55 Replies
posting my soln
@Gyro Gearloose
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.
,rotate
AB means integral B.dl from A to B
@Opt @SirLancelotDuLac @Real potato help pls
Yeah man, the answer seems correct.
Half of all current is q.omega/4pi
And considering an infinitely big square thingy, it works.
Nah it's wrong
Lol I have forgor physix lmao
Take magnetic moment and do the integral of magnetic dipole field.
Magnetic moment is just (q/2m)(angular momentum)
ok so first of all what i did is wrong
what i did here isnt symmetrical since the current in wires would distort the magnetic fields
yeh sab toh nahi padha abhi
can u tell using amperes law
Amperes law applies for a closed loop
yeah
we can close the loop
?
Just do the integration my good man
i dont understand your integration method tho
You got field expression na?
ik how to find magnetic field at centre of given system
nothing else
Electric dipole ka axial field formula patha hai na?
Replace 1/ε with μ,
Replace p with m, and that's magnetic field along axis of magnetic dipole
nhi padha abhi...
Electrostatics nahi kiya hai kya???
Slight doubt: Don't we approximate the dipole formula strictly for large distances?
True. That might be an issue.
We get zero if we follow my method I think. Lol.
My thinking was ki take a half plane with the edge coinciding with z axis, take a big square with one of the sides as z-axis and making the square infinite, we just get the value as mu_0 times I
Which would give mu_0 times q.omega/2pi
Oh right its 2pi not 4pi sorry 😭
We should also be able to take a circular loop tangent to z axis at origin and then take limit as R tends to infinity then?
Ye, but we have to find the value of line integral along the z-axis.
As the square tends to infinity, contributions by other sides of square tend to zero
I have forgotten how to solve questions🙏
This is why I retired
Here the decrease in contribution occurs as a gradient as we move away
sorry meine magnetic dipole padh lia
Same man. I only read about cs now. 🙏
Maine bhi galat answer de diya
isnt that what ive done?
Yes, I think you just took the current or something wrong.
The current passing through the plane is q.omega/2pi
yeh thats what ive taken
ignoring my working
how will u proceed with your idea
So the answer should be mu_0 times q.omega/2pi na?
.
.
yep this is the answer
but acc to your working the B.dl would for the entire loop
But again the other sides are infinitely far away from the ring, so B and hence integral B.dl for the other sides tends to zero very quickly.
the side of square directly infront of sphere will have B=0 . u cant say the same for others ig
Imagine here ki AO is already infinite, toh line AB pe field bohot hi zyada tezi se zero tend karegi na
green side pe B=0
And similarly for other sides, so there contribution tends to zero as AO grows larger
here story is diff ig
Nah man, think over it this way: All the portions of the horizontal line are infinitely away from the sphere
Which makes the B tend to zero
achaaa
makes sense now
why is amperes law so weird
Wait actually I'm second guessing myself rn, don't close the thread :/
alr alr
Hogya kya solve?
tell your method
n equals 2 a rha he using bio savart law integration (didnt learn ampere law )
Share your method
Def this method not recommened for this que but still i didnt learn amperes law so used this long ass one
First integral u can evaluate using some u sub etc and seconld wrt l wala integral diverges but use caucy principal value(dont know what that means exactly but main idea is if a integral is diverging at a point then break the integral in 2 parts and integrate everywhere except that point)