21 Replies
@Gyro Gearloose
Note for OP
+solved @user to close the thread when your doubt is solved. Mention the user who helped you solve the doubt. This will be added to their stats.Did you consider force on PQ and PR too?
oh shit yeah 💀
no wonder i was missing a 3
but its still wrong, 2 nahi hai 4 hai
I think components mein galti hui hogi
kidar yaar ðŸ˜, mai apne working mei mistake dhundh hi ni pata
Mein try krta
Bhai torque eqn apply kr
Direct nahi ayega
dikha bhai thoda
anyone?
@Gyro Gearloose any inputs, folks?
@Gyro Gearloose ping!
I'm
confused.
Likewise... @the_nub should i ask my teachers and get back to you?
yeah, try that
The condition for options A and B is that you need to equate work done by magnetic field with gravitational potential energy
But here I suppose you’ll have to assume the triangle’s axis is parallel to CD
Use torque = m B sin theta, integrate with d theta to find work done (m is I x A here)
equate with mass x g x vertical displacement of centre of mass
yeah that's given in the question
oh right i didn't think of that, i made another attempt jisme weight ka torque and dipole moment ke torque se try kiya, can you see how that would work?
About the axis of rotation?
That can definitely work
You just need to equate the final result to some energy value
yeah but then im getting torque of mg = 0 because its in the plane of the loop
don't know why im going wrong here, must be some rookie error
well it's not always in the plane of the loop
when the loop rotates the plane varies while mg is always downward
oh right yeah, that makes sense
now what should i do for the C and D options?
At equilibrium net force and net torque about the axis of rotation is 0
you just have to use that and directly add all torques
oh okay, gotcha
+solved @Comrade Rock Astley
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