Hey guys,
The only question I have with this is when the area is time varying, how do you know whether to choose cos(wt) or sin(wt)? My guess is it shouldn't matter, but one of the choices had a tan1 in it. Do you think they would be that cruel that they would have one choice be Eo/B(Pi)R and tan1[Eo/B(Pi)R]? If so, then who decides if it's cos or sin (since it doesn't specify that it starts at x=0 or something like that)?
Thanks guys,
Ari
92, Problem 46

 Posts: 80
 Joined: Tue Oct 12, 2010 8:00 am
Re: 92, Problem 46
I assume you are talking about GR9677 Prob. 46, not GR9277.
There is no ambiguity here:
$$\Phi = \int_{\mathcal{S}} \mathbf{B} \cdot d\mathbf{S} = \int_{\mathcal{S}} ( \mathbf{B} \cdot \mathbf{n})\,dS = B \cos {\alpha} \int_{\mathcal{S}} dS = B \pi R^2 \cos {\alpha}$$,
where $$\mathbf{n}$$ is a unit vector (an "outward normal") perpendicular to the infinitesimal area $$d\mathbf{S}$$ and $$\alpha$$ is the angle between $$\mathbf{n}$$ and $$d\mathbf{S}$$. Now,
$$\alpha(t) \equiv \alpha = \alpha_0 + \omega t$$,
where
$$\alpha_0 = \alpha(0)$$
and
$$\mathcal{E} =  \frac{d \Phi}{dt} =  B \pi R^2 \frac{d}{dt} \left[ \cos (\alpha_0 + \omega t)\right] = +B \pi R^2 \omega \sin (\alpha_0 + \omega t)$$
According to problem statement $$\mathcal{E} = \mathcal{E}_0 \sin \omega t$$, so by the initial condition for $$t=0$$ we have $$\mathcal{E} = B \pi R^2 \omega \sin (\alpha_0) = 0$$ and $$\alpha_0 = 0$$, therefore:
$$\mathcal{E} = B \pi R^2 \omega \sin \omega t$$ and
$$\mathcal{E}_0 = B \pi R^2 \omega$$
Finally,
$$\omega = \frac{\mathcal{E}_0}{B \pi R^2}$$
as in choice (C).
There is no ambiguity here:
$$\Phi = \int_{\mathcal{S}} \mathbf{B} \cdot d\mathbf{S} = \int_{\mathcal{S}} ( \mathbf{B} \cdot \mathbf{n})\,dS = B \cos {\alpha} \int_{\mathcal{S}} dS = B \pi R^2 \cos {\alpha}$$,
where $$\mathbf{n}$$ is a unit vector (an "outward normal") perpendicular to the infinitesimal area $$d\mathbf{S}$$ and $$\alpha$$ is the angle between $$\mathbf{n}$$ and $$d\mathbf{S}$$. Now,
$$\alpha(t) \equiv \alpha = \alpha_0 + \omega t$$,
where
$$\alpha_0 = \alpha(0)$$
and
$$\mathcal{E} =  \frac{d \Phi}{dt} =  B \pi R^2 \frac{d}{dt} \left[ \cos (\alpha_0 + \omega t)\right] = +B \pi R^2 \omega \sin (\alpha_0 + \omega t)$$
According to problem statement $$\mathcal{E} = \mathcal{E}_0 \sin \omega t$$, so by the initial condition for $$t=0$$ we have $$\mathcal{E} = B \pi R^2 \omega \sin (\alpha_0) = 0$$ and $$\alpha_0 = 0$$, therefore:
$$\mathcal{E} = B \pi R^2 \omega \sin \omega t$$ and
$$\mathcal{E}_0 = B \pi R^2 \omega$$
Finally,
$$\omega = \frac{\mathcal{E}_0}{B \pi R^2}$$
as in choice (C).

 Posts: 76
 Joined: Thu May 12, 2011 4:53 pm
Re: 92, Problem 46
Haha, yeah I was talking about 96. Thanks alot physicsworks. This forum as always has been a life saver.