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!
@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..
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
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
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?
Just wondering, does this implication of the magnetic field being 0 outside the outer loop bypass the limitations of really small computers? I remember hearing somewhere that there's a limitation of how small circuit elements can get because their magnetic field would induce currents in nearby circuit elements.
Good question! I'm not educated enough to answer it, but I would guess that for these small scales the classical theory presented here does not longer hold.
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
Great explanation! Thank you. Can we also calculate the B-Field for high frequency currents taking the skin effect into consideration with this method?
Because coaxial cables work so that the magnetic flux density cancels and doesn’t interfere with its surroundings. To do that the currents must be opposite
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.
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.
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!
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.
i have a few videos on Gauss' Law. Here's the first of the series. ruclips.net/video/fGB1csk7tX8/видео.html I'll try to finish the one on magnetic moments soon. Physics Ninja is a busy guy!
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.
Amperes law states that it’s only the current enclosed in the loop that will contribute. You only include the outer shell if you are calculating the field outside the coaxial cable.
Beautifully explained - better than my book! Thanks
john stanford thanks!
*Cries in Jackson*
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!
Thanks for the kind words!
Hope you made it bud
An outstanding presentation. It has cleared up all the problems I had understanding the physics behind coaxial cables.
Thank you
Brian Rook
Hey dude I guess you made a mistake. It will be good for you if you check the b
Without watching about 2 minutes of the video, i can already tell that there's good teaching in this video
@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..
You've just saved an anonymous student on the internet! This explanation is awesome!
Congratulations from Brazil, your video help me a lot!!
Thanks for watching and the support. It's great to hear from students.
I think you just saved me before my midterm tomorrow. 😂 Thanks so much!!!
Thank you so much! Not even Griffith’s could properly go over these types of problems
In my experience RUclips tutorials work better than books. At least better than griffiths, purcell and thermodynamics books
This is the best video on the god damn internet
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
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
Excellent video, thank you for creating this. Studying for my E&M final and this has been extremely helpful
Would a
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.
Amazing! I really understood the exercise. Thank you!
Thanks Ninja... Just got it at the right time for my presentation the next morning
for b < r < c situation, why is it (r^2 - b^2) but not (r - b)^2 , same as (c^2 - b^2)
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
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
Marc thanks, looks like an extra I got in there from my cut and paste.
The world's best teacher thanks sir
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?
Same current, different direction will result in 0 field outside of cable because total current enclosed is 0.
Audio mixing could be improved, but the content is very helpful.
Thanks for the clear and smooth explanation
It is the best video for a college student studying electromagnetism. It was very helpful. Thank you.
Literally saved my life
God bless your soul
This was explained so well thank you so much!
simple explanation...helped me a lot.. Thank you Sir
Just wondering, does this implication of the magnetic field being 0 outside the outer loop bypass the limitations of really small computers? I remember hearing somewhere that there's a limitation of how small circuit elements can get because their magnetic field would induce currents in nearby circuit elements.
Good question! I'm not educated enough to answer it, but I would guess that for these small scales the classical theory presented here does not longer hold.
Thx from Brazil!
is there any possibility of electric field being produced by this coaxial cable (atleast in a
Thank you so much
Perfect! I was always confused by these things. Thank you very much! Saved me!
thank for your lectures, I am a teacher, I would like to know exactly the divide you (everyone) used to draw lectures beautifully?
So well explained!
This was beautifully done. Thank you.
it's a good video and good explanation but the microphone popping makes it quite difficult to listen to
Thanks. Ninja has a new microphone now.
better explanation than books..
what would happen if the currents aren't flowing in the opposite directions? would that mean the current is doubled when r is greater than c?
Yes you are correct, the field would be be different inside the outer shell and outside of both the field would that of a wire with current =2I
@@PhysicsNinja thank you for the quick response!!
Thanks for great explanation.
Really helpful
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
Thanks for going into the details! You have a nice voice as well!
Hi ninja ninja leader. Cheers!!. I have a question, what is the difference between magnetic field and magnetic field intensity? Thank you.
Thank you so much .you explained so easily
Thanks a lot man, it was wonderful.
Great Explanation !! TY
thank you
Do you know why it matters to take the fraction of the AREA of the cylinders when finding the B field in situations 1 and 3?
Why ratio of areas?
simple explanation! I love it!
Thanks so much, this clarified my Delima about the magnetic field everywhere in space.
Now your only dilemma is learning how to spell dilemma 😂
Thanks man! Was having trouble understanding this question!
Great explanation!
Sound needs to be fixed but other than that lecture is okay
You’re better than my professor, thank you
Could you please also calculate the current density for each interval?
Great explanation! Thank you. Can we also calculate the B-Field for high frequency currents taking the skin effect into consideration with this method?
Thank you for this clear solution.
Great explanation! I wish I had you teaching my physics class!
This really helped me. Thanks!
Tq very much man...save my day 🙌🙏
i don't think you have to include the interior of I at 13:20 minutes in region (b
Yes you do, it’s the total current enclosed by the loop.
Superposition of fields
@@PhysicsNinja Thank you. You clear up all my confusion.
thanks a ton!! i couldnt understand the solution given in my text, they didnt clearly write the exolanation.
Why there is current in outer conducter current oppsite to inner ? Please answer
Because coaxial cables work so that the magnetic flux density cancels and doesn’t interfere with its surroundings. To do that the currents must be opposite
13:23 does the direction of the current matter in terms of the sign for B? would dl and b be in the same direction?
GREAT HELP SIR !!!
💞INDIA
thanks a lot
ur a legend
Thank you Physics Ninja!!!!
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.
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.
Thanks a lot!
what if the equation was asking for r
I think the solution cover all values of r. Maybe i don't understand your question
wow. you nailed it,, well explained.. Wonder if you could make videos about magnetic moments for rotating charge disk, spheres, etc.
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!
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.
i have a few videos on Gauss' Law. Here's the first of the series. ruclips.net/video/fGB1csk7tX8/видео.html
I'll try to finish the one on magnetic moments soon. Physics Ninja is a busy guy!
Here's the first part of magnetic moment series. I'll add the other 2 parts tonight. Happy Learning! ruclips.net/video/4F0p_VvYvMc/видео.html
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.
man you are the best, I tried to find the answer to the same question
This was so helpful! Thank you!
Glad the video was helpful for you.
Great good teacher
Why do we ignore the outside thin hollow cylindrical conductor in which the current goes into the page?
Amperes law states that it’s only the current enclosed in the loop that will contribute. You only include the outer shell if you are calculating the field outside the coaxial cable.
Thank you :3
This is awesome!
these are great please make more
Very helpful video!
Thank you so much! This was very helpful!
Fantastic work
thanks but too many ads interrupting
You choose the premium and no advertising
tnx a lot! very well !
Thank you 🙏🏼
Great video, thank you!
Thanks a lot, this video helps me a lot anytime i need :)
Beautiful explanation! You're great! :)
Thanks for this great help. 😃
Brasil love's you!! grateful!!
love you man
Thanks man
Happy to help!
That video saved my life
You're awesome. This helped me a lot.
thank you well explained
Thanks brother!
Thank you.
Thank you ♥️♥️
dude, thank you!