29 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.Tf is this question? It looks nice, but it also looks like someone was asked to cram as many topics as possible into one question
I mean yeah but the doubt was how do you find the energy of the system.
Just a moment, what's that q/ω?
Ig the potential difference*effective current would give power but don't see how to find energy and stuff
Bro whar😭😭
what is this monstrosity
I forgot the rotation thing. How do you figure out how much it rotates again?
Amplitude?
Mhm
I mean there are two shms right? So amplitude ke liye we somehow need energy and for energy (I think) we need amplitude.
No no, I mean the simplest case. There's a charged disk and magnetic field changes.
I forgot that
And I don't have notes for that
That would just give rotation no?
Yeah.
How much?
The angular velocity
I just remember i = ωq/2π
How to relate it, I forgot.
Omega would be qB/m ig
😰
? Would it?
What?
qvB=mvomega for a particle and then for entire disk omega=qB/m ig,
why is it even rotating
ik that tension hoti hai closed loop mein in a magnetic field
but switching off the magnetic field will just induce an emf?
Yeah, wait that makes sense.
More like ω= B(dq/dm) i suppose
Working radially
oh ok
got it
I got q²B²R²/4M as the total rotational kinetic energy
Ah. How?
0.5Iω²
But this omega is in case of disc rotating under influence of constant field right? While in this case, we are dealing with shm and motion due to changing flux?
A disk would not rotate because of a constant field in the first place.
It would be an impulse I think
∆ω = (dq/dm) ∆B
Ah I see.
dq/dm ?
Can you explain what it is i don't understand
Idk either