EMI+Mechanics Doubt

How to go about option (3) and (4)?
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@Gyro Gearloose
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Opt
Opt2mo ago
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
SirLancelotDuLac
SirLancelotDuLacOP2mo ago
I mean yeah but the doubt was how do you find the energy of the system.
Opt
Opt2mo ago
Just a moment, what's that q/ω?
SirLancelotDuLac
SirLancelotDuLacOP2mo ago
Ig the potential difference*effective current would give power but don't see how to find energy and stuff
Dexter
Dexter2mo ago
Bro whar😭😭 what is this monstrosity
Opt
Opt2mo ago
I forgot the rotation thing. How do you figure out how much it rotates again?
SirLancelotDuLac
SirLancelotDuLacOP2mo ago
Amplitude?
Opt
Opt2mo ago
Mhm
SirLancelotDuLac
SirLancelotDuLacOP2mo ago
I mean there are two shms right? So amplitude ke liye we somehow need energy and for energy (I think) we need amplitude.
Opt
Opt2mo ago
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
SirLancelotDuLac
SirLancelotDuLacOP2mo ago
That would just give rotation no?
Opt
Opt2mo ago
Yeah. How much? The angular velocity I just remember i = ωq/2π How to relate it, I forgot.
SirLancelotDuLac
SirLancelotDuLacOP2mo ago
Omega would be qB/m ig
Dexter
Dexter2mo ago
😰
Opt
Opt2mo ago
? Would it? What?
SirLancelotDuLac
SirLancelotDuLacOP2mo ago
qvB=mvomega for a particle and then for entire disk omega=qB/m ig,
Dexter
Dexter2mo ago
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?
Opt
Opt2mo ago
Yeah, wait that makes sense. More like ω= B(dq/dm) i suppose Working radially
Dexter
Dexter2mo ago
oh ok got it
Opt
Opt2mo ago
I got q²B²R²/4M as the total rotational kinetic energy
SirLancelotDuLac
SirLancelotDuLacOP2mo ago
Ah. How?
Opt
Opt2mo ago
0.5Iω²
SirLancelotDuLac
SirLancelotDuLacOP2mo ago
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?
Opt
Opt2mo ago
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
SirLancelotDuLac
SirLancelotDuLacOP2mo ago
Ah I see.
hithav
hithav2mo ago
dq/dm ? Can you explain what it is i don't understand
Opt
Opt2mo ago
Idk either
SirLancelotDuLac
SirLancelotDuLacOP5w ago
Oh right I forgor about this. This make sense. Rotational energy would be q²B²R²/4M so max P.E. in spring would be the same. But what about max. P.E. in string?
Opt
Opt5w ago
Wouldn't it be half of that energy? No wait Nvm When the string is stretched taut, the spring is at original length
SirLancelotDuLac
SirLancelotDuLacOP5w ago
Oh right. So neither 3 nor 4 would be correct?
Opt
Opt5w ago
Nope The time period part was fascinating. It took me a while to realise that half the period is controlled only by the spring and there is no contribution from the string
SirLancelotDuLac
SirLancelotDuLacOP5w ago
Ah I see +solved @Opt
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