interaction force
why cant i assume the semi-circle as another infinitely long wire and solve
43 Replies
@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.Why would you assume that btw
Like this
Eduniti - Physics by Mohit Goenka_IITKGP
YouTube
JEE Main 2025 - Trick for Charged ARCπ² | +4 Marks in 2 Min | Ele...
+4 Marks Series Playlist : https://bit.ly/3aamBrh
Want to solve Complete Physics PYQs from 2019-2022 π
JEE Main 2022 July Attempt - https://bit.ly/3LeU4zB
JEE Main 2022 June Attempt - https://bit.ly/3AHXgjJ
JEE Main 2021 Solutions Series - https://bit.ly/3dFA1dB
JEE Main 2020 Solution Series - https://bit.ly/3cucEmE
JEE Main 2019 PYQs - h...
also my sir explained why this is true
Then sure you can
I've heard of this but have never applied it
results dont match
Oh
this is a W results
somehow very few people know thi
ok i tried on this even though i have no idea but what i think is if we do take the ring and the thread then we can get to there through 2 ways
1. what force does dl gives on the ring and then integrate to what force the whole thread will give
2. what force does the thread give on dl of ring and integrate to the whole ring
i tried to solve wasnt able to
im getting stuck
its easy to find force on the ring due to field of wire
wouldve shared the working but disc images down
anyways
in the semicircle , theta angle pe dtheta angle lo
uspe pe bahar k taraf ke force lagega E*dq where E is electric field due to wire = 2klambda/r and dq is charge in element
symmetrically opposite ek aur element hoga so cos component neutralizes and only sin remains
dF = dqEsintheta
integrate from 0 to pi
so second one right
1st
A
This is field on axis of semi ring and parallel to ring (imagine semi ring in xy plane)
This due to wire on axis perpendicular to it (horizontal component)
There is an extra pi in the denominator cause if you think about it, lambda ring= Q/ΟR, introduces extra Ο
Anyways ig due to the fact that ring is a finite body it won't work
Don't know the reason for sure
Integration always provides the exact solution
The wire is infinite. Acc to your diagram it should be 2klambda/a perpendicular to wire
look carefully
Can u elaborate
@Opt @SirLancelotDuLac
Your insights would be well appreciated
That only applies if you're calculating field at the centre of the loop right?
We are doing that only ig?
But the rod extends beyond the centre right?
Nope
It passes thru the centre of half ring
Yeah, but the field expression is for a single point
Not for a line passing through center
thats why this calc is done
(its B right)
its A
this is the integral u made right?
oh yh mb
yeah just took different angle ig
I didn't see this thread....
But how good is that "E"
Oooooofffff
Wdym
It's of the long thread
Are handwriting pe compliment kiya tha yaar π
That was looked really good
Ohh
ECos(theta) vaala
I didn't even read the question ngl
Damnn first time for everything π π
I was too exhausted at that time π
And now I'm caught up some other question
It's a freakin cycle
Real
frfr
hua solve?
nahi π