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..
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?
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
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. ua-cam.com/video/fGB1csk7tX8/v-deo.html I'll try to finish the one on magnetic moments soon. Physics Ninja is a busy guy!
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.
Great explanation! Thank you. Can we also calculate the B-Field for high frequency currents taking the skin effect into consideration with this method?
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
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.
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
5+ years later, how are you doing now?
An outstanding presentation. It has cleared up all the problems I had understanding the physics behind coaxial cables.
Thank you
Brian Rook
Without watching about 2 minutes of the video, i can already tell that there's good teaching in this video
Audio mixing could be improved, but the content is very helpful.
You've just saved an anonymous student on the internet! This explanation is awesome!
This is the best video on the god damn internet
Congratulations from Brazil, your video help me a lot!!
Thanks for watching and the support. It's great to hear from students.
Hey dude I guess you made a mistake. It will be good for you if you check the b
Thanks Ninja... Just got it at the right time for my presentation the next morning
Excellent video, thank you for creating this. Studying for my E&M final and this has been extremely helpful
Thank you so much! Not even Griffith’s could properly go over these types of problems
In my experience UA-cam tutorials work better than books. At least better than griffiths, purcell and thermodynamics books
I think you just saved me before my midterm tomorrow. 😂 Thanks so much!!!
@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..
Amazing! I really understood the exercise. Thank you!
I have question, in the b
Thanks for the clear and smooth explanation
It is the best video for a college student studying electromagnetism. It was very helpful. Thank you.
This was explained so well thank you so much!
simple explanation...helped me a lot.. Thank you Sir
So well explained!
Thx from Brazil!
Perfect! I was always confused by these things. Thank you very much! Saved me!
Great Explanation !! TY
Thanks for great explanation.
Thanks for going into the details! You have a nice voice as well!
This was beautifully done. Thank you.
simple explanation! I love it!
Great explanation
Glad you think so!
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.
The world's best teacher thanks sir
Thanks a lot man, it was wonderful.
Great explanation!
Really helpful
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
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
Literally saved my life
God bless your soul
these are great please make more
Thank you so much .you explained so easily
Thanks man! Was having trouble understanding this question!
Why cant my professors make videos like these!!! Thank you
This really helped me. Thanks!
Tq very much man...save my day 🙌🙏
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.
GREAT HELP SIR !!!
💞INDIA
Thank you Physics Ninja!!!!
Very helpful video!
Thank you for this clear solution.
This was so helpful! Thank you!
Glad the video was helpful for you.
You’re better than my professor, thank you
thank for your lectures, I am a teacher, I would like to know exactly the divide you (everyone) used to draw lectures beautifully?
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
thanks so much
better explanation than books..
Great explanation! I wish I had you teaching my physics class!
Thanks for this great help. 😃
Beautiful explanation! You're great! :)
Fantastic work
Thank you so much! This was very helpful!
thanks but too many ads interrupting
You choose the premium and no advertising
Brasil love's you!! grateful!!
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.
Thank you 🙏🏼
You're awesome. This helped me a lot.
Great video, thank you!
Hi ninja ninja leader. Cheers!!. I have a question, what is the difference between magnetic field and magnetic field intensity? Thank you.
Thanks a lot, this video helps me a lot anytime i need :)
This is awesome!
is there any possibility of electric field being produced by this coaxial cable (atleast in a
You are special❤❤❤❤❤
Thank you ♥️♥️
Sound needs to be fixed but other than that lecture is okay
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. ua-cam.com/video/fGB1csk7tX8/v-deo.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! ua-cam.com/video/4F0p_VvYvMc/v-deo.html
Thanks a lot!
man you are the best, I tried to find the answer to the same question
Great good teacher
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?
thank you well explained
thanks a lot
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.
tnx a lot! very well !
Nice! Thanks alot :)
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.
Great explanation! Thank you. Can we also calculate the B-Field for high frequency currents taking the skin effect into consideration with this method?
Could you please also calculate the current density for each interval?
what if the equation was asking for r
I think the solution cover all values of r. Maybe i don't understand your question
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 so much, this clarified my Delima about the magnetic field everywhere in space.
Now your only dilemma is learning how to spell dilemma 😂
Thank you :3
Thanks brother!
God Bless You
THANK SO MUCH!
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?
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 man
Happy to help!
Thank you.
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.
ur a legend
Melhor explicação que encontrei
dude, thank you!
Nice one