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GR9677 #46
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Electromagnetism$\Rightarrow$}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 $B.dA$?
 Chungha16272017-08-25 01:48:45 I think that dot product between B and A makes the cosine term.
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???
 AER2009-03-31 15:25:31 They want an expression for $\omega$. So the answer is: $\omega=\epsilon_0 / (B \pi R^2)$
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).
 bpat2007-10-30 19:28:01 w = [rads/s] which is the same thing as [1/s]
 Camoph2010-03-30 18:50:20 Is faster..thanks
 flyboy6212010-10-24 16:14:51 nicely done!
 RusFortunat2015-10-20 17:21:52 брат братан братишка
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, $\omega=V_0/{\pi{R^2}B$
Richard
2007-09-23 22:53:49
I don't think you meant to write "angular momentum."

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