iTeachChem•6mo ago

amperes law

i am getting an ans but am not sure if it correct ( key nahi hai)

CorrodedCoffinOP•7/11/25, 8:50 AM

posting my soln

iteachchemhelperv2•7/11/25, 8:50 AM

@Gyro Gearloose

iteachchemhelperv2•7/11/25, 8:50 AM

Note for OP

+solved @user1 @user2...

+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.

CorrodedCoffinOP•7/11/25, 8:52 AM

CorrodedCoffinOP•7/11/25, 8:52 AM

,rotate

TeXit•7/11/25, 8:53 AM

CorrodedCoffinOP•7/11/25, 8:53 AM

CorrodedCoffinOP•7/11/25, 8:54 AM

AB means integral B.dl from A to B

CorrodedCoffinOP•7/12/25, 10:13 AM

@Opt @SirLancelotDuLac @Real potato help pls

CCorrodedCoffin Click to see attachment

SirLancelotDuLac•7/12/25, 10:22 AM

Yeah man, the answer seems correct.

SirLancelotDuLac•7/12/25, 10:22 AM

Half of all current is q.omega/4pi

SirLancelotDuLac•7/12/25, 10:22 AM

And considering an infinitely big square thingy, it works.

CorrodedCoffinOP•7/12/25, 10:24 AM

Nah it's wrong

CCorrodedCoffin Nah it's wrong

SirLancelotDuLac•7/12/25, 10:45 AM

Lol I have forgor physix lmao

CCorrodedCoffin i am getting an ans but am not sure if it correct ( key nahi hai)

Opt•7/12/25, 11:34 AM

Take magnetic moment and do the integral of magnetic dipole field.

Opt•7/12/25, 11:35 AM

Magnetic moment is just (q/2m)(angular momentum)

CorrodedCoffinOP•7/12/25, 11:59 AM

ok so first of all what i did is wrong

CorrodedCoffinOP•7/12/25, 12:00 PM

what i did here isnt symmetrical since the current in wires would distort the magnetic fields

OOpt Take magnetic moment and do the integral of magnetic dipole field.

CorrodedCoffinOP•7/12/25, 12:00 PM

yeh sab toh nahi padha abhi

CorrodedCoffinOP•7/12/25, 12:00 PM

can u tell using amperes law

Opt•7/12/25, 12:01 PM

Amperes law applies for a closed loop

CorrodedCoffinOP•7/12/25, 12:01 PM

yeah

CorrodedCoffinOP•7/12/25, 12:01 PM

we can close the loop

Opt•7/12/25, 12:01 PM

Opt•7/12/25, 12:02 PM

Just do the integration my good man

CorrodedCoffinOP•7/12/25, 12:02 PM

i dont understand your integration method tho

Opt•7/12/25, 12:02 PM

You got field expression na?

CorrodedCoffinOP•7/12/25, 12:04 PM

ik how to find magnetic field at centre of given system

CorrodedCoffinOP•7/12/25, 12:04 PM

nothing else

Opt•7/12/25, 12:04 PM

Electric dipole ka axial field formula patha hai na?

Opt•7/12/25, 12:05 PM

Replace 1/ε with μ,
Replace p with m, and that's magnetic field along axis of magnetic dipole

OOpt Electric dipole ka axial field formula patha hai na?

CorrodedCoffinOP•7/12/25, 12:40 PM

nhi padha abhi...

CCorrodedCoffin nhi padha abhi...

Opt•7/12/25, 12:44 PM

Electrostatics nahi kiya hai kya???

OOpt Take magnetic moment and do the integral of magnetic dipole field.

SirLancelotDuLac•7/12/25, 12:47 PM

Slight doubt: Don't we approximate the dipole formula strictly for large distances?

SSirLancelotDuLac Slight doubt: Don't we approximate the dipole formula strictly for large distanc...

Opt•7/12/25, 12:48 PM

True. That might be an issue.

Opt•7/12/25, 12:49 PM

We get zero if we follow my method I think. Lol.

OOpt We get zero if we follow my method I think. Lol.

SirLancelotDuLac•7/12/25, 12:51 PM

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

SirLancelotDuLac•7/12/25, 12:51 PM

Which would give mu_0 times q.omega/2pi

CCorrodedCoffin Nah it's wrong

SirLancelotDuLac•7/12/25, 12:51 PM

Oh right its 2pi not 4pi sorry

SSirLancelotDuLac My thinking was ki take a half plane with the edge coinciding with z axis, take ...

Opt•7/12/25, 12:52 PM

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?

OOpt We should also be able to take a circular loop tangent to z axis at origin and t...

SirLancelotDuLac•7/12/25, 12:52 PM

Ye, but we have to find the value of line integral along the z-axis.

SirLancelotDuLac•7/12/25, 12:53 PM

As the square tends to infinity, contributions by other sides of square tend to zero

Opt•7/12/25, 12:53 PM

I have forgotten how to solve questions

Opt•7/12/25, 12:53 PM

This is why I retired

OOpt We should also be able to take a circular loop tangent to z axis at origin and t...

SirLancelotDuLac•7/12/25, 12:53 PM

Here the decrease in contribution occurs as a gradient as we move away

OOpt Electrostatics nahi kiya hai kya???

CorrodedCoffinOP•7/12/25, 12:54 PM

sorry meine magnetic dipole padh lia

OOpt I have forgotten how to solve questions🙏

SirLancelotDuLac•7/12/25, 12:54 PM

Same man. I only read about cs now.

CCorrodedCoffin sorry meine magnetic dipole padh lia

Opt•7/12/25, 12:54 PM

Maine bhi galat answer de diya

SSirLancelotDuLac My thinking was ki take a half plane with the edge coinciding with z axis, take ...

CorrodedCoffinOP•7/12/25, 12:55 PM

isnt that what ive done?

CCorrodedCoffin isnt that what ive done?

SirLancelotDuLac•7/12/25, 12:55 PM

Yes, I think you just took the current or something wrong.

amperes law

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