GREPhysics.NET: GR9677 #46

GR9677 #46

Problem

GREPhysics.NET Official Solution

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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 $B.dA$ ?

Chungha1627
2017-08-25 01:48:45

I think that dot product between B and A makes the cosine term.

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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: $\omega=\epsilon_0 / (B \pi R^2)$

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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

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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, $\omega=V_0/{\pi{R^2}B$

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Richard
2007-09-23 22:53:49

I don't think you meant to write "angular momentum."

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

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