Magnetic Field Above a Square Loop

[@mariahhasfuntimeshere]

Thank you so much for this!!! Normally I have a studio class to do this problem where I can ask questions and work in a group, but school has gone online and this is a huge help!

[@cosmolearner5562]

kzbin.info

[@amiyancandol4499]

I derived a general formula for this: nuoi/8π(a²/√x²+(a²/4)cosec²π/n.(x²+a²/4cot²π/n)

[@hlo8214]

orz

[@Omeruy]

Sir thank you very much this helped me a lot. I used to try to memorize that complicated integral that also encounter in the eletric field calculations. Greetings

[@vishakhasasuke]

Thanks a lot for this. This derivation is of great help for my research project on Square Helmholtz Coils. 😭❤️

[@yibowang197]

Subscribed! What more challenging is to calculate the magnetic flux on another loop antenna, then we need to integral the dS, can you do it?

[@PhysicsNinja]

I agree, flux is hard especially if the field is not uniform across the surface.

[@AhmedKhaleelAhmedAhmed]

that is so good please keep update

[@serdaryavuz9697]

Thanks for the video its very helpful ! If I was flowing clockwise would there be changes ?

[@serdaryavuz9697]

Hey me again. I also did the Integral and got 1/(a²) * x/(sqrt(x²+a²))

[@anandannapurna4507]

I solved this before seeing your video a I checked, got the same answer. 😊

[@JuanhongHai]

I have one doubt toward your work, when it comes to the simplification of trigonometrics, I saw that in the 17:50 section of the video, why you simplify into 1/(a^2) rather than 1/a? Could you please interpret a bit? Thank you very much!

[@mina6647]

Thank you so much That really helped❤️

[@mazinshamshad3838]

Sir why are you using x prime instead of just simply writing x?

[@tamuno-omiegogo2835]

Hi, Thanks for your time trying to decimate the problem with your solution. My question is, "What if P is away from the origin" could you please provide a hint? Thanks

[@PhysicsNinja]

Good question: Take the solution at 22:55 in the blue box. Look at the 2 terms in the denominator and if z>> L then you can simplify the term in the first bracket to z^2 (forget about the L^4/4 term) and the term in the square root with simplify to simply z.. At the end the field Bz will fall off as 1/z^3. Hope this helps.

[@tamuno-omiegogo2835]

@@PhysicsNinja I have tried doing a lot but not sure. How can I send you the diagram I have drawn ?

[@tamuno-omiegogo2835]

I want to have a general expression that takes care of different points on the xyz axis

[@amiyancandol4499]

Hey found you again! But I got answer half of this, in denominator why is it 2π not 4π

[@akshaykumarsingh5143]

Great explanation! Thumbs up 👍.

[@shiranbarbalat8311]

hey! you are the best:) its possible to solve this problem with amper law?

[@PhysicsNinja]

No, not possible.

[@diegonavia1404]

does someone know how to integrate this result to obtain the expression of a very large solenoid of square loops

[@GamerTheTurtle]

multiply it by N loops

[@gvincent3246]

sir, could you do this for isosceles trapezium . using same cooncept. please

[@Harmonicanalysis54]

I've seen certain formulas for magnetic field that have the denominator set as r squared instead of r cubed? What's the difference between the two and when would you know when to apply which

[@PhysicsNinja]

Good question: The Biot-Savart law is the most general formula that can be used to calculate the field due to wires (long straight, loops, etc). For example if it's used to calculate the magnetic field from a long wire you will find that after doing some math you will get B=u0*i/(2*pi*r) where r is the distance from the wire. So we started with Biot-Savart and ended with a 1/r dependence. If you look at my video on the circular loop you will also find that after the calculation you end up finding that the field is proportional to 1/R where R is the radius of the current loop. You can always use Biot-Savart but you have to do some math to find the field produced by that current or if you know the the geometry (in the problem statement) you can use the final result that someone has already found.

[@curator_of_leptons2857]

I had the same confusion, and found Physics Ninja's answer a bit unsatisfactory. The answer that I found that helped was that he was using r as being the full vector. Whenever you see the denominator as r^2 it's because they're using hat{r} were the r in the numerator is the unit vector

[@PhysicsNinja]

@@curator_of_leptons2857 Yes, if it's r^3 in denominator you use the full vector r to point to current element. If you write it as r^2 then you use the unit vector r^hat to point in the direction of the current element. I always find it easier to use the full vector to point to the current element. Either way you just need to be careful. Have fun!

[@cosmolearner5562]

Whether the same illustration can be solved by the Ampere's law...?

[@PhysicsNinja]

No, this problem can’t be solved using Amperes law. To solve problems using Amperes law you need to be able to simplify the integral of B.dl. This can be done where the field is constant along a certain section. In this case it would be impossible to pick a loop to get b out of the integral.

[@mariahhasfuntimeshere]

Is it possible to use the definition of a cross product and then the opposite side over the hypotenuse?

[@PhysicsNinja]

The only thing to remember is the direction of the unit vector for the cross product. At the end you need take the z-component. It ends up being identical to the way I did it.

[@srisidvicious]

Excellent explanation. Can I ask what the field at the center of an infinitely long square solenoid would be? Is it simply 2sqrt(2)*mu*I*(N)/(Pi*a) where N = turns/meter?

[@anonymouslearner845]

❤️from India.

[@ryanlee6876]

at 15:34 how is the distance from the wire a when we know the z axis is a distance a/2 away

[@PhysicsNinja]

This is just a substitution to complete the integral over x. I’m simply grouping all the terms that don’t depend on x together and am calling that a.