This was a Question on the last paper of a subject at college. I needed all points of the paper to be able to write the exam and thanks to you I was able to get them! Thanks so much!!!
A current I flows in a square shaped conductor as shown in Figure below. Calculate the (i) magnetic flux density and (ii) Magnetic field intensity at the center of gravity of the square (where the four normal to the sides meet). need solution of this
well done !!!.. a complete video !!.. thank you!!.. is it possible to have done one Quarter of the square, the lower right quadrant, and then multiplied times 8 ?? rather than 4?... thank you!!.. actually... I should try the problem and answer my own question... lol.. but THANK YOU for a perfect video in regards to the Biot Savart Law...
@@PhysicsNinja Thank you for your Repsonse, Physics Ninja... .. when I studied physics, back in college, I was too busy with other courses and other activities . I always wanted to learn physics and the Calculus. Now that I"m 64 years old.. I have the time to study physics.. It's amazing how knowing the laws of nature can be fun and so enlightening... Thank you for contributing to that Dissemination of the Truth that physics reveals....
I am doing a version similar to this problem. In my case, the point p is outside of the square at a point (0,r,0), however r >>> a, I feel like this means I can approximate all side contributions as equal in a similar way to if the point was at the centre, this would still allow me to use symmetry to help. Am I correct?
Hii so i hat cross j hat has to equal to 1 right because in the coordinate system the i hat and the j hat is orthogonal to each other. And if I am wrong so feel free to correct me.
@Physics Ninja Could you kindly show how to setup the same problem when the loop is a rectangle and point P is on the z axis at a height H away from where you show it now in the video? I have tried it and I am making a setup mistake as I can't match the book exact equation. I get close but I don't match it. Please help.
Why not solving with theta? or better, his complementary angle there, isn't it possible to do it? I tried that way before coming here and the difference was that I couldn't get that 2 in the final solution, even tho was way more simple
There are a lot of ways to crack an egg. I recently solved for the field at a point above the plane of the square loop. In that problem I use trig substitution.
Calculate the field due to one segment of wire. Consider the 4 segment separately. Depending on the direction of the currents you may find that 2 of the segments cancel because the field they produce is in opposite directions.
Doubt you'll still need this, however if anyone else is interested in this: For any N-sided (N larger or equal to 3) symmetric shape, given all sides are the same length this answer works. You just have to multiply by N (number of sides, 4 in our case here) and the angle within the sine becomes pi/N
Can you now derive it at a height h from the center in the z-direction? I have tried deriving it, and I end up with Bt = [muo*I / [2*pi*((a/2)^2+(b/2)^2+Z^2)^0.5)] ] * [(a/((b/2)^2+Z^2)^0.5) + (b/((a/2)^2+Z^2)^0.5 ]. However the book I am reading has a the answer as Bt = (muo*I*a*b/4*pi*[(a/2)^2+(b/2)^2+Z^2]^0.5) * [1/((a/2)^2 + Z^2) + 1/((b/2)^2+Z^2)]. I put the calculations in excel computing the strength of the field (H = B/mu0). My answer is higher than the answer in the book. I am sure I am setting it up wrong. Could you do a video showing how to derive it? It would be this video with the B field at the center of the field elevated by a height h. I have tried multiple times, and I can't seem to match the book answer.
@@rjrodrig I wish i knew how to calculate it, that's my issue as well. I was given only the answer by my teacher and have to work out the problem. I can't seem to figure out how after multiple times trying.
Thorough, methodical, and well explained. I love it!
Very helpful! I learned a lot from this example! Thank you so much!❤❤❤
very nice, i love the mathematical reasoning not just throwing around formulas
Your videos are a great contribution for us students preparing for the upcoming exam!
This was a Question on the last paper of a subject at college. I needed all points of the paper to be able to write the exam and thanks to you I was able to get them! Thanks so much!!!
A current I flows in a square shaped conductor as shown in Figure below. Calculate the (i)
magnetic flux density and (ii) Magnetic field intensity at the center of gravity of the square
(where the four normal to the sides meet). need solution of this
That was a pretty rough one. Thanks for the tutorial :)
I really appreciate the way you explain this, thx u so much, it is more easier to understanf
Waouh, your explication is lit brother 🔥❤️. Keep it up
Great videos, great channel, thanks!!
well done !!!.. a complete video !!.. thank you!!.. is it possible to have done one Quarter of the square, the lower right quadrant, and then multiplied times 8 ?? rather than 4?... thank you!!.. actually... I should try the problem and answer my own question... lol.. but THANK YOU for a perfect video in regards to the Biot Savart Law...
