Magnetic Field from a Circular loop
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- Опубліковано 12 вер 2024
- Physics Ninja looks at the magnetic field produced by a circular loop.
The Biot Savart law is used to find the total field produced by the loop at a point above the plane of the loop.
What happens to the angle that come from the cross product in the beginning?
This video is excellent, please don't stop making tutorials! Maybe some Lagrangian and Hamiltonian mechanics also?
only thing I have to say is that the cross product can be defined better. It has to be perpendicular to both vectors but there are two solutions to this, and you did not specify why you took the vector pointing up right instead of down left, which would also fit the parameters
Excellent video as usual. How about a video on the magnetic fields for spiral square and circular coils?
Comparison between the two is a hot topic in wireless power transfer.
Hi this is vikram from India,
Please post more videos as this is lock down time due to Corona virus.
I love your videos.
What about B at a point offset from the center?
Great Videos...
Love this question. I have to constantly refer to this example to learn Biot-Savart law.
Thank you ! Have a final in T-11 hours and this helped explain the law and mathematics of it well.
You’ve got this. Good luck on your final!
This is the best explanation out there, thank you!
Nice video, Thanks. Could you make another at points of off axis?
Hi,
This video was very useful. Would you be able to do a video of the same principle but if the point P is not on the central axis. So still in the same parallel plane of the loop but displace by an angle?
That would be incredible
Great video! Super helpful.
im sorry but isnt the magnitude missing the sin of the angle between dl and r - r' ?
Both are perpendicular to each other so thats 1
Thank you so much😭❤
hi, I know it's been two years since you posted the video but I hope you can help me. You explain perfectly and I understood everything, except why at the very beginning the formula says that the magnetic field is inversely proportional to 4π instead of 2π
Biot savart law has a factor of 4Pi in the denominator.
@@PhysicsNinja hello i love your works so much and u have no idea how much u help me everyday thank u so much
but isnt that r^2 instead of r^3 in the denominator at first equation? im sorry if im not gettin it but the formula on the book says so
@@ズンドウイチンギスオッド it depends how the numerator is written. You can write Biot-Savart with r^2 in the denominator if you have a unit vector r in the numerator. If you write it with r^3 in the denominator you must have the full vector r in the numerator.
@@PhysicsNinja i see, r on the top was unit vector meaning the length is one thank u so much keep going with those awesome vids 💯
🔥🔥Crystal clear clarity....!🔥🔥
Which axis is the phi measured from?
It's typically measured from the x-axis.
great video thanks man
Thank you for this explanantion!!
Great video :)
thank you so much sir!
Great video; the only thing I don’t quite understand is why theta for dBz is the same as for the triangle with r-r1, z and R. Would you mind explaining that, please?
late, but should be because of the triangle Geometry (dB is perpendicular to dL and r-r' by cross product) which then makes the angles mentioned =
I understand the math however, id like to ask if you integrated upon dB without multiplying by costheta.
you need the cos(theta) term because you only want the vertical component of the field. The other components cancel out due to the symmetry of the problem.
@@PhysicsNinja thank you for replying. Moreover i understand why we used costheta however i wanted to understand what would obtain if we just went on and made the integral without multiplying by costheta. I understand that the horizantal component would cancel out while the vertical remains thus multiplying by dB by cos theta so it becomes dBvertical answers the question in the text books correctly. Thus what im really intrested in is what if we made the integral for dB expression right away what would that represent
@@enejedhddhd6882 If you look at the integral at 7:00 you see that you're going to add a bunch of dB's but they are all in different directions. You can get to an answer but i don't know what it means because the adding of all the dB's like this wrong because the vectors are all in different directions. As you know, for vector addition the direction is really important and can't be neglected.
@@PhysicsNinja well yeah i see your point thanks man.
thank you
09:25 I think the Z component is should be dB.cos(theta), not sinuse.
Not the way I’ve defined the angle. The way I have it, it’s sin, but you could also setup the problem defining it so it would be cosine.
@Physics Ninja Isn't the leftside dB component in the same direction with (r-r') vector? If it is, then the angle between leftside dB and z-direction is theta. So, the angle between the rightside dB and z-direction is also theta from the symmetry, isn't it?
youre the best.
Find the exact magnetic field along the z-axis of a hexagonal loop with side a, which carries a steady current I. Find the approximate magnetic field at large distances from the loop by computing the dominant term in the multipole expansion of the magnetic vector potential of the loop. Show that for large values of z along the axis the two approaches give the same result.
I cant understand why is sin and no cos theta?? I put theta where you put it... the oposite angle from theta is theta again on the other side... Method of apex angles gives me that theta is upper angle on the right side... Wish I understand this...