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Electromagnetism}Faraday Law


Recall Faraday Law, \epsilon = -\frac{d\Phi}{dt}, where \Phi=B\cdot dA. Since the magnetic field is constant, the equation simplifies to \epsilon = -B \cdot \frac{dA}{dt} for this case.

B \cdot A = B \cos(\omega t) \pi r^2, and thus d\Phi/dt = -\omega \sin(\omega t) \pi r^2 = -\epsilon = -\epsilon_0 \sin(\omega t). Solving for angular momentum, one has \omega = \epsilon_0/(B\pi R^2).

Alternatively, one has A=\pi r r(t) \Rightarrow \dot{A} = \pi r \dot{r} = \pi r v. Since v=\omega r\sin(\omega t), one has A=\pi r \omega r\sin(\omega t). Plug it into Faraday Law and solve for angular velocity.

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Comments
dipanshugupta
2017-03-29 10:45:02
How do you get the B cos(\\omega t) term for B.dA?
Chungha1627
2017-08-25 01:48:45
I think that dot product between B and A makes the cosine term.
NEC
Poop Loops
2008-11-07 20:35:25
This is such a load of crap. In the question they say the loop rotates with angular speed \omega and then they ask you for the angular speed of the loop. What the hell does that mean???
AER
2009-03-31 15:25:31
They want an expression for \omega. So the answer is: \omega=\epsilon_0 / (B \pi R^2)
NEC
bpat
2007-10-30 19:25:37
I think dimensional analysis works fastest.

w = [rads/s]
emf = B*Area/time

so: w = emf/(B*Area)

Only one choice that looks like that and it is (C).
bpat
2007-10-30 19:28:01
w = [rads/s] which is the same thing as [1/s]
Camoph
2010-03-30 18:50:20
Is faster..thanks
flyboy621
2010-10-24 16:14:51
nicely done!
RusFortunat
2015-10-20 17:21:52
брат братан братишка
NEC
Richard
2007-09-23 23:07:51
V_{0}{\sin(\omega{t})}=V
Integrate this to get the flux as a function of time, with an extra negative one (Faraday's law):

\Phi_{b}(t)=\frac{V_{0}}{\omega}{\cos(\omega{t})}

At t=0 you have the maximum flux which is \pi{R^2}B by the diagram.

Thus, \omega=V_0/{\pi{R^2}B
NEC
Richard
2007-09-23 22:53:49
I don't think you meant to write "angular momentum."NEC

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How do you get the B cos(\\omega t) term for B.dA?

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