Yes you can! Using symmetry to solve problems is a powerful tool that can greatly simplify your life.
@@PhysicsNinja Thank you for your Repsonse, Physics Ninja... .. when I studied physics, back in college, I was too busy with other courses and other activities . I always wanted to learn physics and the Calculus. Now that I"m 64 years old.. I have the time to study physics.. It's amazing how knowing the laws of nature can be fun and so enlightening... Thank you for contributing to that Dissemination of the Truth that physics reveals....
I am doing a version similar to this problem. In my case, the point p is outside of the square at a point (0,r,0), however r >>> a, I feel like this means I can approximate all side contributions as equal in a similar way to if the point was at the centre, this would still allow me to use symmetry to help. Am I correct?
ruclips.net/video/Oyz2XOEDg6E/видео.html
Is it possible to solve it without symmetry and how is it possible to do it in one go without breaking it up in pieces?
Why did you break r hat into two components? In the previous example you didn't. In the circular loop and infinite rod. Why now did you break it?
Hii so i hat cross j hat has to equal to 1 right because in the coordinate system the i hat and the j hat is orthogonal to each other. And if I am wrong so feel free to correct me.
Thank you sir..
@Physics Ninja Could you kindly show how to setup the same problem when the loop is a rectangle and point P is on the z axis at a height H away from where you show it now in the video? I have tried it and I am making a setup mistake as I can't match the book exact equation. I get close but I don't match it. Please help.
Email me a picture of your solution
Onlinephysicsninja@gmail.com
lmk if you get it figured out
I will put a video together later today. Stay tuned
Mariah Kar new video is up
Could you tell me why you have it over r^3 rather than over r^2?
If it’s over r^3 it gets multiplied by a vector r in numerator. If it’s over r^2 it gets multiplied by the unit vector r.
thank you
Thank you So Much!!!!
had to adapt it slightly for a rectangular wire but thanks my guy :)
How did you do this? Please, help me. I need to figure it out for my university work
Please do the same but now for a rectangular loop
Thaks alot
Awesome
Why not solving with theta? or better, his complementary angle there, isn't it possible to do it? I tried that way before coming here and the difference was that I couldn't get that 2 in the final solution, even tho was way more simple
There are a lot of ways to crack an egg. I recently solved for the field at a point above the plane of the square loop. In that problem I use trig substitution.
What happened if the direction of electricity is different(not behave like a loop) on square
Calculate the field due to one segment of wire. Consider the 4 segment separately. Depending on the direction of the currents you may find that 2 of the segments cancel because the field they produce is in opposite directions.
thanks
Hello, may i ask how about triangle and hexagon loop??
Doubt you'll still need this, however if anyone else is interested in this: For any N-sided (N larger or equal to 3) symmetric shape, given all sides are the same length this answer works. You just have to multiply by N (number of sides, 4 in our case here) and the angle within the sine becomes pi/N
Can you now derive it at a height h from the center in the z-direction? I have tried deriving it, and I end up with Bt = [muo*I / [2*pi*((a/2)^2+(b/2)^2+Z^2)^0.5)] ] * [(a/((b/2)^2+Z^2)^0.5) + (b/((a/2)^2+Z^2)^0.5 ]. However the book I am reading has a the answer as Bt = (muo*I*a*b/4*pi*[(a/2)^2+(b/2)^2+Z^2]^0.5) * [1/((a/2)^2 + Z^2) + 1/((b/2)^2+Z^2)]. I put the calculations in excel computing the strength of the field (H = B/mu0). My answer is higher than the answer in the book. I am sure I am setting it up wrong. Could you do a video showing how to derive it? It would be this video with the B field at the center of the field elevated by a height h. I have tried multiple times, and I can't seem to match the book answer.
If you figured out how to do that please lmk, that's the question I wondered.
I just wanted to know what I was doing wrong... any chances you show how to calculate it?
@@mariahhasfuntimeshere I will... I was hoping he can set it up and I can solve it the integral. He has not acknowledged the request.
Physics Ninja can you set it up the calculation when p is at a height H along the z axis? I have tried it, and I can't match the book answer.
@@rjrodrig I wish i knew how to calculate it, that's my issue as well. I was given only the answer by my teacher and have to work out the problem. I can't seem to figure out how after multiple times trying.
orhan hocaya selam olsun
I love u
quick math
super figure
:)