

Now your only dilemma is learning how to spell dilemma 😂

[@hughrountry1281]

Audio mixing could be improved, but the content is very helpful.

[@cnnn7885]

for b < r < c situation, why is it (r^2 - b^2) but not (r - b)^2 , same as (c^2 - b^2)

[@PhysicsNinja]

Because the area will be the difference between r and b. Pi- r^2 will be area. Of entire circle then take away area of inside part Pi-b^2. The area is not pi- ( r-b)^2

[@k12h6]

Thanks man

[@PhysicsNinja]

Happy to help!

[@thuonghoang530]

thank for your lectures, I am a teacher, I would like to know exactly the divide you (everyone) used to draw lectures beautifully?

[@km-sc4kz]

is there any possibility of electric field being produced by this coaxial cable (atleast in a

[@asemshaat6672]

That video saved my life

[@JamBergMann]

Great explanation! Thank you. Can we also calculate the B-Field for high frequency currents taking the skin effect into consideration with this method?

[@danilovtoria]

Melhor explicação que encontrei

[@Dube_lab01]

Hi ninja ninja leader. Cheers!!. I have a question, what is the difference between magnetic field and magnetic field intensity? Thank you.

[@rodolfogarcia6263]

Would a

[@PhysicsNinja]

This is done @ 10:44. In this case the field is not 0 because the Amperian loop will enclose all of the current within the small wire in the middle.

[@salmamahameed2160]

this isn't an empty cable right? like it's not a shell of a coaxial cylinder cable? and if that was the case shouldn't the magnetic field in the r

[@allendeguzman6771]

wow. you nailed it,, well explained.. Wonder if you could make videos about magnetic moments for rotating charge disk, spheres, etc.

[@PhysicsNinja]

Thanks for the comment, i'll make a video tomorrow night comparing the magnetic moments of a disk, spherical shell, and a uniformly charged sphere. Physics Ninja loves requests!

[@allendeguzman6771]

cant wait to learn form that video..
,,,
this might be too much but it will makes sense if there is also a great video about electric fields, (gauss law) or gaussian surfaces to see connections to amperes law(amperian surfaces.

[@PhysicsNinja]

i have a few videos on Gauss' Law. Here's the first of the series. th-cam.com/video/fGB1csk7tX8/w-d-xo.html
I'll try to finish the one on magnetic moments soon. Physics Ninja is a busy guy!

[@PhysicsNinja]

Here's the first part of magnetic moment series. I'll add the other 2 parts tonight. Happy Learning! th-cam.com/video/4F0p_VvYvMc/w-d-xo.html

[@fatmanur5409]

Could you please also calculate the current density for each interval?

[@teambananakinns]

what if the equation was asking for r

[@PhysicsNinja]

I think the solution cover all values of r. Maybe i don't understand your question

[@joepaschal6426]

How could you do this same problem butt starting from the magnetic energy density of a ideal solenoid and combining that with amperes law to find the density as a function of distance radial out from the axis and use this to find the inductance in the cable. thank you.

[@Arif.mohmand]

Very nicely presented, at the end you said, it was a special coaxial cable with the both the same currents, thats is why B is zero, wasn't it zero because they were different currents in the direction?

[@PhysicsNinja]

Same current, different direction will result in 0 field outside of cable because total current enclosed is 0.

[@adityasrivastava4063]

do same for dielectric

[@eckardbezuidenhout]

Hello sir, thank you for your thorough explanation. I just need to understand something please. Let us for now consider only the small conductor (gold). Seeing that it is a conductor, will all the charge/current not be located on it's surface only? Ergo, a proper Amperian loop would be such that it goes around the entire conductor and not inside the conductor. This would then change the outcome of this part and the rest of the problem. I need to understand this because there are so many people on the internet doing their own thing and in the end the information conflicts in the small but very necessary details.
Thank you again for this explanation.

[@braceyourselvesfortruth2492]

Not 100% sure but you may be confusing current (moving charges) with static charges. Current is based on cross-sectional density so the amount of the wire we draw our loop around determines the total enclosed current.

[@helldi2blo]

thanks a ton!! i couldnt understand the solution given in my text, they didnt clearly write the exolanation.