When Magnetic field is suddenly switched off

AI Thread Summary
When the magnetic field is suddenly switched off, the uniform non-conducting ring experiences a change in magnetic flux, leading to induced electric fields that cause the ring to rotate. The angular speed of the ring immediately after the switch-off is calculated to be 1/8 Rads/sec. The discussion highlights the importance of understanding induced electric fields and their relationship to torque and rotation in this context. Participants express their challenges in grasping these concepts, emphasizing the learning process involved in solving such physics problems. Overall, the conversation underscores the application of Faraday's law and the dynamics of charged objects in changing magnetic fields.
AGNuke
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A uniform non conducting ring of mass 1 kg, radius 1 m and having charge 1 mC distributed uniformly is free to rotate only about its central axis. Initially, a uniform magnetic field of 103 T is applied in a circular region of radius 0.5 m with centre on axis of ring. The ring was initially stationary. Now the magnetic field is suddenly switched off.

Now, the angular speed of the ring just after switch-off the magnetic field is? (1/8 Rads/sec)

I thought that since magnetic field is suddenly switched off, the flux (in the small circular region) would also become zero suddenly. So, in order to counter the change, the charged ring will rotate in such a manner so as to conserve the flux enclosed by the ring. ∴
B_{initial}A_{region}=B_{ring}A_{ring}
10^3\times \pi(0.5)^2=\frac{\mu_0i}{2R}\times \pi(1)^2
i=\frac{dq}{dt}=\frac{Q}{2\pi R}\frac{Rd\theta}{dt}=\frac{Q\omega}{2\pi}
I must be going horrendously wrong somewhere because I am not even in remote to the options mentioned, let alone the answer. Please help. Seems like I am missing quite something I am unable to point myself at.
 
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AGNuke said:
I thought that since magnetic field is suddenly switched off, the flux (in the small circular region) would also become zero suddenly. So, in order to counter the change, the charged ring will rotate in such a manner so as to conserve the flux enclosed by the ring. ∴
B_{initial}A_{region}=B_{ring}A_{ring}
There is no such law of nature. Consider the extreme case where the ring is very massive while the charge on the ring is infinitesimal. Could switching off the B field cause the ring to rotate fast enough to generate a flux that would equal the original flux?

Faraday's law (or Lenz's law) implies only that the induced rotation of the ring will be in a direction such as to "oppose" the change in flux caused by switching off B. But "oppose" doesn't mean "maintain a constant value of flux".

Think about what physical quantity acts on the ring to make it start to rotate. Can you think of a way to relate that physical quantity to the changing B field?
 
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You mean to say I need something which can apply a Torque on the ring? Sorry to say but I am at my wits' end.

UPDATE Are you implying Induced Electric field which is due to changed Magnetic field?
 
Last edited:
AGNuke said:
UPDATE Are you implying Induced Electric field which is due to changed Magnetic field?

Yes.

ehild
 
My Solution

Alright. Let's see... I think I got it. Here it is:
Induced Electric Field is given by:\oint \vec{E}.\mathrm{d}\vec{l}=-\frac{\mathrm{d} \Phi_B}{\mathrm{d} t}
\Rightarrow E.2\pi R=-\frac{dB\times \pi(R/2)^2}{dt}
Force on elemental charge on the ring will be tangential to the ring (since Electric field is also circular). Therefore, the Torque is given by\tau=\oint dq.\vec{E}\times \vec{R}\Rightarrow MR^2\alpha=EQR
\Rightarrow MR^2\int_{0}^{t}\alpha dt=-\int_{B}^{0} dB\frac{\pi R^2/4}{2\pi R}QR
\Rightarrow M\omega=\frac{1}{8}BQ
Now B = 103T, Q = 1 mC and M = 1 kg, the answer comes out to be 1/8 Rads/sec.

I hope that this solution is correct.
 
It looks correct. Nice work!

ehild
 
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Never would had guessed about Induced Electric field if hadn't asked here. Each time I solve a question, I realize I need to know this and that. Someone really said, Increaseth knowledge, Increaseth Sorrow. (Now we are not supposed to be happy are we? :wink:)
 
Never would had guessed about Induced Electric field if hadn't asked here. Each time I solve a question, I realize I need to know this and that. Someone really said, Increaseth knowledge, Increaseth Sorrow. (Now we are not supposed to be happy are we? :wink:)
 
Don't you feel happy when you understand something? That things happen in the way they should? That there are simple laws governing the Universe... :smile:

ehild
 
  • #10
I enjoy things at "my" pace and things at the moment are well beyond it. Like I was very amused to read about Quantum Physics, MWI and stuff, even though I didn't understood one bit of it.

So right now, all that knowledge is pressing hard against me. There is happiness in discovering facts, but unhapiness over how less we know...:mad:
 
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You never can know everything. Enjoy what you know and be ready to learn new things.

ehild
 
  • #12
AGNuke said:
Force on elemental charge on the ring will be tangential to the ring (since Electric field is also circular). Therefore, the Torque is given by\tau=\oint dq.\vec{E}\times \vec{R}\Rightarrow MR^2\alpha=EQR

I know that you are done with this thread but could you please tell me from where you got this equation? :)

Thanks!
 
  • #13
Pranav-Arora said:
I know that you are done with this thread but could you please tell me from where you got this equation? :)

Thanks!

What force would act on charge dq in field E? And what torque would it produce?
 
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  • #14
voko said:
What force would act on charge dq in field E? And what torque would it produce?

##dq \cdot E##? About CM of ring, it produces a torque of dqER. Looks like I got it. Thanks voko! :smile:
 
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